MAT4760/MAT9760
4.1 Risk Measures [FS 153]
Date
A risk measure, denoted by ρ, is a function that quantifies the risk associated 17.01.11
with some financial position X, where X : Ω 7→ R. Ω is the set of all possible
outcomes, and X(ω) is the discounted value of the financial position in the
outcome ω ∈ Ω. We define X to be the set of financial positions. X is a linear
space of bounded functions containing the constants. The norm we use is the
supremum norm: kXk := supω∈Ω |X(ω)|.
Definition 4.1 [FS 153]
A mapping ρ : X 7→ R is called a monetary measure of risk if it satisfies the
following conditions for all X, Y ∈ X .
Monotonicity: X ≤ Y =⇒ ρ(X) ≥ ρ(Y ),
X(ω) ≤ Y (ω), ∀ω ∈ Ω
Cash Invariance: m ∈ R =⇒ ρ(X + m) = ρ(X) − m,
∀m ∈ R.
Consequences of cash invariance:
ρ X + ρ(X) = ρ(X) − ρ(X) = 0
(4.1)
ρ(m) = ρ(0) − m
When we do not have a financial position, we do not have any risk. In this
case we say the risk measure is normalized.
Normalization: ρ(0) = 0
Lemma 4.3 [FS 154]
Any monetary measure of risk ρ is Lipschitz continuous wrt the supremum
norm k · k. Lipschitz continuity implies continuity.
ρ(X) − ρ(Y ) ≤ kX − Y k
Proof
X = X ± Y ≤ Y + kX − Y k =⇒ X ≤ Y + kX − Y k
By monotonicity and cash invariance:
ρ(X) ≥ ρ Y + kX − Y k = ρ(Y ) − kX − Y k
1
kX − Y k ≥ ρ(Y ) − ρ(X)
(⋆)
The same reasoning with Y instead of X.
Y = Y ± X ≤ X + kX − Y k =⇒ Y ≤ X + kX − Y k
By monotonicity and cash invariance:
ρ(Y ) ≥ ρ X + kX − Y k = ρ(X) − kX − Y k
ρ(Y ) − ρ(X) ≥ −kX − Y k
(⋆⋆)
By combining (⋆) and (⋆⋆) we get:
−kX − Y k ≤ ρ(Y ) − ρ(X) ≤ kX − Y k
|ρ(Y ) − ρ(X)| ≤ kX − Y k
Definition 4.4 [FS 154]
A monetary measure of risk ρ : X 7→ R is called a convex risk measure if it
satisfies the convexity property: For some 0 ≤ λ ≤ 1,
ρ λX + (1 − λ)Y ≤ λρ(X) + (1 − λ)ρ(Y )
Convexity is a way to implement a risk reduction for diversifying the
investment.
Lemma [FS 154]
If ρ is convex and normalized, then we have the following properties.
ρ(aX) ≤ aρ(X),
ρ(aX) ≥ aρ(X),
∀a ∈ [0, 1], X ∈ X
(i)
∀a ∈ [1, ∞), X ∈ X
(ii)
Proof
(i) By convexity and normalization:
ρ(aX) = ρ aX + (1 − a)0 ≤ aρ(X) + (1 − a) ρ(0) = aρ(X)
|{z}
=0
(ii) Using (i):
aX 1
≤ ρ(aX) =⇒ aρ(X) ≤ ρ(aX)
ρ(X) = ρ
a
a
2
Definition 4.5 [FS 155]
A convex measure of risk ρ is called a coherent risk measure if it satisfies
positive homogeneity.
For λ ≥ 0,
ρ(λX) = λρ(X)
Lemma
A coherent risk measure is always normalized.
Proof
Take any m ≥ 0, m 6= 1.
ρ(m) = ρ(0 + m) = ρ(0) − m
(⋆)
Using positive homogeneity.
ρ(m) = ρ(m · 1) = mρ(1) = m ρ(0) − 1 = mρ(0) − m
(⋆⋆)
Combining (⋆) and (⋆⋆): ρ(m) = ρ(m), which gives ρ(0) − m = mρ(0) − m,
⇒ mρ(0) − ρ(0) = 0 ⇒ ρ(0) m − 1 = 0 ⇒ ρ(0) = 0
| {z }
6=0
Lemma [FS 155]
For a coherent risk measures ρ, convexity ⇔ sub-additivity, defined as:
ρ(X + Y ) ≤ ρ(X) + ρ(Y ),
Proof
e =
⇒) For a, b > 0 define X
positive homogeneity:
a+b
X
a
∀X, Y ∈ X .
and Ye =
a+b
Y
b
. Using convexity and
a
b e
a
e
e + b ρ(Ye ) =
ρ(X + Y ) = ρ
X+
Y ≤
ρ(X)
a+b
a+b
a+b
a+b
b a + b a a + b ρ
X +
ρ
Y = ρ(X) + ρ(Y )
a+b
a
a+b
b
3
e = λX and Ye = (1 − λ)Y . Using sub-additivity and positive
⇐) Define X
homogeneity:
e + Ye ≤ ρ(X)
e + ρ(Ye ) = λρ(X) + (1 − λ)ρ(Y ) ρ λX + (1 − λ)Y = ρ X
Positive homogeneity is a restrictive assumption as it requires risk measures
to be linear.
From the m.r.m (monetary risk measure) we get its acceptance set.
Aρ := X ∈ X | ρ(X) ≤ 0
The set of all acceptable positions in the sense that “they do not need any
additional capital”. Here are two results that summarize the relations between
m.r.m and their acceptance set.
Proposition 4.6 [FS 155]
Assume ρ is a m.r.m with acceptance set A := Aρ .
(a) A is non-empty, A 6= ∅, and satisfies the following conditions.
inf m ∈ R | m ∈ A > −∞
X ∈ A, Y ∈ X , Y ≥ X =⇒ Y ∈ A
(4.3)
(4.4)
Moreover, A has the following closure properties. For X ∈ A and Y ∈ X ,
n
o
S := λ ∈ [0, 1] λX + (1 − λ)Y ∈ A is closed in [0, 1]
(4.5)
(b) We can get ρ just by knowing A.
ρ(X) = inf m ∈ R | m + X ∈ A
(4.6)
(c) ρ is a convex risk measure ⇐⇒ A is convex.
(d) ρ is positively homogeneous ⇐⇒ A is a cone. In particular,
ρ is coherent ⇐⇒ A is a convex cone.
Proof
20.01.11
(a) (A 6= ∅)
Define z := ρ(0), then ρ(0) ∈ R. Choose some m ∈ R such that z − m ≤ 0.
Then, by cash invariance, ρ(m) = ρ(0) − m ≤ 0, so m ∈ A, and A is nonempty.
4
(4.3)
Consider any constant m ∈ A. By cash invariance, ρ(m) = ρ(0) − m ≤ 0. If
we set z := ρ(0) this means we have z ≤ m, which implies - since the left
side is independent of m,
z ≤ inf m ∈ R | ρ(m) ≤ 0
z is a number, z ∈ R, so z > −∞.
(4.4)
X ∈ A =⇒ ρ(X) ≤ 0
Y ≥ X =⇒ ρ(Y ) ≤ ρ(X)
Combining these inequalities means ρ(Y ) ≤ ρ(X) ≤ 0, hence Y ∈ A.
(4.5)
By assumption, X ∈ A and Y ∈ X . For all λ ∈ S,
λX + (1 − λ)Y ∈ A =⇒ ρ λX + (1 − λ)Y ≤ 0.
By lemma 4.3, ρ is continuous. For fixed X, Y
fX,Y : λ 7→ λX + (1 − λ)Y
is also a continuous function, which means that the composite function
ρ ◦ fX,Y is also continuous,
ρ ◦ fX,Y (λ) = ρ λX + (1 − λ)Y
−1
Now we set S = ρ ◦ fX,Y
, and use the result that the counter image of
a closed set is still closed, thus S is closed.
(b) By the defining property of A:
inf m ∈ R | m + X ∈ A = inf m ∈ R | ρ m + X ≤ 0
Using cash invariance and the definition of the infimum:
inf m ∈ R | ρ X ≤ m = ρ(X)
(c)
⇒) Assume first that ρ is convex. Assume X, Y ∈ A and λ ∈ [0, 1].
ρ λX + (1 − λ)Y ≤ λρ(X) + (1 − λ)ρ(Y ) ≤ 0 =⇒ λX + (1 − λ)Y ∈ A
which means A is a convex set.
5
⇐) Assume A is convex. We choose arbitrary X1 , X2 ∈ X and let λ ∈ [0, 1].
We now consider m1 , m2 ∈ R such that, by cash invariance, ρ(mi + Xi ) =
ρ(Xi ) − mi ≤ 0 for i = 1, 2. By the construction of A, both m1 + X1 and
m2 + X2 are elements in A. Using that A is a convex set and the property
for all elements in A:
λ(m1 +X1 )+(1−λ)(m2 +X2 ) ∈ A =⇒ ρ λ(m1 +X1 )+(1−λ)(m2 +X2 ) ≤ 0
By cash invariance,
ρ λX1 + (1 − λ)X2 − [λm2 + (1 − λ)m2 ] ≤ 0
ρ λX1 + (1 − λ)X2 ≤ λm2 + (1 − λ)m2 =
λ inf m1 | m1 + X1 ∈ A + (1 − λ) inf m2 | m2 + X2 ∈ A =
λρ(X) + (1 − λ)ρ(Y ) =⇒ ρ is convex.
(d)
⇒) Assume ρ is positively homogeneous. Recall that X ∈ A =⇒ ρ(X) ≤ 0
and for any λ ≥ 0 we have the positive homogeneity property: ρ(λX) =
λρ(X). Since ρ(X) ≤ 0 then λρ(X) ≤ 0, so λρ(X) ∈ A for all λ ≥ 0, so A is
a cone.
⇐) Now suppose A is a cone and suppose X ∈ X . From part (b), ∃m ∈ R
such that ρ(X) ≤ m, or equivalently, m + X ∈ A. Since A is a cone, we have
for λ ≥ 0 that λ(m + X) ∈ A.
λm + λX ∈ A =⇒ 0 ≥ ρ λm + λX = ρ(λX) − λm =⇒ ρ(λX) ≤ λm
ρ(λX) ≤ λ · inf m ∈ R | m + X ∈ A = λρ(X) =⇒ ρ(λX) ≤ λρ(X)
We are going to show that ρ is positively homogeneous, so now we only need
the reverse inequality. We consider m0 ∈ R such that m0 < ρ(X). By cash
invariance (and using λ ≥ 0):
m0 < ρ(X) =⇒ 0 < ρ(X + m0 ) =⇒ X + m0 6∈ A =⇒ λ(X + m0 ) 6∈ A
0 < ρ λ(m0 + X) = ρ(λX) − λm0 =⇒ λm0 < ρ(λX)
=⇒ λ · sup m0 ∈ R | X + m0 6∈ A ≤ ρ(λX)
(⋆)
|
{z
}
=ρ(X)?
If we can show that the underbraced sup-term in the proof is equal to ρ(X)
the proof will be concluded. We know from above that:
sup{M0 } := sup m0 ∈ R | X + m0 6∈ A ≤ ρ(X)
6
and we know
ρ(X) = inf m ∈ R | ρ(X) ≤ m =: inf{M}
Strategy: show that the sup cannot be less than ρ(X), and therefore must
be equal. If sup{M0 } < ρ(X) = inf{M}, then we can define:
ε=
inf{M} − sup{M0 }
> 0.
2
}
< ρ(X),
We also define m
e :=
e = sup{M0 }+inf{M
2
sup{M0 } + ε. We note that m
and that ρ X + m
e = ρ(X) − m
e > 0, so X + m
e 6∈ A, however m
e > sup{M0 },
so we have a contradiction! This means sup{M0 } cannot be strictly less than
ρ(X) which means they are equal. Returning to (⋆) we have λρ(X) ≤ ρ(λX),
and with the first inequality we have shown that λρ(X) = ρ(λX).
From a m.r.m we can find an acceptance set, and conversly, from a set A ⊆ X
we can define a m.r.m ρA :
ρA := inf m ∈ R : m + X ∈ A , ∀X ∈ X
and we note that ρAρ = ρ.
Proposition 4.7 [FS 156]
Assume that A ⊆ X satisfies:
(i)
(ii)
(iii)
A 6= ∅
inf m ∈ R | m ∈ A > −∞
X ∈ A, Y ∈ X , Y ≥ X =⇒ Y ∈ A
then ρA satisfies:
(a) ρA is a monetary measure of risk.
(b) A is convex =⇒ ρA is a convex risk measure.
(c) A is a cone =⇒ ρA is positively homogeneous.
(Note that property b and c combined imply that ρA is coherent).
(d) Define
AρA := X ∈ X | ρA (X) ≤ 0
Then A ⊆ Aρ and if we have the closure property, i.e for X ∈ A, Y ∈ X ,
S := λ ∈ [0, 1] | λX + (1 − λ)Y ∈ A
then A = Aρ .
7
Lemma
(A) If X ∈ A, then ρA (X) ≤ 0. (Follows from the definition).
(B) A ⊆ AρA . (Follows from (a)).
(C) If (i), (ii) and (iii) from prop. 4.7 are satisfied, then X 6∈ A implies
ρA ≥ 0. In particular
X 6∈ AρA
=⇒ ρA (X) > 0
(C1 )
X ∈ AρA \ A =⇒ ρA (X) = 0
(C2 )
Proof
A, B and C1 are obvious. Proving C2 : We know that,
X ∈ AρA \ A =⇒ ρA ≤ 0.
Suppose for a contradiction that ρA (X) < 0, then ∃m < 0 such that
m + X ∈ A. Note that
m + X < |{z}
X
| {z }
6∈A
∈A
but by (iii) this is absurd! Strictly less is impossible, so we have equality. Proof of Proposition 4.7
27.01.11
(a) To prove that ρA is a m.r.m, we must verify that it is monotone and cash
invariant.
Monotonicity
Let X, Y ∈ X and assume X ≤ Y . Of course, m + X ≤ m + Y . Choose
m such that
(iii)
m + X ∈ A =⇒ m + Y ∈ A
inf m ∈ R | m + X ∈ A ≥ inf m
e ∈R|m
e +Y ∈A
ρA (X) ≥ ρA (Y )
Cash invariance
Let a ∈ R.
ρA (X + a) = inf m | m + X + a ∈ A = inf m
e −a|m
e +X ∈A
= inf m
e |m
e + X ∈ A − a = ρA (X) − a
8
Verifying two additional properties.
The risk measure always has an upper bound: ρA (X) ∈ R, ∀X ∈ X .
Take X ∈ X and Y ∈ A ⊆ X . Since X is the set of bounded functions,
∃m ∈ R such that X + m > Y .
Y ∈A
ρA (X) − m = ρA (X + m) ≤ ρA (Y ) ≤ 0 =⇒
ρA (X) ≤ m < ∞.
Risk measures are not minus infinity. For X ∈ X we can find a ∈ R
such that a + X ≤ 0. By monotonicity and cash invariance:
ρA (a + X) ≥ ρA (0)
(ii)
ρA (X) − a ≥ ρA (0) = inf m | m + 0 ∈ A} > −∞ =⇒
ρA (X) ≥ ρA (0) + a > −∞
(b) We assume A is convex, and show that ρA is convex.
We assume X1 , X2 ∈ X and take λ ∈ [0, 1]. We then consider
m1 , m2 ∈ R such that m1 + X1 ∈ A and m2 + X2 ∈ A. Since we
have assumed A is convex:
λ m1 + X1 + (1 − λ) m2 + X2 ∈ A
By definition of A, and then cash invariance.
0 ≥ ρA λ m1 + X1 + (1 − λ) m2 + X2
= ρA λX1 + (1 − λ)X2 − λm1 + (1 − λ)m2
We get,
ρA
ρA λX1 + (1 − λ)X2 ) ≤ λm1 + (1 − λ)m2
λX1 +(1−λ)X2 ) ≤ λ·inf m1 | m1 + X1 ∈ A +(1−λ)·inf m2 | m2 + X2 ∈ A
|
{z
}
{z
}
|
ρA (X1 ) by 4.6(b)
ρA (X2 ) by 4.6(b)
=⇒ ρA λX1 + (1 − λ)X2 ) ≤ λρA (X1 ) + (1 − λ)ρA (X2 )
With this we have proved that ρA is a convex function.
9
(c) We assume A is a cone, and show that ρA is positively homogeneous.
We consider some X ∈ X and m ∈ R : m + X ∈ A. Choose a λ > 0.
Since A is a cone, m + X ∈ A ⇒ λ(m + X) ∈ A. By definition of A
and cash invarianc:
0 ≥ ρA λ(m + X) = ρA (λX) − λm
In general, ρA (λX) ≤ λm, and in particular
ρA (λX) ≤ λ inf m | m + X ∈ A
≤λ·m
ρA (λX) ≤ λρA (X).
If we can show the opposite inequality as well, we have shown that ρA
is positively homogeneous. We consider m ∈ R such that we have the
strict inequality m < ρA (X).
m < ρA (X) =⇒ 0 < ρA (X) − m =⇒ 0 < ρA (X + m) =⇒ m + X 6∈ A
For λ > 0, and using that A is a cone, we also have λ(m + X) 6∈ A.
0 < ρA λX − λm =⇒ λm < ρA (λX) =⇒
λ · sup m ∈ R | m + X 6∈ A ≤ ρA (λX)
From this inequality, we can derive the following:
sup m ∈ R | m + X 6∈ A ≤ inf m ∈ R | m + X ∈ A
We show, as we did in the proof by contradiction of Proposition 4.6(d),
that the left side can not be strictly less, and therefore we must have
equality. This results in λρA (X) ≤ ρA (λX) which is the required
inequality, and we have proved (c).
(d) We assume
A ⊆ AρA := X ∈ X | ρA (X) ≤ 0
(i)
and that X has the “closure property”, i.e for every X ∈ A, Y ∈ X and
[0, 1] ⊇ S := λ ∈ [0, 1] | λX + (1 − λ)Y ∈ A
(ii)
(where S is closed in [0, 1]), and show that (i) and (ii) imply that A = Aρ .
10
By (i), we have A ⊆ Aρ which implies Acρ ⊆ Ac . To show that these
sets are equal we only have to show that Ac ⊆ Acρ .
We observe that λ = 1 ∈ S since we know that X ∈ A, but for λ = 0
we do not know, since we don’t know wether or not Y is in A.
Consider a new X ∈ Ac = X \ A, which we know is bounded since X
is bounded. We can find a m ∈ R such that
0 ≤ sup |X(ω)| = kXk < m.
ω∈Ω
Choose an ε ∈ (0, 1) such that
ε · m + (1 − ε) |{z}
X 6∈ A
6∈A
X is not in A by assumption, but we don’t know if m is.
0 ≤ ρA εm + (1 − ε)X = ρA (1 − ε)X − εm
=⇒ εm ≤ ρA (1 − ε)X
4.3 0 ≤ ρA (1 − ε)X − ρA (X) ≤ (1 − ε)X − X = εkXk < εm
Rearranging:
ρA (X) ≥ ρA (1 − ε)X − εkXk ≥ εm − ε = ε(m − kXk) > 0
so ρA (X) > 0 which means X 6∈ Aρ which is equivalent to saying
X ∈ Acρ . For an arbitrary X ∈ Ac we have X ∈ Acρ , so Ac ⊆ Acρ , and
the proof is completed.
Examples
As usual X denotes the linear space of all bounded, measurable functions on
some probability space (Ω, F ), and M1 = M1 (Ω, F ) denotes the class of all
probability measures on (Ω, F ).
Ex 4.8 Worst-case Risk measure [FS 157]
We define the worst-case r.m by
ρmax (X) = − inf X(ω)
ω∈Ω
The r.m ρmax is in fact a coherent r.m, which we can see by verifying
monotonicity, cash invariance and checking that Aρmax is a convex cone.
11
Monotonicity
If X ≤ Y , or more precisely, X(ω) ≤ Y (ω) ∀ω ∈ Ω, then
inf X(ω) ≤ inf Y (ω) =⇒
ω∈Ω
ω∈Ω
− inf X(ω) ≥ − inf Y (ω) =⇒
ω∈Ω
ω∈Ω
ρmax (X) ≥ ρmax (Y )
Cash invariance
For some m ∈ R we need to show that ρmax (X + m) = ρmax (X) − m.
ρmax (X + m) = − inf X(ω) + m
ω∈Ω
= − inf X(ω) − m
ω∈Ω
= ρmax (X) − m
Aρmax is a convex cone
Aρmax = X ∈ X such that ρmax (X) ≤ 0
= X|X≥0
From this we see that Aρmax is convex cone, and by Proposition 4.7 (b)
and (c), it follows that ρmax is a coherent risk measure.
Ex 4.11 Value at Risk (V aR) [FS 158]
We assume we have fixed a probabiliy measure P on the measurable space
(Ω, F ). A position X ∈ X is acceptable if P (X < 0) ≤ λ for some λ ∈ (0, 1).
Usually denote this r.m by V aRλ . The acceptance set for V aRλ is:
A = X ∈ X | P (X < 0) ≤ λ
(X does not necessarily have to be less than 0, could also be e.g a small
constant). The corresponding monetary risk measure is
V aRλ (X) = ρvar (X) = inf m ∈ R | P (m + X < 0) ≤ λ
We verify that this actually is a m.r.m by verifying monotonicity and cash
invariance.
12
Monotonicity
Assume X ≤ Y and derive the inequality V aRλ (Y ) ≥ V aRλ (X).
Writing out the risk measures.
V aRλ (X) = inf m ∈ R | P (m + X < 0) ≤ λ
V aRλ (Y ) = inf n ∈ R | P (n + Y < 0) ≤ λ
Since Y is larger than X, we need a bigger number to make it negative,
hence n ≥ m, so
X ≤Y
=⇒ V aRλ (Y ) ≥ V aRλ (X).
Cash invariance
V aRλ (X + a) = inf m ∈ R | P (m + a + X < 0) ≤ λ
= inf x − a ∈ R | P (x + X < 0) ≤ λ m=x−a
= inf x | P (x + X < 0) ≤ λ − a
= V aRλ (X) − a
Positive homogeneity
We assume a > 0.
V aRλ (aX) = inf m ∈ R | P (aX + m < 0) ≤ λ
= inf ax ∈ R | P (X + x < 0) ≤ λ x=m/a
= a inf x ∈ R | P (X + x < 0) ≤ λ
= a · V aRλ (X)
Ex 4.41 Value at Risk 2 [FS 158]
We consider the case where we invest in two corporate bonds, each with
payoff re, where r is the return on a riskless investment and re > r ≥ 0.
These bonds are defaultable, i.e they can default and become worthless. We
assume the bonds are independently defaulting (e.g they belong to dofferent
non-connected market sectors). We introduce w > 0 as the size of the initial
investment.
The discounted net gain for bond i is given by (probabilities denoted by pi )

 −w
default
pi
re − r Xi =
w
otherwise (1 − pi )
1+r
13
Now we have enough information to find the V aR for the first bond, which
defaults with probability p1 . We choose some λ ∈ (0, 1) so p1 ≤ λ.
V aRλ (Y ) = inf a | P (a + Y < 0) ≤ λ
a = w =⇒ P X1 < −w = 0 ≤ λ
(It is impossible to loose more than the initial investment).
a = −w
So,
re − r 1+r
V aRλ (X1 ) = −w
re − r =⇒ P X1 < w
= p1 ≤ λ
1+r
re − r < 0 =⇒ X1 is acceptable.
1+r
Let us see what happens if we diversify our investment, by investing w/2 in
each bond, so we have Z = 12 X1 + 21 X2 , and if both default, one defaults or
none default we get (with the corresponding probabilities):

p1 p2

 −w




 w re − r
−
1
<0
p1 (1 − p2 ) ∨ p2 (1 − p1 )
Z=
2 1+r






 w re − r
(1 − p1 )(1 − p2 )
1+r
In the second column we used
w w re − r w re − r
− +
=
−1
2
2 1+r
2 1+r
and to establish the last, strict inequality,
0 ≤ r < re < 2r + 1.
a = w =⇒ P (Z < −w) = 0 ≤ λ
w
re − r w re − r
a=
1−
=⇒ P Z <
− 1 = p1 p2
2
1+r
2 1+r
a = −w ·
re − r re − r
= p1 p2 + p1 (1 − p2 ) + p2 (1 − p1 )
=⇒ P Z < w
1+r
1+r
= p1 + p2 − p1 p2
14
To simplify we assume p1 = p2 = p, so p1 p2 = p2 and p1 + p2 − p1 p2 = 2p − p2 .
Since p ≤ λ, we get p2 ≤ λ ≤ 2p − p2 , so we get
V aRλ (Z) =
w
re − r 1−
> 0 =⇒ V aRλ (Z) > V aRλ (X1 )
2
1+r
so V aRλ is not a convex measure which subsequently means it punishes
diversification. This means that the acceptance set, A is not convex, but
that is not obvious:
A = X ∈ X | P (X < 0) ≤ λ .
4.2 Robust Representations of Risk Measures [FS 161]
Linear functions vs Additive Set functions
Definition A.49 [FS 426]
Let (Ω, F ) be a measurable space. A mapping µ : F 7→ R is called a finitely
additive set function if (i) and (ii) are satisfied. It is called a finitely additive
probability measure if we also have (iii):
(i)
(ii)
µ(∅) = 0
Pairwise disjoint A1 , . . . , AN ∈ F , implies µ
(iii)
µ(Ω) = 1.
N
[
i=1
Ai =
N
X
µ(Ai )
i=1
We use M1,f := M1,f (Ω, F ) to denote the set of all finitely additive set
functions µ : F 7→ [0, 1] where µ(Ω) = 1, so M1,f is the set of all finitely
additive probability measures. The total variation of a finitely additive set
function µ is defined, where A1 , . . . , AN are mutually disjoint sets in F and
N ∈N
N
X
kµkvar := sup
|µ(Ai)|
A1 ,...,AN
i=1
With ba := ba(Ω, F ) we denote the collection of set functions with finite
total variation. We note that M1,f ⊂ ba, since for some Q ∈ M1,f ,
(A := {A1 , . . . , AN }),
kQkvar
N
N
[
X
≤ sup
Q(Ai ) = sup Q
Ai ≤ Q(Ω) = 1 < ∞
A
i=1
A
15
i=1
Integration Theory
This is a brief outline of the integration theory wrt a measure µ ∈ ba. The
space X endowed with the supremum norm k · k,
kF k := sup |F (ω)|, F ∈ X
ω∈Ω
is a Banach space. This is a space with a nice structure where we can define
a topology, Cauchy sequences converge and we can define a limit. For an
element F ∈ X , we can define the simple (or elementary) representation:
F ∈ X =⇒ F =
N
X
αi 1Ai
i=1
for some constants αi . From this representation, we can define the integral.
Z
F dµ :=
Ω
N
X
αi µ(Ai ) =: ℓµ (F )
i=1
We have
N
N
X
Z
X
|αi ||µ(Ai )| ≤ sup |F (ω)| ·
|µ(Ai )| ≤ kF k · kµkvar
F dµ ≤
ω
i=1
i=1
For any F ∈ X can be approximated by simple functions:
k·k
Fn −→ F, n → ∞.
Z
Z
Z
=
(F
−
F
)dµ
F
dµ
−
F
dµ
≤ kFn − Fm k · kµkvar −→ 0
n
m
n
m
m,n→0
R
som
Fn dµ n is Cauchy. By setting µ = Q ∈ M1,f ⊂ ba, we denote the
integral by
Z
F dQ = EQ [F ].
Theorem A.50 [FS 427]
The integral
ℓ(F ) =
Z
F dµ,
F ∈X
defines a one-to-one correspondence with the finitely additive set functions
µ ∈ ba, and the linear continuous functionals ℓ on X . We have a one-to-one
correspondence ℓ ↔ µ.
16
Proof
←) From µ ∈ ba to the definition of ℓµ we simply follow the construction of
the integral.
→) We define µ(A) = ℓ(1A ) for A ∈ F , where 1 is the indicator function.
For pairwise disjoint sets A1 , . . . , AN , which we abbreviate as A, we get
kµkvar = sup
A
N
X
µ(Ai )
i=1
= ℓ(1Ai )
We divide the sum into negative and positive terms.
= sup
A
= sup
A
= sup
A
The norm of ℓ is given by
N
X
µ+ (Ai ) + µ− (Ai )
i=1
X
X
ℓ(1Ai ) + ℓ(1Bj )
ℓ(1Ai + 1Bj ) ≤ kℓk
kℓk := sup kℓ(X)k
kXk≤1
For linear, continuous functions they are bounded.
µ ∪ Ai = ℓ 1∪Ai = ℓ ΣAi = Σℓ 1Ai = Σµ(Ai )
For finite measures, in the sense that the total variation is finite, we have
a one-to-one correspondence to linear, continuous functionals, The mapping
that links them is an integral.
Summary
03.02.11
R
The integral ℓ(F ) = F dµ for F ∈ X defines a one-to-one relationship
between the spaces
X = ℓ | X 7→ R where ℓ is continuous and linear
and
ba = ba(Ω, F ) = µ : F 7→ R | signed additive bounded tot. variation
σ-additivity is equivalent to An ր Ω, that is, An ⊆ An+1 ⇒ ∪n An = Ω.
For the measure Q : Q(An ) ≤ Q(An+1 ), a monotonic sequence, Q(An ) ր
Q(Ω) = 1 (for a probability measure). ℓ is continuous and linear: ℓ(X) ≥
0, X ≥ 0, ℓ(1) = 1.
17
Theorem 4.15 [FS 162] - Rep. Theorem for Risk Measures
Any convex risk measure ρ : X 7→ R admits representation as:
ρ(X) = sup EQ [−X] − αmin (Q)
Q∈M1,f
where
αmin (Q) = sup EQ [−X],
X∈Aρ
Q ∈ M1,f .
If there is any other α penalty function such that
ρ(X) = sup EQ [−X] − α(Q)
Q∈M1,f
then α(Q) ≥ αmin (Q), for all Q ∈ M1,f . (The penalty function isn’t unique).
We can use the max instead of sup. The EQ [−X] is the linear part, and αmin
is an element in the family of penalty functions, and it penalizes linearity.
In general:
α : M1,f 7→ R ∪ {∞},
such that
inf
Q∈M1,f
α(Q) ∈ R.
The function
X 7→ EQ [−X] − α(Q)
is convex, cash invariant and convex, and these properties are preserved when
we take the sup (we call this function ρ for now).
ρ(0) = − inf α(Q)
ρ(X) : X 7→ sup EQ [−X] − α(Q) ,
Q∈M1,f
Q∈M1,f
Proof of Theorem 4.15
We show that ρ(X) both dominates and is dominated by the representation
in theorem 4.15.
Step 1
ρ(X) ≥ sup
Q∈M1,f
EQ [−X] − αmin (Q)
If we define X ′ := X + ρ(X) for X ∈ X , we get, by cash invariance,
ρ(X ′ ) = ρ X + ρ(X) = ρ(X) − ρ(X) = 0 =⇒ X ′ ∈ Aρ .
18
By the definition of αmin we have that for all Q ∈ M1,f ,
αmin (Q) = sup EQ [−Z] ≥ EQ [−X ′ ] = EQ [−X − ρ(X)] = EQ [−X] − ρ(X)
Z∈Aρ
αmin (Q) ≥ EQ [−X] − ρ(X) =⇒ ρ(X) ≥ EQ [−X] − αmin (Q)
The left side of the inequality is independent of Q, so we can take the sup
on the right side, and we have derived the first inequality.
ρ(X) ≥ sup EQ [−X] − αmin (Q)
Q∈M1,f
Step 2
ρ(X) ≤ sup
Q∈M1,f
EQ [−X] − αmin (Q)
To derive this inequality, we find a measure QX such that
ρ(X) ≤ EQX [−X] − αmin (QX ) .
When this is true, it is also true when we take the sup over all Q, and this
will subsequently result in the required inequality. We need a result from
functional analysis:
Theorem A.54 [FS 429]
In a topological vector space E, any two disjoint convex sets B and C,
where at least one of them has an interior point, can be separated by a
non-zero continuous linear functional ℓ on E:
ℓ(X) ≤ ℓ(Y ) ∀X ∈ C, ∀Y ∈ B.
We define the set X0 ⊆ X by
X0 := X ∈ X | ρ(X) = 0
Suppose Y ∈ X \ X0 , and define X := Y + ρ(Y ) If ρ(X) = 0, then
Y = X − ρ(Y ) [?]. Without loss of generality we can assume ρ(0) = 0,
since if ρb(0) = c which implies, by cash invariance, ρb(−c) = 0, then we can
set ρ(X) := ρb(X − c) for X ∈ X .
Take X ∈ X0 and define the set
B := Y ∈ X | ρ(Y ) < 0 ,
19
then we have X 6∈ B and X ∈ C = Bc , which are complementary sets, and
one of them is non-empty. These are the necessary conditions of Theorem
A.54, and we have the existence of a non-zero, continuous ℓ:
∃ℓ : X 7→ R =⇒ ℓ(Z) ≤ ℓ(Y ),
In fact,
∀ |Z {z
∈ C}, ∀Y ∈ B
∀X∈X0
ℓ(X) ≤ inf ℓ(Y ) := b ∈ R.
Y ∈B
By properties of ℓ we have:
ℓ(X)
(1) ℓ(X) ≥ 0, X ≥ 0
b
which means we can normalize: ℓ(X)
=
(2)
ℓ(1) > 0
ℓ(1)
Verifying these two properties.
(1) Y ≥ 0 ⇒ ℓ(Y ) ≥ 0 We first claim that:
Y ≥ 0 =⇒ 1 + λY ∈ B, ∀λ > 0
By monotonicity,
Y ≥ 0 =⇒ λY ≥ 0 =⇒ ρ(λY ) ≤ ρ(0)
We can exploit the following relationship: 1 + λY > λY .
ρ(1 + λY ) = ρ(λY ) − 1 < ρ(λY ) ≤ ρ(0) = 0
So 1+λY ∈ B. Now, for X 6∈ B (with ρ(X) = 0) we apply the inequality
for ℓ, and also use that ℓ is a linear functional:
ℓ(X) ≤ ℓ(1 + λY ) = ℓ(1) + λℓ(Y ), ∀λ > 0.
This holds for all λ, so this inequality would not be possible if ℓ(Y ) < 0,
hence ℓ(Y ) ≥ 0.
(2) ℓ(1) > 0 For some Y ∈ X we have
0 < ℓ(Y ) = ℓ(Y + ) − ℓ(Y − )
Also,
kY k := sup |Y (ω)| = sup Y + (ω) + Y − (ω) < 1
ω∈Ω
ω∈Ω
20
From these equations, we know that
ℓ(Y + ) > 0,
and 1 > Y + + Y − ≥ Y + ≥ 0
=⇒ 1 − Y + > 0 =⇒ ℓ(1 − Y + ) ≥ 0
Now we can use linearity of ℓ again:
ℓ(1) = ℓ(1 − Y + + Y + ) = ℓ(1 − Y + ) + ℓ(Y + ) > 0 =⇒ ℓ(1) > 0
| {z } | {z }
≥0
>0
b
The normalized ℓ(X)
:= ℓ(X)/ℓ(1) for X ∈ X is in a one-to-one relationship
b
with some Q.
b
EQb [X] = ℓ(X),
X∈X
With this we can derive the following inequality (in (⋆) we are reducing the
set, which we will verify)
(⋆)
b )=− b
b
b = sup E b [−Y ] ≥ sup E b [−Y ] = sup ℓ(−Y
) = − inf ℓ(Y
αmin (Q)
Q
Q
Y ∈B
ℓ(1)
Y ∈B
Y ∈B
Y ∈Aρ
Now we verify that we can do the step indicated by (⋆). This is true if B ⊆ Aρ ,
which is true if any arbitrary element Y ∈ B is automatically in Aρ , but by
the defining properties of each set this is true:
Y ∈ B =⇒ ρ(Y ) < 0 =⇒ ρ(Y ) ≤ 0 =⇒ Y ∈ Aρ , =⇒ B ⊆ Aρ.
To derive the other inequality,
Y ∈ Aρ , ∀ε > 0 =⇒ Y + ε ∈ B, since ρ(Y + ε) = ρ(Y ) − ε < 0
Now, for all ε > 0 and all Y ∈ Aρ ,
b
sup EQb [−Z] ≥ EQb [−(Y + ε)] =⇒ sup EQb [−Z] ≥ sup EQb [−Y ] = αmin (Q)
Z∈B
Z∈B
Y ∈Aρ
b =
Combining these inequalities, we have αmin (Q)
b
b = ℓ(−X)
EQb [−X] − αmin (Q)
−
−
−b
.
ℓ(1)
b b − ℓ(X)
=
≥ 0 = ρ(X)
ℓ(1)
ℓ(1)
b ∈ M1,f ,
which is true for all X ∈ X0 . We have, for some Q
b
ρ(X) ≤ EQb [−X] − αmin (Q)
21
b so we can take the sup and keep the
The left side is independent of Q,
inequality.
ρ(X) ≤ sup EQ [−X] − αmin (Q)
Q∈M1,f
and we have proved the representation of ρ from Theorem 4.15.
Now we verify that αmin is the smallest penalty function. From our
representation theorem and for some α:
ρ(X) = sup EQ (−X) − α(Q) ≥ EQ [−X] − α(Q)
Q∈M1,f
Shifting terms, we get:
α(Q) ≥ EQ [−X] − ρ(X)
We get:
α(Q)
≥ sup EQ [−X] − ρ(X)
X∈X ≥ sup EQ [−X] − ρ(X)
since Aρ ⊆ X
X∈Aρ
≥ sup EQ [−X]
X∈Aρ
since X ∈ Aρ ⇒ ρ(X) ≤ 0
so α(Q) ≥ αmin (Q).
= αmin (Q)
Remarks: From Theorem 4.15 we have:
αmin (Q) = sup EQ [−X]
ρ(X) ≥ EQ [−X] − αmin (Q), ∀X
and
Aρ
So,
αmin (Q) ≥ EQ [−X] − ρ(X) =⇒
αmin (Q) ≥ sup
X∈Aρ
EQ [−X] − ρ(X)
≥ sup EQ [−X] = αmin (Q) =⇒
X∈Aρ
αmin (Q) = sup
X∈Aρ
EQ [−X] − ρ(X)
22
(♥)
Definition A.59 [FS 430] - Fenchel-Lengendre Transform
For a function f : X 7→ R ∪ {∞}, we define the Fenchel-Legendre transform,
f ∗ : X ′ 7→ R ∪ {±∞} by
f ∗ (ℓ) := sup ℓ(X) − f (X) , ℓ ∈ X ′ .
X∈X
If f 6≡ ∞, then f ∗ is a proper convex (true for risk measures) and lower
semicontinuous function as the supremum of affine functions.
lim inf f ∗ (ℓ) ≥ f ∗ (ℓ0 )
ℓ→ℓ0
(ℓ converges to ℓ0 in X ′ in the weak star sense). If f is itself a proper convex
function, then f ∗ is proper convex and lower semicontinuous. We call f ∗
the conjugate functional. Theorem A.61 [FS 431] states that if f is lower
semicontinuous, then f ∗∗ = f .
Remark [FS 164]: In view of this, we have that the equation (♥) corresponds
to the Fenchel-Legendre transform (or conjugate function) of the convex
function ρ on the Banach space X . More precisely:
αmin (Q) = ρ∗ (ℓQ )
where ρ∗ : X ′ → R ∪ {+∞} is defined on the dual X ′ of X by
ρ∗ (ℓ) = sup ℓ(X) − ρ(X)
X∈X
and where ℓQ (X) = EQ [−X] for X ∈ X and Q ∈ M1,f .
Given a set A, we can find a risk measure:
ρA (X) = inf m ∈ R | m + X ∈ A
By Theorem 4.15, this r.m has the representation:
ρA (X) = max EQ [−X] − αmin (Q)
Q∈M1,f
where:
αmin (Q) = sup EQ [−X] ≥ sup EQ [−X].
X∈AρA
X∈A
23
We want to have an expression for αmin with A and not AρA . We know the
following inequality is true, but now we want to derive the opposite inequality
and get the expression with only A.
Y ∈ AρA =⇒ ρ(Y ) ≤ 0
∀ε > 0, ρ(Y + ε) = ρ(Y ) − ε < 0 =⇒ Y + ε ∈ A
sup EQ [−X] ≥ EQ [−(Y + ε)]
X∈A
sup EQ [−X] ≥ sup EQ [−Y ].
X∈A
Y ∈AρA
This shows that we have αmin (Q) = supX∈A EQ [−X].
Corollary 4.18 [FS 165]
If ρ is a coherent risk measure, then
0
in Qmax ⊆ M1,f
αmin (Q) =
c
+∞ in Qmax
Qmax is a convex set and it is the largest set where you can find a
representation
ρ(X) = sup EQ [−X] − αmin (Q) = max EQ [−X].
Q∈Qmax
Q∈M1,f
We define
Qmax := Q ∈ M1,f | αmin (Q) = 0
Proof
From Proposition 4.6 we know that the acceptance set Aρ of a coherent r.m
is a cone. For λ > 0 we apply positive homogeneity:
αmin (Q) = sup EQ [−X] = sup EQ [−λX] = λ· sup EQ [−X] = λ·αmin (Q).
X∈Aρ
λX∈Aρ
λX∈Aρ
This is true for all Q ∈ M1,f and λ >. Hence αmin can only take values 0
and +∞.
Now we focus our attention to the convex risk measures that admit a
representation in terms of the σ-additive probability measures. Such measures
24
ρ can be represented by a penalty function α which is infinite outside the set
M1 := M1 (Ω, F ).
ρ(X) = sup EQ [−X] − α(Q)
(⋆)
Q∈M1
Lemma 4.20 [FS 166]
If a convex risk measure admits the representation (⋆), then it is
(a) continuous from above:
Xn ց X =⇒ ρ(Xn ) ր ρ(X).
(b) Property (a) is equivalent to lower semicontinuity with respect to
bounded pointwise conergence. If {Xn } is a bounded sequence in X
which converges pointwise to X ∈ X , then
ρ(X) ≤ lim inf ρ(Xn ).
n→∞
Proof
We show that (⋆) ⇒ (a), then (a) ⇒ (b) and finally (b) ⇒ (a).
(⋆) ⇒ (a)
Xn → X is a bounded sequence, so by dominated convergence, E[Xn ] →
E[X]. For all Q ∈ M1
lim EQ [−Xn ] − α(Q)
Q∈M1 n→∞
≤ lim inf sup EQ [−Xn ] − α(Q)
(⋆)
ρ(X) = sup
n→∞
Q∈M1
= lim inf ρ(Xn )
n→∞
We use the liminf because we don’t know if we have convergence.
(b) ⇒ (a)
For Xn ց X we have ρ(Xn ) ≤ ρ(X), since ρ(Xn ) ≤ ρ(Xn+1 ) etc. This
means: limn→∞ ρ(Xn ) ≤ ρ(X) but as we verified in the first step: ρ(X) ≤
lim inf ρ(Xn ), thus
n→∞
ρ(X) = lim ρ(Xn )
n→∞
(a) ⇒ (b)
Xn → X is a bounded sequence. If we define Ym := supn≥m Xn we have
25
a decreasing sequence (and we know that Ym ց X). By monotonicity,
Yn ≥ Xn ⇒ ρ(Yn ) ≤ ρ(Xn ), so since Yn ց X:
ρ(X) = lim ρ(Yn ) ≤ lim inf ρ(Xn )
n→∞
n→∞
Proposition 4.21 [FS 167]
Let ρ be a convex risk measure which is continuous from below:
Xn ր X =⇒ ρ(Xn ) ց ρ(X).
Let α be some penalty function in
ρ(X) = max EQ [−X] − α(Q) .
M1,f
Then α is concentrated on the class M1 of probability measures, i.e
α(Q) < ∞ =⇒ Q is additive.
Proof
Result follows since we know Q ∈ M1,f , which is all we need.
Q is σ-additive ⇐⇒ An ր Ω : Q(An ) ր Q(Ω) = 1
Lemma 4.22 [FS 167] (Not proved)
Let ρ be a convex risk measure with the usual representation:
ρ(X) = max EQ [−X] − α(Q) .
Q∈M1,f
Now consider the set:
Λc := Q ∈ M1,f | α(Q) ≤ c ,
[
Q ∈ M1,f | α(Q) < ∞ =
Λc
c
for c > −ρ(0) = inf Q∈M1,f α(Q) > −∞. Then for any sequence {Xn } in X
where 0 ≤ Xn ≤ 1, the following two statements are equivalent.
ρ(λX) → ρ(λ), ∀λ ≥ 1
26
inf EQ [Xn ] → 1, ∀c > −ρ(0).
Q∈Λc
Sketch of proof
For all λ ≥ 0,
ρ(λ1An ) → ρ(λ1Ω ) = ρ(λ).
By monotonicity, Q(An ) ≤ Q(An+1 ), so
inf Q(An ) ≤ inf Q(An+1 ),
Q∈Λc
Q∈Λc
and we have a growing sequence. By monotone convergence,
inf Q(An ) ր 1, ∀c =⇒ Q(An ) ր 1.
Q∈Λc
Remark: If ρ is convex and continuous from below, then, by Proposition 4.21:
ρ(X) = sup EQ [−X] − α(Q)
Q∈M1
and then, by Lemma 4.20, ρ is continuous from above, which implies that for
all bounded sequences, we can use the dominated convergence theorem, so
Xn → X means that ρ(Xn ) → ρ(X).
From now on, we make some assumptions:
Ω
F = B(Ω)
X
Cb (Ω) ⊂ X
Separable metric space
the Borel σ-algebra on Ω.
linear space of bounded functions on (Ω, F )
Bounded, continuous functions on Ω.
Proposition 4.25 [FS 168]
If ρ : X 7→ R is a convex risk measure on X such that ρ(Xn ) ց ρ(λ) for any
pointwise convergence Xn ր X, where Xn ∈ Cb (Ω) for all n and λ > 0, then
there exists a penalty function α : M1 7→ R ∪ {∞} such that
ρ(X) = sup EQ [−X] − α(Q) , X ∈ Cb (Ω).
Q∈M1
General representation (true for a larger set)
ρ(X) = max EQ [−X] − αmin (Q) , X ∈ X
Q∈M1
We can choose α:
α(Q) =
inf
e
Q∈M
1,f
e ,
αmin (Q)
EQe [Y ] = EQ [Y ], ∀Y ∈ Cb (Ω)
27
10.02.11
The equality between the expectations is a special kind of equality, similar
to equality under distributions.
Recall:
αmin (Q) = sup EQ [−X] = sup EQ [−X] − ρ(X)
X∈X
X∈Aρ
4.3 Convex Risk Measures on L∞ [FS 171]
From now on we assume we have a fixed probability measure P on (Ω, F ),
identify X as L∞ and consider risk measures ρ such that
ρ(X) = ρ(Y ),
if X = Y P − a.s.
(4.28)
Lemma 4.30 [FS 172]
If ρ is a convex risk measure satisfying (4.28), and it has the representation
ρ(X) = sup EQ [−X] − α(Q) ,
Q∈M1,f
then α(Q) = +∞ for any Q ∈ M1,f (Ω, F ) which is not absolutely continuous
wrt P . (Q is abs. cont. to P , Q ≪ P if P (A) = 0 =⇒ Q(A) = 0).
Proof
Assume Q 6≪ P , then ∃A ∈ F such that P (A) = 0 and Q(A) > 0. Take some
X ∈ Aρ = X ∈ X | ρ(X) = 0 ,
and construct a sequence Xn := X − n · 1A , so Xn = X P -a.s, which means
ρ(Xn ) = ρ(X).
e
0 ≥ ρ(Xn ) = sup EQ [−Xn ] − α(Q) ≥ EQe [−Xn ] − α(Q)
Q∈M1,f
e
e =⇒
= EQe [−X] + n Q(A)
−αQ
| {z }
>0
e ≥ E e [−X] + nQ(A)
e
α(Q)
Q
The same argument applies to the representation of ρ via αmin
e
e = sup EQ [−X] ≥ E e [−X] + nQ(A)
=⇒
αmin (Q)
Q
X∈Aρ
e ≥ αmin (Q)
e ≥ E e [−X] + n · Q(A)
e
α(Q)
−→ ∞
Q
n→∞
28
Theorem 4.31 [FS 172]
Suppose ρ : L∞ 7→ R is a convex risk measure. Then the following statements
are equivalent.
(a) ρ can be represented by some penalty function on M1 (P ) ⊆ M1 , where
M1 (P ) = Q | Q ≪ P
(b) ρ admits the representation
ρ(X) = sup
EQ [−X] − αmin (Q) , X ∈ L∞
Q∈M1,f (P )
where αmin M1 (P ) = αmin .
(c) ρ is continuous from above: if ∀Xn ց X P -a.s, then ρ(Xn ) ր ρ(X).
(d) The “Fatou property”. For any bounded sequence Xn → X (P -a.s
pointwise convergence):
ρ(X) ≤ lim inf ρ(Xn )
n→∞
(e) ρ is lower semicontinuous for the weak∗ topology on L∞ .
(f) The set Aρ ⊂ L∞ is weak∗ closed.
Proof
We will prove (b) ⇒ (a) ⇒ (c) ⇒ (d) ⇒ (e) ⇒ (f ) ⇒ (b).
(b) ⇒ (a). This is obvious.
(a) ⇒ (c). Recall Lemma 4.20 part (a). For ρ : X 7→ R and ρ(X) given as in
Lemma 4.20, we have continuity from above.
(c) ⇒ (d). Lemma 4.20 part (b). Continuity from above is equivalent to
ρ(X) ≤ limn→∞ ρ(Xn ) for all Xn → X where {Xn } is bounded. P -a.s.
(d) ⇒ (e). We have lower semi-continuity for ρ if ∀c ∈ R the set C ,
{X ∈ L∞ | ρ(X) ≤ c} := C
is closed (in the weak∗ sense). For weak∗ convergence we must specify
closedness. Note that C is convex, since for X, Y ∈ C and some λ ∈ [0, 1]
(using that ρ is convex):
ρ λX + (1 − λ)Y ≤ λρ(X) + (1 − λ)ρ(Y ) ≤ λc + (1 − λ)c = c
29
=⇒ λX + (1 − λ)Y ∈ C
The set C is weak∗ closed if ∀r > 0, the set Cr ,
Cr := C ∩ Y ∈ L∞ kY k∞ ≤ r
L
1
is closed in L1 . Take a sequence Xn ∈ Cr such that for all n, Xn −→
X. We
want to show that X ∈ Cr , which means it is closed. Every Xn is bounded,
so the limit X is also bounded, and hence
X ∈ Y ∈ L∞ kY k∞ ≤ r .
Now for L1 convergence, we have a subsequence that converges P -a.s.in a
pointwise sense: ∃{Xnk }k ⊆ {Xn }n such that Xnk −→ X. Then property (d)
k→∞
tells us:
ρ(X) ≤ lim inf ρ(Xnk ) ≤ c =⇒ X ∈ C .
k→∞
This proves that C is a closed set.
(e) ⇒ (f ). For Aρ : X | ρ(X) ≤ 0 just take c = 0 in C and we are done.
(f ) ⇒ (b). Given a X ∈ L∞ , and defining
m :=
sup
Q∈M1 (O)
EQ [−X] − αmin (Q) .
We want to show that m = ρ(X) and do that by showing that ρ(X) ≥ m
and then ρ(X) ≤ m.
For the first inequality:
ρ(X) = max
Q∈M1,f
≥
EQ [−X] − αmin (Q)
sup {EQ [−X] − αmin (Q) = m.
Q∈M1 (P )
For the other inequality, we use the representation:
ρ(X) = inf a ∈ R | X + a ∈ Aρ .
With this we want to show that ρ(X) ≤ m, which by cash invariance amounts
to ρ(X + m) ≤ 0 which is equivalent to showing that X + m ∈ Aρ . For a
contradiction we assume m + X 6∈ Aρ . Since ρ is a convex measure, we know
that Aρ is a convex set. By our assumption that (d) is true, we also have
that Aρ is weak∗ closed.
30
We define the single point set B = {X + m}, and by our assumption B 6∈ Aρ ,
so B ∩Aρ = ∅. By applying the Hahn-Banach Theorem (Thm A.56 [FS 429]),
∃ℓ : L∞ 7→ R which is non-zero, linear and continuous such that
sup ℓ(X + m) < inf ℓ(Y ) abbreviated as γ < β.
Y ∈Aρ
B
(⋆)
we also know that −∞ < γ. We always have the representation ℓ(Y ) = E[Y Z]
for Y ∈ L∞ and Z ∈ L1 , where Z ≥ 0, P -a.s., and P (Z > 0) > 1.
Now take Y ≥ 0 and λ > 0, which means:
λY ≥ λ·0 =⇒ ρ(λY ) ≤ ρ(0) =⇒ ρ λY +ρ(0) ≤ 0 =⇒ λY +ρ(0) ∈ Aρ
By (⋆),
γ
ℓ(ρ(0))
γ < ℓ λY + ρ(0) = λℓ(Y ) + ℓ(ρ(0)) =⇒
= ℓ(y) +
λ
λ
and when passing to the limit: λ → ∞, 0 ≤ ℓ(Y ) = E[Y Z]. Now define
Z
0
Q0 ∈ M1 (P ), given by dQ
= E[Z]
, which implies
dP
h
Z i
=⇒
EQ0 [Y ] = E Y ·
E[Z]
β = inf ℓ(Y ) = inf EQ0 [Y ] · E[Z] =⇒
Y ∈Aρ
Y ∈Aρ
β
= sup EQ [−Y ] = αmin (Q0 )
E[Z] Y ∈Aρ
Hence, (with the strict inequality a consequence of (⋆)):
EQ0 [X] + m =
ℓ(X + m)
β
<
= −αmin (Q0 ) =⇒
E[Z]
E[Z]
m < EQ0 [−X] − αmin (Q0 )
This is a contradiction, since by definition:
m = sup
EQ [−X] − αmin (Q) .
Q∈M1 (P )
Our assumption that m + X 6∈ Aρ was false, so ρ(X) ≤ m and we have
proved ρ(X) = m.
31
Entropic Risk Measure
For risk measures Q and P we have the relative entropy:

h
i
dQ
dQ
dQ

 EQ log
=E
if Q ≪ P
log
dP
dP
dP
H(Q | P ) =


+∞
otherwise
(⋆)
We have H(Q|P ) = 0 if Q = P . (For f (x) = x·log x ⇒ Jensen’s inequality).
For β > 0 we set:
α(Q) =
1
H(Q|P ).
β
With this we have risk measure for when we strongly believe that P is the
correct measure.
ρ(X) =
1
EQ [−X] − H(Q|P )
β
Q∈M1 (P )
sup
(⋆⋆)
An alternative representation of the expression in (⋆), (but one that isn’t
very important to us) is
H(Q|P ) = (⋆) = sup EQ [−Z] − log E[eZ ] .
Z∈L∞
If we set Z = −βX in this expression, and remove the supremum, we have:
H(Q|P ) ≥ EQ [−βX] − log E[e−βX ]
1
1
1
H(Q|P ) ≥ EQ [−βX] − log E[e−βX ]
β
β
β
1
1
H(Q|P ) ≥ EQ [−X] − log E[e−βX ]
β
β
1
1
log E[e−βX ] ≥ EQ [−X] − H(Q|P )
β
β
The left side is now independent of Q, so we can take the supremum on the
right.
1
1
log E[e−βX ] ≥ sup
EQ [−X] − H(Q|P ) = ρ(X)
β
β
Q∈M1 (P )
For a particular choice:
b
dQ
e−βX
=
dP
E[e−βX ]
32
(♠)
we derive from (⋆):
and
so,
b
b
d
Q
d
Q
b )=E
H(Q|P
log
dP
dP
−βX
e−βX
e
log
=E
E[e−βX ]
E[e−βX ]
−βX e
−βX
−βX
=E
log e
− log E[e
]
E[e−βX ]
−βX
e
−βX
− βX − log E[e
]
=E
E[e−βX ]
h
bi
e−βX
dQ
= E −X ·
EQb [−X] = E − X ·
dP
E[e−βX ]
1
b )=
EQb [−X] − H(Q|P
β
−βX
1
e
e−βX
−βX
− E
− βX − log E[e
] =
E −X ·
E[e−βX ]
β E[e−βX ]
−βX
e−βX
e
1
−βX
E −X ·
−E
− X − log E[e
] =
E[e−βX ]
E[e−βX ]
β
−βX
e−βX 1
e−βX
e
−βX
log E[e
] =
E −X
· −βX + X · −βX +
] E[e
] E[e−βX ] β
E[e
h e−βX 1
i
i
h1
1
−βX
−βX
E
e
=
E
log
E[e
]
= E[e−βX ]
b
Q
−βX
E[e
]β
β
β
b
The Q-expectation
disappears because the interior is just a number.
Summarising, we have:
EQb [−X] −
1
b ) = 1 E[e−βX ]
H(Q|P
β
β
Taking the supremum over all Q on the left side, we get the reverse inequality
from (♠), so:
1
ρ(X) = log E[e−βX ]
β
The logarithm is not a linear function, so for λ ≥ 0 we have ρ(λX) 6= λρ(X).
The measure is in fact normalized, but it is not coherent since it is not
homogeneous.
33
For the representation:
ρ(X) =
sup
Q∈M1 (P )
EQ [−X] − αmin (Q)
we see from equation (⋆⋆) that αmin = β1 H(Q|P ).
By definition (X = L∞ ):
αmin (Q) = sup EQ [−X] = sup EQ [−X] − ρ(X)
X∈X
X∈Aρ
1
1
EQ [−X] − log E[e−βX ] = H(Q|P )
β
β
X∈L∞
= sup
Definition 4.32 [FS 173]
A convex risk measure is sensitive (or relevant) w.r.t P if:
ρ(−X) > ρ(0) ∀X ∈ L∞ such that X ≥ 0 P -a.s., P (X > 0) > 0.
Corollary 4.35 [FS 175]
For a coherent risk measure ρ on L∞ , the following conditions are equivalent.
(a) ρ is continuous from below:
∀Xn ր X P -a.s. =⇒ ρ(Xn ) ց ρ(X).
(b) There exists a set Q :
Q ⊂ M1 (P ) such that ρ(X) = sup EQ [−X].
Q⊂Q
(c) There exists a set Q ⊂ M1 (P ) representing ρ such that the set of densities:
dQ D :=
Q∈Q
dP
is weakly compact in L1 (P ).
So a coherent risk measure on L∞ can be represented on Q ⊂ M1 (P ) if, and
only if the conditions of this corollary are met. We can use:
Qmax := Q ∈ M1 (P ) | αmin (Q) = 0 .
34
Recall: M1,f denotes the set of measures that sum to 1, and are finitely 17.02.11
additive. M1 denotes the set of measures that sum to 1 and are σ-additive.
X dentes the set of bounded random variables.
If ρ is a convex risk measure
ρ(X) = max
Q∈M1,f
EQ [−X] − αmin (Q)
with αmin (Q) = supX∈Aρ E[−X] for finitely additive probability measures.
ρ(X) = sup EQ [−X] − α(Q)
Q∈M1
for σ-additive probability measures. This implies continuity above, which
actually is equivalent to the Fatou property: ρ(X) ≤ lim inf ρ(Xn ), and since
n→∞
{Xn } is a bounded sequence, we have P -a.s convergence Xn → X. It is
important to know what kind of convergence you have!
Convergence
In the box, the probability measure P is fixed. For the ⋆-implication
we use dominated convergence or monotone convergence. For the ⋆⋆implication we use uniform integrability. In general we do not have
equivalence between the convergence types. For our applications, convergence
in measure is convergence in probability measures. Finally, the weakest kind
of convergence: convergence in distribution, is on that does not depend on
the measure space.
For a risk measure X = {bounded r.v}. ρX → R that is convex (P equiv.?)
we have the representation
ρ(X) = sup
EQ [−X] − α(Q)
Q∈M1 (P )
35
where
M1 (P ) = Q prob. measure on F | Q ≪ P .
For convex risk measures on L∞ we have a representation property, lower
semicontinuity:
ρ(X) ≤ lim⋆ inf ρ(Y )
Y →X
for a weak ⋆ convergence.
We will now consider spaces X = Lp for p ∈ [1, ∞] and its dual space:
X ′ = ℓ | X → R are linear continuous .
For X = Lp for p ∈ (1, ∞), the dual is X ′ = Lq where
1
p
+
1
q
= 1.
For X = L1 , the dual is X ′ = L∞ .
For X = L∞ , the dual is X ′ ≃ ba(P ), which are the bounded, signed,
finite, additive measures that are absolutely continuous wrt P . In general
L1 ⊆ ba(P ).
Riesz Representation Theorem
For p ∈ [1, ∞). Let ℓ : Lp → R be linearly continuous (then ℓ ∈ L′p , the dual
space). Then there exists a unique g ∈ Lq (Lq = L′p ) such that
Z
ell(X) = gXdP, ∀X ∈ Lp = E[gX]
where g is unique P -a.s. (All linear functionals have the form of the
expectations).
For p = ∞. Let ℓ : L∞ → R be linearly continuous. Then ∃µ ∈ ba(P ) such
that
Z
ℓ(X) = Xdµ.
(we will use the weak ⋆ topology as a convention).
Expectations can be written as linear function and vice versa.
Result
The (convex?) measure ρ : X → R admits the representation
ρ(X) = sup ℓ(X) − α(ℓ)
ℓ∈P
where α : X ′ → R ∪ {∞} is convex, and P ⊆ X ′ is convex.
36
Theorem
a) If ρ : X → R is
convex and lower semicontinuous, i.e for all c ∈ R,
C = X : ρ(X) ≤ c is a closed set. Then
ρ(X) = sup ℓ(X) − α(ℓ) , X ∈ X .
ℓ∈X ′
b) The converse is true.
Proof Sketch
Let c ∈ R, and C = X ∈ Lp = X ρ(X) ≤ c ⊆ Lp . Since ρ is convex,
then so is C .
n
C = X ∈ Lp
n
= X ∈ Lp
o
sup ℓ(X) − α(ℓ)
′
ℓ∈X o
sup E[gX] − α(ℓ)
g∈Lq
Lp
Take a sequence {Xn } ∈ C for all n and we consider if Xn → X. If X ∈ C ,
then C is closed.
E[gXn ] − α(ℓ) ≤ ρ(Xn ) ≤ c,
∀g∈Lq
∀n
Passing to the limit:
E[gX] − α(g) ≤ c
∀g
Take the sup and we are done.
Proposition
If ρ : X → R is convex, lower semicontinuous and
ρ(X) ≤ ρ(0), ∀X ≥ 0 (monotonicity),
then
ρ(X) = sup ℓ(−X) − α(ℓ) ,
′
X+
where
X∈X
X+′ = ℓ ∈ X ′ | ℓ(X) > 0, X ≥ 0
(ℓ(0) = 0).
37
(♣)
Remark: If ρ is convex and lower semicontinuous, then (♣) is equivalent to
monotone:
ρ(X) ≤ ρ(Y ), ∀X ≥ Y.
The converse is also true.
From now on, we will regard X as Lp spaces for p ∈ [1, ∞].
Proposition
1) If ρ : X → R is convex, lower semicontinuous and ∀X ≥ 0; ρ(X) ≤ ρ(0),
then
ρ(X) = sup ℓ(−X) − α(ℓ) , X ∈ X .
ℓ∈P
P = X+′ = ℓ ∈ X ′ ℓ(X) ≥ 0, X ≥ 0 .
The converse is also true. (X+′ is the dual. The ℓ’s are linear functionals which
are related to the expectation. ℓ(−X) = E[Y (−X)] = EQ [−X]. See Riesz’
representation theorem, and the Radon-Nikodym theorem).
2) If ρ : X → R is convex, lower semi-continuous and ρ(0) = 0, then
= sup E[−Y X]
ρ(X) = sup ℓ(−X)
ℓ∈X ′
where
Y ∈Lq
X ′ = ℓX → R linear and cont. .
The converse is also true.
Revision:
We write L′p ≃ Lq , which is identifiable equality. From the definition of the
dual space, we have Lq and its dual is: L′p = Lq when q is the conjugate of p.
For ℓ ∈ Lq , we have ℓ : Lp → R. For the random variable Y : Ω → R, where
E[Y ] < ∞, we have ℓ(X) = E[Y X]. By a representation theorem, there is a
one-to-one correspondene: ℓ(·) = E[Y ·].
Proof of Proposition
1) ⇒) Since ρ is convex and lsc (lower semicontinuous) we have the general
representation
ρ(X) = sup ℓ(−X) − α(ℓ) , X ∈ X .
X∈X ′
We restrict this to finite α’s.
ρ(X) =
sup
X∈X ′ ,α(ℓ)<∞
ℓ(−X) − α(ℓ) = (⋆)
38
24.02.11
From X ≥ 0, ρ(X) ≤ ρ(0), we get, for all ℓ ∈ X ′ (and using that α(ℓ) < ∞):
ρ(X) ≤ ρ(0) =⇒
ℓ(−X) − α(ℓ) ≤ ρ(0) =⇒
ℓ(−X) ≤ α(ℓ) + ρ(0) < ∞ ∀X ≥ 0 =⇒
ℓ(−X) ≤ 0 ∀X ≥ 0
(Must be negative for this to be true for all X).
=⇒ ℓ(X) ≥ 0 =⇒ ℓ ∈ X+′ =⇒
(⋆) = sup ℓ(−X) − α(ℓ)
′
ℓ∈X+
⇐) From the shape of the representation, we know that ρ : X → R is convex
and lsc. It remains to show that ρ(X) ≤ ρ(0).
X ≥ 0 =⇒ ℓ(X) ≥ 0 =⇒ ℓ(−X) ≤ 0 =⇒
ρ(X) = sup ℓ(−X) − α(ℓ) ≤ sup 0 − α(ℓ) = − inf′ α(ℓ) =: ρ(0).
ℓ∈X+
′
ℓ∈X+
′
ℓ∈X+
2)
ρ(X) =
sup
ℓ∈X ′ ,α(ℓ)<∞
ℓ(−X) − α(ℓ) = sup{ℓ(−X)}
(Used 0 = ρ(0) = sup{−α(ℓ)}. Recall that ρ(0) is always represented by the
penalty function). The equality is verified by showing ≤ and ≥. Since we are
working with the sup, this is quite obvious.
ρ(X) =
sup
(⋆⋆)
ℓ(−X)
ℓ∈X ′ ,α(ℓ)=0
If ρ is lsc, convex and positive homogeneous, we get (⋆⋆) and ρ(0) = 0.
If ρ is lsc, convex, ρ(0) = 0 and sub additive, we get (⋆⋆) and positive
homogeneity.
If ρ is lsc and sub-linear (i.e sub-aditive and positive homogeneous), then we
get (⋆⋆), ρ(0) = 0 and convexity.
For the properties convex, positive homogeneous and sub-additivity, two of
these combined with ρ(0) = 0 imply the third one. Consider
ρ(X) ≤ ρ(Y ),
X ≥Y
and
ρ(X) ≤ ρ(0),
X ≥0
We immediately have ⇒, but for ⇐ we need convexity and lsc.
39
6 Backwards Stochastic Differential Equations [Ph 139]
Preliminaries/Revision
A complete probability space (Ω, F , P ), is a space where ∀A ∈ F , where
P (A) = 0, and ∀C ⊂ A then C ∈ F (and an obvious observation is that
P (C) = 0.
An probability space is P -augmented if all the P -null sets belong to F0 .
For a general space (Ω, F , P ), if A ∈ F and P (A) = 0, then A ∈ F0
but C ⊂ A 6⇒ C ∈ F0 . An example: Ω = {a, b, c, d}, we can have
F = ∅, {a, b}, {c, d}, Ω , then {a, b} ∈ F but a 6∈ F . Augmented and
completeness are not the same concepts.
For a measure space (Ω, F , P ), with the filtration F and the index set T (e.g
T = [0, T ]), we have the process {Xt }t∈T , where we write X : Ω × T → Rd or
X : T → {r.v. : Ω → R}.
Important properties the process can have:
1) Measurability: ∀A ∈ B(Rd ), X −1 (A) ∈ F × BT .
2) Adaptedness: ∀t ∈ T, Xt is Ft -measurable. (How randomness is moving
with time).
3) Progressive Measurability: ∀A ∈ Rd and ∀t ∈ T,
(ω, s) ∈ Ω × [0, t] X(ω, s) ∈ A = X −1 (A) ∩ (Ω × [0, t]) ∈ Ft × B[0, t]
Another way to see this is:
X [0,t] : Ω × [0, t] → Rd .
(Note: for (R, τ ) → (R, τ ), the topology τ is X −1 (O) ∈ τ . The σ-algebra is
generated by the topology, and is bigger than the topology. B(R) = σ{τ }).
Property 3) implies 1),2), but 1),2) does not imply 3)! However if X satisfies
1),2) then there exists a Y , which is a modification of X, that satsifies 3). If
X is 2) and is either right or left continuous, then it satisfies 3).
Predictability (used in integration)
P = A × (s, t] ∀A ∈ Fs , ∀s, t : 0 ≤ s ≤ t
For 1A×(s,t] (ω, u) = 1A (ω) × 1(s,t] (u), then A ∈ Fs . For simple functions
φ(u) =
N
X
ei (ω)1(ti−1 ,ti ) (u)
i=1
40
03.03.11
If ei (ω) ∈ Fti−1 -measurable, ei (ω) is left continuous.
Random Time
We define the concept of random time as a random variable τ : Ω → [0, ∞].
We call random time a stopping time if:
∀t, {τ ≤ t} ∈ Ft
(the stopping depends on the filtration). We call random time for optional
time if:
∀t, {τ < t} ∈ Ft
An optional time implies it is a stopping time. A stopping time is optional if
the filtration F is right continuous.
We say that F is right continuous if
Ft = Ft+ :=
\
u>t
Fu .
This is in general true for the filtration, since the information does not make
“jumps”.
If τ and σ are stopping times, then min(τ, σ) and sup(τ, σ) are stopping
times, and it is left as an exercise to verify that τ + σ is a stopping time.
Local Martingales - (terminology and notation from Karatzas & Shreve).
Definition: A local martingale, denoted with Mt for a finite or infinite time
horizon T is such that ∃{τn }n of F-stopping times, τn ր ∞ P -a.s as n → ∞
such that ∀n
Mt
0 ≤ t < τn is an F-martingale.
Mt∧τn :=
Mτn t ≥ τn
(A local martingale produces an infinite amount of local martingales).
From the definition of Mt , it follows that it is (1) adapted, (2) Mt ∈ L1 (P )
for all t and (3) E[Mt |Fs ] = Ms . (Martingales or Local Martingales?)
41
Uniform Integrability (K&S, Pham and Øksendal appendix)
For {Yt }t∈T ,
lim sup E |Yt |1{|Yt |>c} = 0
c→∞
t
Is E[|Yt |1{|Y −t|≤c} ≤ 0?
For Mt , t ∈ [0, T ] we can write Mt = E[MT |Ft]. For Mt with t ∈ [0, ∞)
we can write Ms = E[Mt |Fs ] for any future t, but in general we can
not write Ms = E[M∞ |Fs ], since M∞ may not exist. If M∞ exists and
limt→∞ Mt = M∞ , we can write it.
If the martingale converges P -a.s and in L1 (P ), it has right continuous
trajectories with left side limits, and we have uniform integrability.
(This isn’t true for the Girsanov transform Zt ? Zt → Z∞ as t → ∞, but
Zt 6= E[Z∞ |Ft ]).
- If Mt is a local martingale and
E[sup |Ms |] < ∞,
s≤t|
∀t,
then Mt is also a martingale.
- Mt is a local martingale, M0 ∈ L1 (P ) and Mt is non-negative, thn Mt is a
super martingale.
- If Mt is a continuous local martingale, M0 = 0 and Mt has finite total
variation:
n
X
Mt − Mt < ∞,
sup
i
i−1
t1 ,...,tn
i=1
then Mt ≡ 0. In Ito calculus we have the important equality (dBt )2 = dt,
which is a result from the quadratic variation.
The notation for local martingales is d < Bt >.
Burkholder-Davis-Gundy (BDG)
For all p > 0, there exists constants cp , Cp > 0 such that for all continuous
local martingales Mt , and for all stopping times τ < ∞ P -a.s., we have the
inequalities:
h
h
h
p2 i
p i
p2 i
cp E < Mt >τ
≤ E sup |Ms |
≤ Cp E < Mt >τ
0≤s≤τ
This inequality gives a sufficient condition to determine if Mt is a martingale.
42
Doob
Assume Mt for t ∈ [0, T ] (or t ∈ [0, ∞)) is a non-negative submartingale.
Then ∀τ stopping times, τ < ∞ P -a.s we have
E[M ]
τ
P sup |Mt | ≥ λ ≤
, λ>0
λ
0≤t≤τ
h
p i p p
E[|Mτ |p ], p > 1.
E sup |Mt |
≤
p
−
1
0≤t≤τ
Gives us an upper bound on the supremum, which we do not know.
Simple integrals (Elementary integrals)
1)
X
φn =
ei 1(ti−1 ,ti ]
i
where ei is Fti−1 -measurable and bounded. The set of such functions are
denoted by V[0, T ] in Øksendals book. From the step function we define the
simple integral:
X
It (φn ) =
ei ∆Bti .
i
2) The set {It (φn )}n are all in L2 (P ).
3) ∀φ ∈ V[0, T ], there exists φn such that φn → φ (in L2 (P × dt)).
4) {In (φn )} is a Cauchy sequence. Banach spaces (L2 ) are complete, so the
limit exists.
A common assumption:
hZ
E
T
0
i
φ2t dt < ∞
but for many applications this is too strong, for instance when we work with
local martingales. A way to relax this assumption is:
Z t
2
P
φt dt < ∞ = 1,
∀t
0
(in Øksendal’s notation this is the assumption for the set W[0, T ]). Under this
weaker assumption we loose the Ito isometry. Instead we have ψn → φ, a P a.s. pointwise convergence, and bound it with a stopping time. I(ψn ) → I(φ),
and
Z t
It (φ) =
φs dBs .
0
If φ ∈ V[0, T ] we have the standard Ito construction. If φ ∈ W[0, T ] this is a
local martingale and we don’t know if it is integrable.
43
On a probability space (Ω, F , P ), we have the Brownian motion Wt for
t ∈ [0, T ], T < ∞ and the filtration F = FW which is the filtration generated
by Brownian motion, F = {Ft | 0 ≤ t ≤ T }.
For a martingale Mt ∈ L2 (P ) for all t, we have, by the martingale
representation theorem:
Z t
Mt = M0 +
φs dWs
0
for some φ ∈ V[0, T ] and E[Mt ] = M0 . By the martingale property we also
have:
Mt = E[MT | Ft ],
∀t.
Now consider the equation
−dMt
MT
= 0dt − φt dWt
= ξ
We can find the solution to this with the martingale representation theorem
(MRT).

Z t
φs dWs
 Mt = E[ξ] +
Mt
0
= E[ξ | Ft ]
If we exchange 0 in the equation with f (Mt , φt ) (called the driver), we have
a BSDE. Solutions to BSDE’s consist of a process Mt and the function
φ. We solve BSDE’s with the MRT and stochastic differentiation. For
ξ ∈ L2 (P ) being Ft -measurable, φt = Dt ξ (a non-anticipating derivative).
If ξ ∈ D1,2 ( L2 (P ), then φt = Dt ξ = E[Dt ξ|Ft]. (?)
They are called ’Backwards’ because we know the terminal point MT .
44
BSDE’s [Ph 139]
We have the Brownian motion Wt , the probability space (Ω, F , P ) with the
filtration generated by Brownian motion F = FW . For the theory of BSDE’s,
we always work with a finite, fixed time horizon T = [0, T ], T < ∞. Two
important spaces are:
S2 (0, T ), the set of real valued, progressively measurable processes Y such
that
E sup |Yt |2 < ∞,
t≤T
and H2 (0, T ), the set of Rd -values, progressively measurable processes Z such
that
hZ T
i
E
|Zt |2 dt < ∞.
0
For each BSDE we are given a pair (ξ, g), where ξ is the terminal condition
and the function g is called the driver (or the generator). Assumptions we
make:
(A)
(B)
−
−
−
ξ ∈ L2 (FT )
g : Ω × [0, T ] × R × Rd → such that:
g(·, t, y, z) ∼ g(t, y, z) is progressively measurable for all y,z
g(t, 0, 0) ∈ H2 (0, T )
g satisfies a uniform Lipschitz continuity in (y, z):
There exists a constant Cg such that
g(t, y1 , z1 ) − g(t, y2, z2 ) ≤ Cg y1 − y2 + z1 − z2 P -a.s., dt-a.e
for y1 , y2 ∈ R and z1 , z2 ∈ Rd .
A BSDE is an equation on the form:
−dYt = g(t, Yt , Zt )dt − Zt dWt
YT = ξ
Definition 6.2.1 [Ph 140]
A solution to a BSDE is a pair (Y, Z) ∈ S2 (0, T ) × H2 (0, T ) satisfying:
Z T
Z T
Yt = ξ +
g(s, Ys, Zs )ds −
Zs dWs , 0 ≤ t ≤ T.
t
t
45
Theorem 6.2.1 - Existence & Uniqueness [Ph 140]
Given a BSDE with the pair (ξ, g) satisfying the assumptions above, there
exists a unique solution (Y, Z).
Proof
We have the function Φ : S2 (0, T ) × H2 (0, T ) → S2 (0, T ) × H2 (0, T ), for
(U, V ) → Φ(U, V ). We define the martingale
Z
h
Mt := E ξ +
T
0
i
g(s, Us , Vs )ds Ft
which, under the assumptions in (A),(B), is square integrable: E[|Mt |2 ] < ∞
for all t. By the MRT we have
Z t
Mt = M0 +
Zs dWs , Z ∈ H2 (0, T ).
0
It follows from the MRT that Z exists and is unique. Next we define Yt .
Z T
h
i
Yt := E ξ +
g(s, Us , Vs )ds Ft
t
Z T
Z t
h
i
=E ξ+
g(s, Us , Vs )ds ±
g(s, Us, Vs )ds Ft
t
0
Z T
Z t
h
i
=E ξ+
g(s, Us , Vs ds) −
g(s, Us, Vs )ds Ft
0
0
Z t
Z T
h
i
=E ξ+
g(s, Us , Vs ds) Ft −
g(s, Us , Vs )ds
0
0
Z t
= Mt −
g(s, Us , Vs )ds
0
Z t
Z t
= M0 +
Zs dWs −
g(s, Us , Vs )ds
(⋆)
0
0
Similarly,
YT = ξ = MT −
= M0 +
Z
T
0
Z T
0
g(s, Us , Vs )ds
Z T
Zs dWs −
g(s, Us , Vs )ds
0
46
(⋆⋆)
Now we continue with (⋆):
Z t
Z t
Z T
Z T
(⋆) = M0 +
Zs dWs −
g(s, Us , Vs )ds ±
Zs dWs ±
g(s, Us , Vs )ds
0
0
t
t
Z T
Z T
Z T
Z T
Zs dWs +
g(s, Us , Vs )ds
= M0 +
Zs dWs −
g(s, Us , Vs )ds −
t
t
0
0
{z
}
|
=ξ−
=(⋆⋆)=YT =ξ
Z
T
Zs dWs +
t
Z
T
g(s, Us , Vs )ds
t
which is the general form of the solution to a BSDE. The next step is to show
that Y ∈ S2 (0, T ). Use that |a + b + c|2 ≤ 5(|a|2 + |b|2 + |c|2 ).
Z T
Z T
2 i
h
i
h
2
2
2
E[ sup |Yt | ] ≤ 5E[ξ ]+5E sup
|g(s, Us , Vs )| ds +5E sup Zs dWs 0≤t≤T
0≤t≤T
0≤t≤T
t
t
We want to determine if this is finite. The two first terms are finite, as the
first term is a constant in L2 (P ) and the second term is a continuous function
on a finite interval. The third term is unknown, but we can find an upper
bound by using Doob’s inequality.
2 i 2 2 h Z T
Z T
2 i
hZ T
i
h
·E
Zs dWs
= 4E
|Zs |2 ds < ∞.
E sup Zs dWs ≤
2−1
0≤t≤T
0
0
t
(The square of a martingale is a sub-martingale). The third term is also
finite, so Y ∈ S2 (0, T ). The final step is to verify that we have a contraction.
For the pairs (U, V ) and (U ′ , V ′ ), we define U = U − U ′ , V = V − V ′ and
g = g(t, Ut , Vt ) − g(t, Ut′ , Vt′ ). For some constant β > 0 that we determine
later, we apply Ito’s formula (product formula with some substitutions) to
2
eβs Y s , for 0 ≤ s ≤ T :
2
2
2
d eβs · Y s = βeβs Y s ds − 2eβs Y s g s ds + 2Y s eβs Z s dWs + eβs Z s ds
βT
2
T
2
0
e Y −Y =
hZ
2
E[Y 0 ]+E
0
T
eβs
Z
T
βs
e
0
βY s − 2Y s g s +
i
hZ
2
2
βY s +Z s ds = E
0
T
2
Z s ds
+2
Z
T
eβs Y s Z s dWs
0
i h Z T
i
βs
2e Y s g s ds +E 2
eβs Y s Z s dWs
0
|
{z
}
=0
The last term should be 0, which is the case if Y s Z s is a martingale, which we
will show separately. For now we simply assume that is 0. For the remaining
47
terms we work with the right hand side of the equality. Since g is constrained
(since it is Lipschitz), we get:
hZ T
i
hZ T
i
βs
E
2e Y s g s ds ≤ 2Cg E
eβs Y s U s + V s ds
0
0
√
2 · 2 · Cg Y s and b = (|U s | + |V s |)/ 2:
i 1 hZ T
hZ T
i
βs
2
βs 2
e U s + V s ds
≤ 4Cg E
e Y s ds + E
2
0
0
Using that ab ≤
a2
2
+
b2
,
2
for a =
√
Now we choose β = 1 + 4Cg2 , and obtain:
Z
hZ T
i
i 1 h T βs 2
2
2
βs
E[Y 0 ] + E
e U s + V s ds
e Y s + Z s ds ≤ E
2
0
0
This shows that Φ is a strict contraction on the Banach space S2 (0, T ) ×
H2 (0, T ) endowed with the norm:
hZ T
1
i 2
βs
2
2
.
k(Y, Z)kβ = E
e Ys + Zs ds
0
For a contraction in a complete space, we immediately get uniqueness and
existence. To conclude the proof, we have to verify that the term
h Z T
i
E 2
eβs Y s Z s dWs
0
is zero, which we do by applying the Burkholder-Davis-Gundy
(BDG)
q R
·
< 0 eβs Y s Z s dWs >t < ∞ for
inequality, and showing that it is finite: E
all t. We study the quadratic variation (by the Doob-Meyer decomposition),
and show that the quadratic variation is finite. We use that Y ∈ S2 (0, T )
(which, incidentally is the reason why we require that assumption), and use
the product to sum inequality ab ≤ (a2 + b2 )/2.
s



s
Z t
Z t
Z T
r
i
eβT h
2 2
2
2
2
2
2βs
βt




E
E sup Y t +
sup e Y t
Z s ds ≤
Z s ds < ∞
e Y s Z s ds ≤ E
2
t≤T
0≤t≤T
0
0
0
| {z } | {z }
∈S2 (0,T )
∈H2 (0,T )
This is finite, and this is a sufficient condition for a local martingale to be a
martingale, and hence the integral is 0 and the proof is concluded.
48
Linear BSDE
10.03.11
We continue using the spaces S2 (0, T ) and H2 (0, T ). In the Peng article, we
generally work with adapted processes Yt , while we work with progressively
measurable spaces in Pham, so Pham is more rigorous than Peng.
As we saw in the existence and uniqueness result, we know that the BSDE
is well defined when the pair (ξ, g) is given (final condition and a driver).
We recall that we have certain assumptions on the driver g, denoted by (B).
Sometimes we write the Lipschitz condition:
|g(t, y1, z1 ) − g(t, y2, z2 )| ≤ Cg |y1 − y2 | + |z1 − z2 |
|g(t, y1, z1 ) − g(t, y2, z2 )| ≤ ν|y1 − y2 | + µ|z1 − z2 |
where Cg = max(ν, µ).
Digression
Consider the BSDE given by:
−dYt = g(t, Yt , Zt )dt − Zt dWt
YT = ξ
where ξ ∈ L2 (Ω) and ξ is FT -measurable. What happens if ξ is Ft measurable
for some t < T ? This means that at some point t we reach the final
condition, which we already know. An exercise: If ξ is Ft -measurable, prove
that the solution applies to the interval [t, T ]. The solution is a couple
(Y· , Z· ) ∈ S2 (0, T ) × H2 (0, T ).
Remember that the solution of a BSDE is on the form:
Z T
h
i
Yt = E ξ +
g(u, Yu, Zu )duFt .
t
Since the integral goes from t to T , this is not a martingale. We get a proper
martingale (not just a local martingale) if we define
Z T
h
i
Mt = E ξ +
g(u, Yu, Zu )duFt .
0
Due to the assumptions we made on g (listed in (A)), g is integrable from 0
to T . From g you can get properties on Yt .
49
Linear BSDEs are BSDEs where we have the time dependent coefficients
At , Bt , Ct , and have the form:
−dYt = At Yt + Bt Zt + Ct dt − Zt dWt
YT = ξ
Both At and Bt are bounded, progressively measurable processes, and Ct ∈
H2 (0, T ).
Proposition 6.2.1 [Ph 142]
The unique solution (Y, Z) to the linear BSDE is given by
Z T
h
i
Γt Y t = E ΓT ξ +
Γu Cu du Ft , t ∈ [0, T ]
t
with the corresponding linear SDE, also called the adjoint or dual SDE.
dΓt = Γt At dt + Γt Bt dWt
Γ0 = 1
By solving the SDE, you also solve the linear BSDE.
Proof
By Ito’s product formula:
d(Γt Yt ) = Yt dΓt + Γt dYt + dΓt dYt
+ B dW − Γ A
Z
= Γt Yt At
dt
t
t
t t Yt + t Bt + Ct dt
+ Γt Zt dWt + Γt
Zt
Bt dt
= −Γt Ct dt + Γt (Yt Bt + Zt )dWt =⇒
Z t
Z t
Γt Yt = Γ0 Y0 −
Γu Cu du +
Γu Yu Bu + Zu dWu
0
0
Now, is this a local martingale? Rewrite the equation:
Z t
Z t
Γt Y t +
Γu Cu du = Γ0 Y0 +
Γu Yu Bu + Zu dWu
0
0
(and recall that Γ0 = 1). Now we apply the BDG-inequality for p = 1, and
get an estimate of the sup. Look at the quadratic variation (think of Ito
isometry. See definitio of quadratic variation which we can use for stochastic
integrals):

s
#
"s Z
Z t
·
2
Γ2u Yu Bu + Zu du ≤
<
Γu Yu Bu + Zu dWu > = E 
E
0
0
50

r

E
sup Γ2u
u
sZ
t
2

Yu Bu + Zu du
0
Since both A and B are bounded, then so is Γ. Now we apply the product to
sum inequality: ab ≤ (a2 + b2 )/2. We also use that (c + d)2 = c2 + 2cd + d2 ≤
2c2 + 2d2 and |Bu | ≤ bmax for all u (it is bounded).
Z T
Z T
1
2
2
bmax Yu du + 2
Zu du < ∞
≤ E sup Γu + 2
2
u
0
0
To see that Γt is finie, just study E[supu Γ2u ]. From here you can show that
the expression for Γt Yt is a martingale, and not just a local martingale, which
concludes the proof.
Theorem 6.2.3 Comparison Theorem [Ph 142]
Let (ξ 1 , g 1) and (ξ 2 , g 2) be two pairs of terminal conditions and generators
satsifying the conditions in (A) and (B), and let (Y·1 , Z·1 ) and (Y·2 , Z·2 ) be
solutions to the corresponding BSDEs. Assume:
− ξ1 ≤ ξ2
P -a.s
1
1
1
2
1
1
− g (t, Yt , Zt ) ≤ g (t, Yt , Zt ) dt × dP a.e
− g 2 (t, Yt1 , Zt1 ) ∈ H2 (0, T )
Then Yt1 ≤ Yt2 for all 0 ≤ t ≤ T a.s.
Also, if Y02 ≤ Y01 , then Yt1 = Yt2 for 0 ≤ t ≤ T . In particular, if
P (ξ 1 < ξ 2 ) > 0, or if g 1 < g 2 (depends on time and ω) on a set of strictly
positive measure dt × dP , then Y01 < Y02 .
Proof
We define the differences:
Y·
Z·
= Y· 2 − Y· 1
= Z·2 − Z·1
The pair (Y · , Z · ) is a solution to the BSDE corresponding to the pair of
terminal condition and driver given by: (ξ 2 − ξ 1 , ∆y Y · + ∆z Z · + g), where
∆yt
g 2 (t, Yt2 , Zt2 ) − g 2(t, Yt1 , Zt2 )
· 1{Yt2 −Yt1 6=0}
:=
Yt2 − Yt1
51
∆zt :=
g 2(t, Yt1 , Zt2 ) − g 2 (t, Yt1 , Zt1 )
· 1{Zt2 −Zt1 6=0}
Zt2 − Zt1
g t := g 2 (t, Yt1 , Zt1 ) − g 1 (t, Yt1 , Zt1 )
Starting from the BSDE:
−dYt = dYt2 − dYt1
h
i
= g 2 (t, Yt2 , Zt2 ) − g 1 (t, Yt1 , Zt1 ) dt − (Zt2 − Zt1 )dWt
(⋆)
we work our way do incorporate ∆y and ∆z . We add and subtract terms.
g 2 (t, Yt2 , Zt2 ) − g 1 (t, Yt1 , Zt1 ) ± g 2 (t, Yt1 , Zt2 ) ± g 2 (t, Yt1 , Zt1 )
g 2 (t, Yt2 , Zt2 )−g 2 (t, Yt1 , Zt2 )+g 2 (t, Yt1 , Zt2 )−g 2 (t, Yt1 , Zt1 )+g 2 (t, Yt1 , Zt1 ) − g 1 (t, Yt1 , Zt1 )
{z
}
|
=g t
g 2(t, Yt1 , Zt2 ) − g 2(t, Yt1 , Zt1 ) 2
g 2 (t, Yt2 , Zt2 ) − g 2 (t, Yt1 , Zt2 ) 2
1
1
Y
−Y
+
Z
−Z
t
t
t
t +g t
Yt2 − Yt1
Zt2 − Zt1
∆yt Yt2 − Yt1 + ∆zt Zt2 − Zt1 + g t =⇒
h
i
(⋆) = ∆yt (Yt2 − Yt1 ) + ∆zt (Zt2 − Zt1 ) + g(t, Yt1 , Zt1 ) dt − Z t dWt
i
h
= ∆yt Y t + ∆zt Z t + g(t, Yt1 , Zt1 ) dt − Z t dWt
Both ∆yt and ∆zt are progressively measurable and bounded, and g ∈
H2 (0, T ). We see that this has the form of a linear BSDE where At = ∆yt ,
Bt = ∆zt and Ct = g t .
g(t, Y 1 , Z 1 ) = g 2 (t, Y 1 , Z 1 ) − g 1 (t, Y 1 , Z 1 ) =
t
t
t
t
t
t
2
g (t, Y 1 , Z 1 ) −g 2 (t, 0, 0) + g 2(t, 0, 0) −g 1(t, Y 1 , Z 1 ) −g 1 (t, 0, 0) + g 1(t, 0, 0) ≤
t
t
t
t
2
g (t, Y 1 , Z 1 )−g 2 (t, 0, 0)+g 2 (t, 0, 0)+g 1 (t, Y 1 , Z 1 )−g 1 (t, 0, 0)+g 1(t, 0, 0) ≤
t
t
t
t
Cg2 |Yt1 | + |Zt1 | + g 2 (t, 0, 0) + Cg1 |Yt1 | + |Zt1 | + g 1 (t, 0, 0)
The solution to the linear BSDE is given by:
Z T
h
i
Γt Y t = E ΓT ξ +
Γs g s ds Ft
t
with the adjoint SDE:
(
dΓt
Γ0
=
=
Γt ∆yt dt + ∆zt dWt
1
52
Since the expression for dΓt contains Γt , we know that it is on the form of
an exponential function, so Y t ≥ 0.
If Y01 = Y02 , we get Y 0 = 0, and since the integral grows to the left:
0 = E[Γ0 Y 0 ] ≥ E[Γt Yt ].
Now we see recall that a non-negative local martingale is a supermartingale.
A supermartingale with constant expectation is a martingale. Γt Y t ≥ 0 for
all t and E[Γt Y t ] ≥ 0, but E[Γt Y t ] = 0 for all t implies Γt Yt = 0, so in this
case we have Yt1 = Yt2 P -a.s.
Corollary
Let (ξ, g) satisfy:
(a)
ξ ≥ 0, P -a.s
(b)
g(t, 0, 0) ≥ 0, dP × dt-a.s.
Then the solution (Y· , Z· ) is such that Yt ≥ 0 for all t, P -a.e. Moreover, if
P (ξ > 0) > 0 or g(t, 0, 0) > 0, dP × dt-a.e, then Y0 > 0 P -a.s.
Proof
This is a simple application of the Comparison Theorem. We apply it to the
BSDE corresponding to (ξ 1 , g 1 ) = (0, 0) and (ξ 2, g 2 ) = (ξ, g). (The solution
to the first BSDE is of course 0). The equation:
−dYt = −Zt dWt
YT = 0
has the solution
Z T
h
i
Yt = E ξ +
g ds Ft
|{z}
t |{z}
=0
=0
(which is a martingale).
53
g-Expectations [Peng]
Let X be Ft -measurable: X ∈ L2 (Ft ), with a BSDE associated with the pair
(ξ, g), where ξ = X and g satisfies:
(i)
(ii)
g(·, y, z) is progressively measurable ∀y ∈ R, z ∈ Rd .
g(·, y, z) ∈ H2 (0, T )
(iii)
g satisfies the Lipschitz condition
(iv)
g(·, y, ·) = 0 for all y ∈ R.
g(·, 0, ·) = 0
(iv)⇒(v)
(v)
(vi)
(vi)⇒(iv)
g does not depend on y, and g(·, 0) = 0.
(In addition to these properties, g still satisfies the assumptions in (B) given
previously as it is a driver to a BSDE).
Definition [Peng Def 3.2]
The g-expectation of X is defined as:
Eg [X] = Y0X
The conditional g-expectation of X given Ft is:
Eg [X | Ft ] = YtX
Obviously:
Eg [X | F0 ] = Eg [X]
This is a way to generalize expectations. With this type of expectation, we
loose linearity, which is a central property to the conventional expectation,
but we retain monotonicity and a type of homogeneity.
Theorem [Peng Thm 3.4 + Prop 3.6]
1) Monotonicity.
∀X1 ≤ X2 =⇒ Eg [X1 | Ft ] ≤ Eg [X2 | Ft ]
2) “Constant” preservation.
Eg [X | Ft ] = X for all X ∈ L2 (Ft )
3) Time consistency.
h
i
Eg Eg [X | Ft ] Fs = Eg [X | Fs ],
54
s≤t
4) “Zero-one” for all A ∈ Ft (a weak sense of Ft -homogeneity that only applies
to the indicator function).
Eg [1A X | Ft ] = 1A Eg [X | Ft ]
Proof
1) Proved by considering two BSDEs associated to (ξ 1 , g 1 ) = (X1 , g) and
(ξ 2 , g 2 ) = (X2 , g) and applying the Comparison Theorem.
2) (Exercise). From the definition of the g-expectation Eg [X | Ft ] = YtX . If X
is Ft -measurable, 2) is proved if you can show YtX = X. (Linked to previous
exercise).
3) Assume s ≤ t:
Eg [X | Fs ] =
Z
YsX
Z
T
T
=X+
g(u, Yu, Zu )du −
ZudWu
s
s
Z T
Z T
h
i
= X+
g(u, Yu, Zu )du −
Zu dWu
t
t
Z t
Z t
+
g(u, Yu, Zu )du −
ZudWu
s
s
Z t
Z t
X
= Yt +
g(u, Yu, Zu )du −
ZudWu
s
s
= Eg [YtX | Fs ]
By the definition of g-expectations: YtX = Eg [X | Ft ], so we get:
h
i
= Eg Eg [X | Ft ] Fs
4) We have the solution to the BSDE, given as:
Z T
Z
Yt = X +
g(s, Ys , Zs )ds −
t
T
Zu dWu
t
Now, take any A ∈ Ft . Then we can multiply both sides:
Z T
Z T
1A Yt = 1A X +
1A g(s, Ys, Zs)ds −
1A Zu dWu
t
1A Yt = 1A X +
Z
T
t
(⋆)
t
g(s, 1A Ys , 1A Zs )ds −
Z
t
T
1A Zu dWu
(⋆⋆)
From assumption (v) for g-expectations, we have g(·, 0, 0), so 1A g(s, Ys, Zs ) =
g(s, 1A Ys , 1A Zs ), so we can set (⋆) = (⋆⋆). Now if we solve each of these
55
BSDEs, we get that the solution of (⋆) is 1A Eg [X | Ft ] and the solution of
(⋆⋆) is Eg [1A X | Ft ], and by uniqueness of solutions for BSDEs, these must
be the same.
Eg [1A X | Ft ] = 1A Eg [X | Ft ]
Lemma [Peng Lemma 3.2]
Consider g satisfying properties (i), (ii), (iii) and (vi) (and recall that
(vi)⇒(iv)⇒(v)). Then for all η ∈ L2 (Ft ),
Eg [X + η | Ft ] = Eg [X | Ft ] + η
(We can incorporate additivity for the g-expectations).
Proof
From the definition of g-expectations, we have Eg [X | Ft ] = YtX , and consider
the BSDE associated to (ξ, g) = (X, g), which has the solution (Y· , Z· ).
We also set Eg [X + η | Ft ] = YetX+η = YtX + η, with BSDE associated to
(ξ, g) = (X + η, g), with solution (Y· + η, Z) = (Ye· , Z). We have:
Z T
h
i
Eg [X | Ft ] = E X +
g(s, Zs )ds Ft
t
We add η.
Z
h
Eg [X | Ft ] + η = E X +
t
T
i
g(s, Zs )ds Ft + η
Z
h
=E X +η+
T
t
= Eg [X + η | Ft ].
i
g(s, Zs )ds Ft
We note that E0 [X + η | F0 ] = E0 [X] + η, which reveals a connection to cash
invariance for risk measures. We will see more of this later.
Proposition [Peng 3.4, Prop 3.6]
g-expectations with different g-functions. Denote by gµ (t, y, z) = µ|z|
(structure from the Lipschitz condition for g), (µ = Cgµ ). Consider Eg and
Egµ =: E µ . If g satisfies (i), (ii), (iii) and (iv), then
−E µ [−X | Ft ] ≤ Eg [X | Ft ] ≤ E µ [X | Ft ],
|{z}
|{z}
a)
b)
56
∀X ∈ L2 (FT )
Proof
g(t, y1, z1 ) − g(t, y2, z2 ) ≤ µ |u1 − y2 | + |z1 − z2 |
Set z2 = 0, and y = y1 = y2 .
•) g(t, y, z) ≤ µ|z| = g − µ(t, y, z). We apply the Comparison Theorem to
(ξ 1 , g 1 ) = (X, g) and (ξ 2 , g 2) = (X, gµ ), and we get inequality (b).
•) −gµ (t, y, z) ≤ g(t, y, z). Use the Comparison Theorem for (ξ 1 , g 1) =
(X, −gµ ) and (ξ 2 , g 2) = (X, g). We get:
Eg [X | Ft ] ≥ E−gµ [X | Ft ]
Z T
h
i
=E X+
−gµ (s1 , Ys , Zs )ds Ft
t
Z T
h
i
= −E − X +
−gµ (s1 , Ys , Zs )ds Ft
t
= Egµ [−X | Ft ]
So by the Comparison Theorem, we have established inequality (a).
As mentioned previously the g-expectations (g is the driver to a BSDE) 17.03.11
are a generalization of normal expectations. We have retained monotonicity
for g-expectations, but we have lost linearity. We also have a weak form of
homogeneity, as we have seen.
Consider the price of a claim X: E[X | Ft ] = rtT (X), which
is the price
operator. For s ≤ t ≤ u we have rsu (X) = rst rtu (X) . Suppose the
conditional operator is unknown, and all we have is this equality. If we
have a fair market with no arbitrage, the only possibility is the conditional
expectation.
If we move away from the typical assumptions of the B& S market (can
always hedge etc.) we loose the linearity of the conditional expectation. We
have to model prices in a convex (or concave) way. (We still have arbitrage, as
this is not linked with the B& S-assumptions). The conditional g-expectation
is an example of pricing in a non-linear market.
Last time we explored the relationship between the g-expectation and a
BSDE. We have the Lipschitz condition:
gµ,ν
}|
{
z
g(t, y1 , z1 ) − g(t, y2, z2 ) ≤ µ|z1 − z2 | + ν|y1 − y2 |
| {z } | {z }
gµ
57
gν
Sometimes g does not depend on y, so we only need gµ . We can use gµ as the
driver to a corresponding BSDE, and then apply the Comparison Theorem.
Proposition [Peng Prop. 3.7]
Let g satisfy (i), (ii), (iii) and (iv) from the previous lecture. Then Eg is
“dominated” by Egµ in the sense that
Eg [X1 | Ft ] − Eg [X2 | Ft ] ≤ Egµ,ν [X1 − X2 | Ft ],
∀X1 , X2 ∈ L2 (FT )
(P -only measure of FT ). Since we don’t have linearity, this isn’t immediately
obvious.
If g does not depend on y and g satisfies (vi), then Eg is “dominated” by Egµ .
Proof
We consider the BSDEs.
−dYt
YT
(
−dYet
YeT
= g(t, Yt , Zt )dt − Zt dWt
= X1
= g(t, Yet , Zet )dt − Zet dWt
= X2
We consider the difference, which is the BSDE with terminal point and driver
(X1 − X2 , g):
g
z
}|
{
(
g(t, Yt, Zt ) − g(t, Yet, Zet ) dt − Zt − Zet dWt
−dY t =
YT
= X 1 − X2
We compare with the following BSDE, with the terminal point and driver
given by: (X1 − X2 , gµ,ν ):
(
−dYbt = gµν (t, Ybt , Zbt )dt − Zbt dWt
YbT = X1 − X2
By the Lipschitz condition, we have g ≤ gµν , and this implies, by the
Comparison Theorem, Y t ≤ Ybt for all t, P -a.s, where Y t = Yt − Yet .
Inequalities are typically proved by applying the Comparison Theorem.
58
Proposition
Let Xn → X as n → ∞ in L2 (FT ). Then
Eg [Xn | Ft ] −→ Eg [X | Ft ]
n→∞
in L2 (Ft )
Sketch of Proof
We have the limit:
h
2 i
lim E Eg (Xn | Ft ) − Eg (X | Ft )
n→∞
(⋆)
Similarly as in the previous proof, we consider the following BSDEs:
(
(n)
(n)
−dYt
= g (n) (t, Yt , Zt )dt − Zt dWt
(n)
YT
= Xn
(
−dYet
YeT
= g(t, Yet , Zet )dt − Zet dWt
= X
(n)
= g (n) − ge dt − (Zt − Zet )dWt
= Xn − X
−dY t
YT
The process Y t is given by:
Z
h
Y t = E (Xn − X) +
T
g u, Y
t
By Jensen’s inequality on (⋆):
(n)
(n) n , Zn
i
Ft
Z
h
i
T
(⋆) ≤ lim E[(Xn − X) ] + lim E 2E[(Xn − X) | Ft ] · E
gds Ft
n→∞
n→∞
t
h Z T
2 i
gds | Ft
+ lim E E
n
n→∞
t
Since Xn → X, we get (Xn − X) → 0, which eliminates the two first terms.
Using the Lipschitz condition with g ≤ gµν :
Z T
h
2 i
≤ lim E E[0 +
gµν ds Ft ]
=0
n→∞
t
We can interpret the interior of this expectation as a BSDE with terminal
point 0 and driver gµν , and with that we can conclude that it is 0.
59
Lemma
When gµ (t, y, z) = µ|z|, we get a form of homogeneity.
∀c > 0,
∀c < 0,
Egµ [cX | Ft ] = cEgµ [X | Ft ],
X ∈ L2 (FT )
Egµ [cX | Ft ] = −cEgµ [−X | Ft ]
Proof
(We get Z from the MRT).
Z T
h
Egµ [cX | Ft ] = E cX +
t
Z T
h
= cE X +
t
= cEgµ [X | Ft ]
i
µ|Zs |ds Ft
Zs i
µ Ft
c
Conditional Expectation as a norm
Now we will define a norm via the conditional expectation.
kXkµ = Egµ [|X|],
X ∈ L2 (FT )
This is a norm and we have the properties:
kX1 + X2 kµ ≤ kX1 kµ + kX2 kµ
c > 0,
kcXkµ = ckXkµ
The last property is verified by the previous lemma. We verify the first
property. Recall that |X1 +X2 | ≤ |X1 |+|X2 | and set (ξ 1 , g 1) = |X1 +X2 |, gµ
and (ξ 2 , g 21) = |X1 + X2 |, gµ and we apply the Comparison Theorem.
If (Y (1) , Z (1) ) solves the first BSDE, and (Y (2) , Z (2) ) solves the second, the
Comparison Theorem says:
Y (1) ≤ Y (2) ,
∀t =⇒
Egµ |X1 + X2 | Ft ≤ Egµ |X1 | + |X2 | Ft
This is true for all t, so we set t = 0 and we have verified the property.
60
Proposition
Under k · kµ , the conditional g-expectation is a contraction.
[X
|
F
]
−
E
[X
|
F
]
∀X1 , X2 ∈ L2 (FT )
Eg 1
t
g
2
t ≤ kX1 − X2 kµ ,
µ
True for any g satisfying (i), (ii), (iii) where (iii) is the Lipschitz condition:
g(t, y1, z1 ) − g(t, y2, z2 ) ≤ µ|y1 − y2 | + ν|y1 − y2 |
Proof
Eg [X1 | Ft ] − Eg [X2 | Ft ] ≤ Egµ [X1 − X2 | Ft ]µ =
µ
i
h
Egµ Egµ [X1 − X2 | Ft ]
(⋆)
We continue with a BSDE argument.
−dYt = µ|Zt|dt − Zt dWt
Y T = X 1 − X2
We know Yt satisfies:
Z
h
Y t = E X1 − X2 +
T
t
Take the absolute value on both sides.
Z
h
Yt = E X1 − X2 +
t
R
i
µ|Zs |ds Ft
T
i
µ|Zs |ds Ft R
From measure theory, we recall that | Adµ| ≤ |A|dµ which we can use
here since the expectation is an integral.
Z T
i
h
≤ E X1 − X2 +
µ|Zs |ds Ft = Egµ [|X1 − X2 | | Ft ]
t
So, we apply time consistency (or the tower property):
h i
(⋆) ≤ Egµ Egµ |X1 − X2 | | Ft = Egµ X1 − X2 = kX1 − X2 kµ
61
Definition: g-Martingales
An F-adapted, stochastic process Xt for t ∈ [0, T ] with E[Xt2 ] < ∞, ∀t (values
in L2 for all t), is a g-martingale (resp. g-supermartingale, g-submartingale)
of for every s ≤ t we have:
Eg [Xt | Fs ] = Xs ,
(resp. ≤,
≥)
g
g
In Peng’s notation, Eg [Xt | Fs ] = Es,t
. And Es,t
= Ys for the BSDE
−dYu = g(u, Yu, Zu )du − Zu dWu ,
u ∈ [s, t)
YT = X
g
So EsT
[X] = Estg on L2 (Ft ) (confer proposition in Peng), which is the solution
to the given exercise.
Theorem - Convergence Result (no proof)
Let g satisfy (i), (ii) and (iii). Let Xt , t ∈ [0, T ] be a g-supermartingale
(normal martingales are also supermartingales). Let D ⊆ [0, T ] be a
countable, dense subset. Then for every t,
lim Xs ,
lim Xs ,
s∈D,sցt
s∈D,sրt
P -a.s limits
exist and are finite. (Since we have pointwise limits, this is a strong result).
Moreover, define
X t :=
P -a.s, t ∈ [0, T ), X T = XT
lim Xs ,
s∈D,sցt
which is Ft -adapted and
E sup |X t |2 < ∞.
t∈[0,T ]
Moreover, if g satisfies (iv), then X t , t ∈ [0, T ] is a g-super martingale.
The limit for X t is not obvious: it is a limit from the right. This is Ft+
measurable due to the continuity of the filtration: Ft+ = Ft , so it is Ft adapted. We have two types of continuity for filtrations:
\
Fu
(right continuous)
Ft+ :=
u>t
Ft− :=
_
s<t
Fs
(left continuous)
These are both continuous, so filtrations apply. (∨ is the σ-algebra union).
62
We use ST to denote the set of F-stopping times that are bounded by T .
Definition
For stopping times σ, τ ∈ ST , then Eg,σ,τ [X] = YσX (BSDE), for X ∈
L2 (Ft , P ).
Z τ
Z τ
X
Yr = X +
g(u, Yu, Zu )du −
Zu dWu
r
r
so YσX = Eg [X | Fσ ].
Proposition [Peng 7.1]
Eg ,σ,τ [·] : L2 (Fτ , P ) → L2 (Fσ , P ) satisfies:
(i) Monotonicity:
Eg ,σ,τ [X1 ] ≤ Eg ,σ,τ [X2 ],
when X1 ≤ X2 , P -a.s
(ii)
Eg ,τ,τ [X] = X
(iii) Time consistency:
h
i
Eg ,ρ,τ Eg ,σ,τ [X] = Eg ,ρ,τ [X],
X ∈ L2 (Ft )
(iv) Weak homogeneity:
∀A ∈ Fτ ,
1A Eg,σ,τ [X] = 1A Eg,σ,τ [1A X] = Eg,σ,τ [1A X]
(Not true for all random variables, since we don’t have linearity)
(v) Majorization:
If gµ (t, y, z) < µ|z| and µµν (t, y, z) = ν|y| + µ|z|, then
Eg ,σ,τ [X1 ] − Eg ,σ,τ [X2 ] ≤ Egµν σ,τ [X1 − X2 ]
63
24.03.11
Theorem: Optional Sampling Theorem [Peng 7.3]
Let g satisfy (i), (ii) and (iii) (measurability, Lipschitz and integrability)
from previous lecture. Let Yt , t ∈ [0, T ] be cadlag in S2 (0, T ) and let Yt be a
g-supermartingale. Then for σ, τ ∈ ST such that σ ≤ τ , we have
Eg,σ,τ [Yτ ] ≤ Yσ
(Also works for submartingales, in which case we get an ≥-inequality).
Monotonic Limit Theorem for Ito Processes
(The limit, and not the convergence, is of interest). We have the Ito processes
Z t
Z t
i
i
i
yt = y0 +
gs ds +
zsi dWs , i = 1, 2, . . .
0
0
where we have the usual assumptions from Ito Calculus:
hZ T
i
hZ T
i
i
E
|gs |ds < ∞,
E
|zsi |2 ds < ∞
0
0
We make the following assumptions:
(i)
hZ
E
0
T
|gsi |ds
i
≤ K1 ,
hZ
E
0
T
i
|zsi |2 ds ≤ K2 ,
K1 , K2 > 0
(ii)
yti ր yt i → ∞ P -a.s (monotone convergence), with E sup |yt |2 < ∞
t∈[0,T ]
From this we can make the observations:
E[ sup |yti|2 ] ≤ C,
C independent of i
0≤t≤T
(By monotonicity, yti ≤ yt . We don’t know the sign, but we know that
|yti| ≤ |y 1 | + |yt |, where |y 1 | is larger than any negative value, and |y t| is
larger than any positive values.)
hZ T
i
i
2
E
|yt − yt | dt → 0 as i → ∞
0
64
The limit yt og the Ito processes, is again an Ito process:
Z t
Z t
0
yt = y0 +
g ds +
zs dWs
0
0
The functions g i → g 0 weakly as i → ∞. For the space Ls (dP × dt) and its
dual L′2 , we write ℓ(g i ) → ℓ(g 0) for all ℓ ∈ L′2 (definition of weak convergence),
or
hZ T
i
hZ T
i
i
E
fs gs ds → E
fs g 0ds , ∀f ∈ L2 (dP × dt).
0
0
Also, z i → z in a weak sense as i → ∞.
Theorem [Peng Theorem 7.2] (No proof)
Under these assumptions, we have a strong convergence for z i .
hZ T
i
lim E
|zsi − zs |p ds = 0,
p ∈ [0, 2).
i→∞
0
Moreover, if yt , t ∈ [0, T ] is continuous, then (for p = 2):
hZ T
i
lim E
|zsi − zs |2 ds = 0
i→∞
0
(Note that the Ito processes we have presented so far, have a continuous
modification/version).
Monotonic Limit Theorem for BSDEs
We consider the pairs (xi , g) for i = 1, 2, . . ., where g is a fixed driver and xi is
a sequence of terminal points. For these pairs we get a sequence of solutions
(y·i , z·i ) satisfying:
Z T
Z T
i
i
i
yt = yT +
g(s, ys, zs )ds −
zsi dWs
t
yTi .
t
yti
where xi =
We assume that ր yt P -a.s for all t as i → ∞ (monotone
convergence), and we assume E[supt |yt |2 ] < ∞.
Theorem [Peng Theorem 3.8]
Under these assumptions, there exists a Z ∈ H2 (0, T ) such that
Z T
Z T
yt = x +
g(s, ys, zs ) −
zs dWs
t
(x = yT ), with x =
limi→∞ yti
t
in L2 (P ).
65
Proof
Recall that
E sup |yti|2 ≤ C,
C > 0, independent of i.
t∈[0,T ]
By the Lipschitz condition: g(t, y, z) − g(t, 0, 0) ≤ ν|y| + µ|z|:
hZ T
hZ T
i
2 i
i
i 2
E
≤E
g(s, ys , zs )
µ|zsi | + ν|ysi | + |g(s, 0, 0)| ds ≤
0
hZ
2E
0
0
T
i
hZ
2 i 2
ν |ys | ds + 2E
T
µ
2
0
|zsi |2 ds
(only on term depends on y):
hZ T
i
2
C1 + 2µ E
|zsi |ds ,
i
hZ
+ 2E
T
0
i
|g(s, 0, 0)|2ds ≤
C1 > 0, independent of i.
0
We see that it depends on the z i ’s. Consider now, using Ito’s formula:
2
2
d yti
= 2ytidyti + dyti
Substituting the dynamics, and using dWt dt = dt2 = 0 and dWt2 = dt:
= 2yti g(t, yti, zti )dt − zti dWt + (zti )2 dt
Integral form, to T :
(yTi )2
−
(y0i )
Z
=
T
0
2ysi g(s, ysi , zsi )ds
+
Z
T
0
(zsi )2
−
Z
T
0
2ysi zsi dWs
The last term is a proper Ito integral, and if we take the expectation it
disappears.
hZ T
i
hZ T
i
i 2
i 2
i 2
E[(y0 ) ] + E
(zs ) ds = E[(yT ) ] − E
2ysi g(s, ysi , zsi )ds
(⋆)
0
0
Majorize the g in the last term:
i
i
i
i
i
i
−2ys g(s, ys, zs ) ≤ 2|ys| ν|ys | + µ|zs | + |g(s, 0, 0)|
Using a trick:
≤ 2ν|ysi |2 + 2µ|ysi ||zsi | + 2|ysi ||g(s, 0, 0)|
= 2ν|ysi |2 +
√
|z i |
2 · 2µ|ysi | √s + 2|ysi ||g(s, 0, 0)|
2
66
Using the product to sum inequality 2ab ≤ a2 + b2 on the two last terms:
|z i |2
≤ 2ν|ysi |2 + 2µ2 |ysi |2 + s + |ysi |2 + |g(s, 0, 0)|2
2
|z i |2
= [2ν + 2µ2 + 1]|ysi |2 + |g(s, 0, 0)|2 + s
2
Now:
hZ
E
0
T
|zsi |2 ds
i
hZ
≤ (⋆) ≤ C+CT [2ν+1+2µ ]+E
T
2
hZ
E
T
0
i
i 1 hZ T
|zsi |2 ds =⇒
|g(s, 0, 0)| ds + E
2
0
2
0
i
|zsi |2 ds ≤ K
for some constant K that does not depend on i. Combining this with the
previous theorem, the proof is completed.
Risk Measures via g-Expectations [RG]
We have the probability space (Ω, F , P ), a fixed future date T and we let
X be the space of all financial positions of interest (e.g an element in X
is the net worth at the maturity T of a financial contract). We assume that
X = Lp (FT ), with the corresponding topological dual L′q (FT ) (the space of all
continuous linear functionals ℓ : X → R). For p = +∞ we set X ′ = L1 (FT ).
As usual F is the filtration generated by BM on the probability space. We
recall we have the relationship ℓ(X) = E[f X] for all X ∈ L2 (FT ), with ℓ ↔ f
(P )
(related to Thm. A.50, page 16). Again we have M1 which denotes the set
of all probability measures Q ∈ L2 (FT ), with Q ≪ P .
The driver g satisfies the following properties:
(i)
(ii)
(iii)
(iv)
g(·, y, z) is progressively measurable ∀(y, z)
g(·, 0, 0) ∈ H2 (0, T )
g(t, y1, z1 ) − g(t, y2, z2 ) ≤ ν|y1 − y2 | + µ|z1 − z2 |
g(·, y, 0) ≡ 0,
∀y
67
31.03.11
Definition [RG Eqn. (14)]
We set the monetary risk measure ρg : X → R as
ρg (X) := Eg [−X],
∀X ∈ X
Revision
Recall that Eg is monotone (by the Comparison Theorem), and since we have
’−X’, we get translation invariance (cash invariance). We have Eg [c] = c
for a constant c, since Eg [c] = Y0c = c (being F0 -measurable). This means
ρg (c) = Eg [−c] = −c, which is a special case for risk measures, and this
implies that ρg (0) = 0.
Positive homogeneity for risk measures is that for α > 0, we have ρ(αX) =
αρ(X). We want to determine if ρg (αX) = αρg (X). As we have seen for
g-expectations, if we have the special driver gµ = gµ (t, y, z) = µ|z|, then we
have Egµ [αX | Ft ] = αEgµ [X | Ft ]. We can extend this property to drivers
that are positive homogeneous, i.e g(t, αy, αz) = αg(t, y, z).
Proposition [RG Prop. 8]
If g is positive homogeneous in (y, z), then for any t ∈ [0, T ],
Eg [αX | Ft ] = αEg [X | Ft ]
Proof
Eg [αX | Ft ] =
YtαX
Now, without α we have:
Eg [X | Ft ] =
YtX
Z
h
= E αX +
t
T
i
g(s, YsαX , ZsαX )ds Ft
Z T
h
i
1
αX
αX
g(s, Ys , Zs )ds Ft
=α·E X +
α
t
Z T
h
i
Y αX Z αX
=α·E X +
g(s, s , s )ds Ft
α
α
t
Z
h
=E X+
t
X
T
i
g(s, YsX , ZsX ) Ft
(⋆)
We compare these expectations: (Y· , Z·X ) is the solution of the BSDE
with (X, g), and as we have seen the solution of BSDEs are unique in
αX
αX
S2 (0, T ) × H2 (0, T ). Thus we can identify ( Y α , Z α ) as Y·X , Z·X ), so (⋆) =
αEg [X | Ft ].
68
Proposition [RG Prop. 7]
If g is convex, then Eg [· | Ft ] is convex (for all t).
Proof
We take an arbitrary λ ∈ (0, 1) and X1 , X2 ∈ L2 (FT ) and consider
Eg λX1 + (1 − λ)X2 | Ft = Yt
which is related to the BSDE (λX1 + (1 − λX2 ), g). The solution has the
representation:
Z T
h
i
Yt = E λX1 + (1 − λ)X2 +
g s, YsλX1 +(1−λX2 ) , XsλX1 +(1−λX2 ) Ft (⋆⋆)
t
Since g is convex, we have the following inequality:
g t, λy1 + (1 − λ)y2 , λz1 + (1 − λ)z2 ≤ λg(t, y1, z1 ) + (1 − λ)g(t, y2, z2 )
We have:
Eg [X1 | Ft ] =
YtX1
Eg [X2 | Ft ] =
YtX2
Z
h
= E X1 +
T
g(s, YsX1 , ZsX1 | Ft
t
Z
h
= E X2 +
T
g(s, YsX2 , ZsX2 | Ft
t
i
i
λ · Eg [X1 | Ft ] + (1 − λ)Eg [X2 | Ft ] =
Z T
h
i
X1
X2
X2
X1
E λX1 + (1 − λ)X2 +
λg(s, Ys , Zs ) + (1 − λ)g(s, Ys , Zs ds Ft ≥
t
By convexity of g:
Z
h
E λX1 + (1 − λ)X2 +
t
T
g(s, λYsX1
+ (1 −
λ)YsX2 , λZsX1
+ (1 −
λ)ZsX2 ds
We can compare this solution to the one in (⋆⋆)
λX +(1−λ)X2
= Yt 1
= Eg λX1 + (1 − λ)X2 | Ft =⇒
h
i
Eg λX1 + (1 − λ)X2 | Ft ≤ λ · Eg X1 | Ft + (1 − λ)Eg X2 | Ft
Hence, Eg is convex if the driver g is convex.
69
i
Ft
Sub-linearity
A function is linear if it satisfies additivity and positive homogeneity:
f (αx + βy) = αf (x) + βf (y). Closely related to this is sub-linearity
which requires the function to be positive homogeneous and sub-additive:
f (αx + βy) ≤ αf (x) + βf (y). If g is sublinear in (y, z) it satisfies:
g(t, ay1 +by2 , az1 +bz2 ) ≤ ag(t, y1, z1 )+bg(t, y2, z2 ),
∀a, b > 0, ∀(y1 , z1 , y2 , z2 )
and in this case g is convex and positive homogeneous. The converse is also
true.
g t, λy1 + (1 − λ)y2 , λz1 + (1 − λ)z2 ≤ λg(t, y1, z1 ) + (1 − λ)g(t, y2, z2 )
g(t, ay1 + by2 , az1 + bz2 ) = (a + b)g t,
a
b
a
b
y1 +
y2 ,
z1 +
z2
a+b
a+b
a+b
a+b
Here λ = a/(a + b) and (1 − λ) = b/(a + b), and we apply convexity:
≤ (a + b)
i
h a
b
· g(t, y1, z1 ) +
· g(t, y2, z2 )
a+b
a+b
= a · g(t, y1, z1 ) + b · g(t, y2, z2 )
So, if g is convex, positive homogeneous and sublinear, this implies/we can
define a g-expectation Eg [· | Ft ].
If we have a g-expectation, this implies that g is convex, positive homogeneous
and sublinear. (So we have an equivalence between these notions).
For risk measures we have the two properties:
(i)
Positive:
X ≥ 0 =⇒ ρ(X) ≤ ρ(0)
(ii)
Monotone:
X≥Y
=⇒ ρ(Y ) ≤ ρ(X)
Monotone implies positivity. Linearity and positivity implies monotonicity.
70
Proposition [RG Prop. 11]
(1) If g is sublinear, ρg (X) = Eg [−X] for X ∈ L2 (FT ) is a coherent risk
measure.
(2) If g is convex, hen ρg is convex, positive (monotone) and cash invariant
(ρg (c) = −c ⇒ ρg (0) = 0).
The “new” property is cash invariance, as we have already proved the others.
Note that coherent: positive (monotone for us), positive homogeneous and
cash invariant.
Proof
First we mention this result for convex functions:
If h : Rn → R is a convex function, and it is bounded from above, then
it must be constant.
Consider now g(t, y, z). Fix z, so g(t, ·, z) : y → g(t, y, z). Then g(t, ·, z) is
convex. Moreover, from the Lipschitz condition:
g(t, y1, y2 ) − g(t, y2 , z2 ) ≤ ν|y1 − y2 | + µ|z1 − z2 |
Take (y1 , z1 ) and (y1, 0), and since (iv) from the property of drivers apply:
|g(t, y1, z1 )| ≤ µ|z1 |.
Since z is fixed, the right side is a constant function, so we have a constant
bound. We apply the result mentioned above, and we conclude that g(t, ·, z)
is a constant function for all z, which in turn means that g does not depend
on y.
We continute with risk measures via g-expectations:
ρg (X) = Eg [−X],
X ∈ X (= L2 (FT ))
(defined for L2 -objects because of BSDEs). Last time we studied monotonicity, and now we will look at “constancy”:
ρg (c) = −c,
ρg (0) = 0 (normalization)
71
07.04.11
Recall the Representation Theorem for risk measures:
ρ(X) = sup
EQ [−X − αmin (Q)
or any α(Q)
Q∈M1 (P )
where
αmin (Q) = sup EQ [−X],
X∈Aρ
X ∈ L∞ (FT ) (= X )
Instead of restricting ourselves to L∞ , we can look at the dual. For X =
Lp (FT ), we get X ′ = L′p (FT ) ≃ Lq (FT ). For X we have:
ρ(X) = sup
EQ [−X − α(Q)
Q∈M1 (P )
For X ′ we have:
ρ(X) = sup ℓ(−X) − α(ℓ) =⇒ sup E[ZX] − α(Z)
Z∈′q
ℓ∈X ′
(Note: in [RG] cash invariance isn’t that important, but lsc (lower
semicontinuity) is).
Corollary [RG Cor. 12]
If g is convex, we have the following representation for ρg :
ρg (X) = sup EQ [−X] − αg (Q)
Q∈Pg
where αg (Q) < ∞, and
dQ
∈ L2 (FT )
Pg = Q prob. measure on (Ω, FT ) Q ≪ P such that
dP
This is a type of M1 (P ). We think of Z = dQ/dP and Q(A) = E[1A Z]
for A ∈ FT , and αg (Q) := Fg (dQ/dP ), a function of the Radon-Nikodym
derivative:
Fg (f ) = sup (FT ) E[−f X] − Eg [−X] , (Eg [−X] = ρg (X))
X∈L2
(Look at the Fenchel transform and its link to the penalty function.
If g is sublinear (positive homogeneous and sub-additive), then
ρg (X) = sup EQ [−X]
Q∈Pg
72
Comment: Monotonicity wrt g
Suppose there are two drivers g1 and g2 and g1 ≤ g2 (a pointwise relationship,
which does not apply to all functions as we would need an ordering).
g1 ≤ g2 ⇐⇒ Eg1 [X] ≤ Eg2 [X],
∀X ∈ L2 (FT ) =⇒ ρg1 (X) ≤ ρg2 (X),
∀X ∈ L2 (FT ).
If something is acceptable wrt g2 , it is immediately acceptable wrt g1 , but
not vice versa. g2 is more “conservative”.
If we add convexity:
Fg1 (f ) = sup
E[−f X]−Eg1 ≥
X∈L2 (FT )
sup
X∈L2 (FT )
Pg1 ⊆ Pg2 .
E[−f X]−Eg2 = Fg2 (f ) =⇒
Comment: Monotonicity wrt X
a)
b)
X1 ≥ X2 =⇒ Eg [X1 ] ≥ Eg [X2 ]
X1 ≥ X2 and Eg [X1 ] = Eg [X2 ] ⇐⇒ X1 = X2
ρg (X1 ) = ρg (X2 ) ⇐⇒ X1 = X2
Important: property b) does not apply for coherent risk measures!
Example
Let Ω = {ω1 , ω2 }, and let F = ∅, {ω1}, {ω2 }, Ω with P (ω1) > 0 and
P (ω2 ) > 0, and where X = L∞ (FT ). We will use the essential supremum:
Essential supremum
ess sup(X) = inf c | X ≤ c P -a.s
Used as the norm for L∞ : kXk∞ = ess sup |X|.
Consider the risk measure on X given by:
ρ(X) = ess sup(−X), X ∈ X
Now take
X1 =
0
−4
ω1
,
ω2
X2 =
−2
−4
ω1
ω2
In this case we have X1 ≥ X2 and X1 6= X2 since P (X1 6= X2 ) = P (ω1) > 0.
Also, ρ(X1 ) = ρ(X2 ) = 4.
73
Consider another space (Ω, F , P ) with L∞ (F ) = X . For the uniform
distribution U, we set:
X1 ∼ U[1, 2]
=⇒ X2 ≥ X1 and X1 6= X2
X2 = X12
In this case, ρ(X1 ) = ρ(X2 ) = −1.
B & S-market: Risk Measures and Pricing
We have the classical setting, with te price dynamics:
dSt = µt St dt + σt St dBt ω1
S0 > 0
interest rate process Rt ≡ 1 (dRt = 0) and value process dVt = πt dSt where
πt is the portfolio. For this market we are interested in the portfolio that
replicates X, for X ∈ L2 (FT ), so we require VT = X. We now have the
elements we need for a BSDE.
From the value process, we insert the dynamics dSt .
dVt = πt µt St dt + πt σt St dBt
with VT = X. We set Yt = Vt . The general form for solutions to the BSDEs
is:
Z T
Z T
Yt = ξ +
g(s, Ys, Zs )ds −
Zs dWs , 0 ≤ t ≤ T,
t
t
and we immediately see that Zt = πt σt St . We also find that g(t, y, z) =
−(µ2 /σ2 ) · z, which is linear and independent of y. From this we get the
BSDE with the pair (X, g).
Y0X = Eg [X] = ρg (−X)
Y0X = V0π = E[X].
(Pricing as a form of risk).
74
Dynamic Risk Measures
14.04.11
Say X ∈ X is your risky position, and you want to monitor the risk of this
position in the entire time interval t ∈ [0, T ]. For this we need a risk measure
that changes with time, which is a dynamic risk measure.
Definition [RG Def. 15]
A dynamic risk measure is a family of mappings ρt , depending on the time
t, 0 ≤ t ≤ T , that satisfies the following three properties:
(a)
∀t, ρt : X → L0 (Ft )
(b)
ρ0 is a static risk measure.
(c)
ρT (X) = −X, when X ∈ X is FT -measurable.
Assumptions for Dynamic Risk Measures
∀t ∈ [0, T ], ρt is convex.
X ≥ 0 =⇒ ρt (X) ≤ ρt (0) ∀t ∈ [0, T ]
∀t ∈ [0, T ] and ∀c ∈ R, ρt (c) = −c
∀t ∈ [0, T ], ρt (X + Y ) = ρt (X) − Y
∀X ∈ X , ∀Y ∈ X , Ft -measurable
(dyn. sub-linearity) ∀t ∈ [0, T ], ∀α ≥ 0, ∀X, Y ∈ X
ρt (αX) = αρt (X), ρt (X + Y ) ≤ ρt (X) + ρt (Y )
(dyn. convexity)
(dyn. positivity)
(dyn. constancy)
(dyn. translability)
Definition [RG Def. 16]
A dynamic measure ρt is time-consistent, if for all t ∈ [0, T ], ∀X ∈ X and
∀A ∈ Ft :
ρ0 (X 1A ) = ρ0 − ρt (X)1A
Alternatively, for s ≤ t:
ρs (X 1A ) = ρs − ρt (X)1A
Time consistency is important, otherwise there are arbitrage opportunities.
75
A dynamic risk measure is coherent if it satisfies positivity, translability and
sub-linearity.
A dynamic risk measure is convex if it satisfies convexity and ρt (0) = 0.
Example 1
ρt (X) = ess sup EQ [−X | Ft ] − αt (Q)
Q∈M1 (P )
This is convex if we have some properties on α: if for all t, αt : M1 (P ) →
[0, ∞] is convex and inf Q∈M1 (P ) (Q) = 0. With these properties on α, the
example 1 dynamic risk measure is a convex.
Example 2
ρt (X) = ess sup EQ [−X | Ft ]
Q∈M1 (P )
This risk measure is a coherent risk measure, as it satisfies positivity, cash
invariance/translability and sublinearity.
Neither of the two risk measures are time consistent.
To verify some of the properties for these risk measures (which we call 1) and
2)), we first check that they satisfy ρT (X) = −X. For a general X ∈ X :
2)
EQ [−X | FT ] = −X
1) The first term is the same as 2).
ρT (X) = −X + sup{−αT (Q)} = −X − inf {αT (Q)} = −X
Q
Q
|
{z
}
=0
(this is why we require ρt (0) = 0 in the definition of convexity).
Why do we need convexity for α?
n o
ρt λX + (1 − λ)Y = ess sup EQ − λX − (1 − λ)Y | Ft − αt (Q)
n
o
= ess sup λ · EQ [−X | Ft ] + (1 − λ) · EQ [−Y | Ft ] − αt (Q)
n
= ess sup λ · EQ [−X | Ft ] + (1 − λ) · EQ [−Y | Ft ]
o
− λ · αt (Q) − (1 − λ) · αt (Q)
≤ λ · ess sup EQ [−X | Ft ] − αt (Q)
(1 − λ) · ess sup EQ [−Y | Ft ] − αt (Q)
76
The last inequality could come from the essential supremum, which is convex
- but does it apply directly to the penalty function?
Risk measure 1) is not positive homogeneous. since c · αt changes the penalty
function. Risk measure 2) is positive homogeneous because of the positive
homogeneity of the conditional expectation.
Time Consistency
(Reference: Delbaen, 2004). The structure of the m-stable sets, and in
particular the set of risk neutral measures.
Since Q ≪ P in M1 (P ), we have the Radon-Nikodym derivative, dQ/dP =
f , where f is an Ft -measurable object. On (Ω, Ft ):
h dQ i
dQ Ft
=E
dP Ft
dP
For all Q1 , Q2 ∈ Mm
1 (P ) (where m means m-stable), a probability measure
Q3 , Q3 ≪ P , has the density:
dQ2
h dQ i
dQ3
1 Ft · dQdP
:= E
dP
dP
E dP2 | Ft
For probability measures Q1 , Q2 , Q3 on (Ω, FT ), we have (measure (2)?)
(⋆)
ρ0 − ρt (X)1A = sup EQ1 [ρt (X)1A ]
Q1
Since
we get:
h dQ
X]
EQ [X] = E
dP
dQ
h
i
dP
X Ft
EQ [X | Ft ] = E dQ
E dP | Ft
dQ2
h
i
dP
Ft
ρt (X) = ess supEQ2 [−X | Ft ] = ess supE − X dQ2
E dP | Ft
Q2
Q2
Now we use: (†: 1A is a constant/measurable wrt Ft )
1ρt (X) = 1A ess sup EQ2 [. . . | Ft ] = ess sup EQ2 [1A . . . | Ft ] = ρt (1A X)
†
77
This means we have:
dQ2
h
i
dQ
dP
Ft
(⋆) = sup E
ess supE − X 1A dQ2
dP Q2
E dP | Ft
Q1
dQ2
h
h dQ i
i
1 dP
Ft
Ft · dQ2
= sup E ess supE − X 1A E
dP
E dP | Ft
Q1
Q2
Where we used the trick (check this):
h i
E ξ · ess sup E[η | G] = E E ξ · ess sup E[η | G] G
= E ess sup E[η | G] · E[ξ | G]
h
i
= E ess sup E ηE[ξ | G] G
Finally, we can “invert” the sup and ess.sup:
h
dQ3 i
= sup sup E − X 1a
dP
Q1 Q2
Dynamic Risk Measures via g-Expectations
We set X = L2 (FT ), we have the Brownian filtration F and we have g, which
is a driver for BSDE satisfying all the conditions that guarantee the existence
of a solution.
Definition [RG Eq. (22)]
We define the risk measure as:
ρgt (X) := Eg [−X | Ft ],
X ∈ L2 (Ft )
We have the following properties:
a) Ft -measurability for all t ∈ [0, T ].
b) ρg0 is a static risk measure.
c) ρgT (X) = −X.
Recall that for a pair (g, X), we have the BSDE:
−dYt = g(t, Yt , Zt )dt − Zt dWt
YT = ξ
which is solved by the couple (Y· , Z· ), with E[supt |Yt |2 ] < ∞, and Eg [X |
Ft ] = YtX .
78
Proposition [RG Prop. 19]
a) Continuous time recursivity. For all s, t where 0 ≤ s ≤ t ≤ T
ρgs (X) = ρgs (−ρgt (X)), X ∈ L2 (FT )
b) If there exists a t such that ρgt (X) ≤ ρgt (Y ), then
∀s ∈ [0, t], ρgs (X) ≤ ρgs (Y )
(but we do not necessarily have X ≤ Y ).
c) If g is sublinear, then ρgt , 0 ≤ t ≤ T is coherent and time-consistent.
d) If g is convex, then ρgt , 0 ≤ t ≤ T is convex and time-consistent.
Moreover, it satisfies positivity, constancy and translativity/cash
invariance.
(Note: a) and b) also hold for stopping times).
Proof
a) s ≤ t, and using the time consistency/tower property of g-expectations:
ρgs (−ρgt (X)) = Eg Eg [X | Ft ] Fs ] = Eg [X | Fs ] = ρgs (X).
b) Take some s ∈ [0, t]. We will compare two BSDEs on [0, t] with pairs
(g, Eg [−X|Ft ]) and (g, Eg [−Y |Ft ]), with Eg [−X|Ft ] ≤ Eg [−Y |Ft ] by applying
the Comparison Theorem.
ρgs (X) = Eg [−X | Fs ] =Eg Eg [−X | Ft ] Fs ≤
Eg Eg [−Y | Ft ] Fs = Eg [−Y | Fs ] = ρg (Y )
s
Properties
We want to verify if the following is true for all t:
ρg0 (X 1A ) = ρ0 − ρt (X)1A .
This follows by the properties of g-expectations:
Eg [−X 1A ] = Eg Eg [−X | Ft ]1A
(already proved).
79
The dynamic risk measures have positivity (or monotonicity), which follows
for g-expectation by the Comparison Theorem, and constancy, which is also
a property from g-expectations: Eg [c | Ft ] = c.
Cash invariance:
ρgt (X + Y ) = ρgt (X) − Y, ∀X ∈ L2 (FT ), ∀Y ∈ L2 (Ft )
From the properties og g-expectations:
Eg [−X − Y | Ft ] = Eg [−X | Ft ] − Y
which is true if g does not depend on y, or equivalently, g is sublinear. This
is true when the driver is on the form: gµ (t, y, z) = µ|z|.
The risk measures:
ρt (X) = ess sup EQ [−X | Ft ]
Q∈M(P )
ρt (X) = ess sup EQ [−X | Ft ] − αt (Q)
Q∈M(P )
are not time consistent, though they are dynamic. The risk measure:
ρgt (X) = Eg [−X | Ft ]
is time consistent, because of the BSDE.
Note, for some suitable functionals g and sets Q, we have a relationship
between the g-expectations and normal expectations.
Eg [X | Ft ] = ess supEQ [X | Ft ]
Q∈Q
Proposition [RG Prop. 20]
Let ρt , for 0 ≤ t ≤ T , be a dynamic, time-consistent and coherent (resp.
convex and cash invariant) risk measure on X = L2 (FT ). If E[X] := ρ0 for
X ∈ L2 (FT ) is:
(i) strictly monotone (by the Comparison Theorem):
X ≥Y
µ
=⇒ E[X] ≥ E[Y ]
X = Y ⇐⇒ E[X] = E[Y ]
(ii) E -dominated: ∃µ > 0 such that
E[X + Y ] − E[X] ≤ Eµ [Y ],
∀X, Y ∈ L2 (FT )
where Eµ is a gµ -expectation, i.e solution of a BSDE with driver
gµ = gµ (t, y, z) = µ|z|, for µ > 0.
80
Then there exists a unique g satisfying:
(A) g is Lipschitz in (y, z).
(B) g(·, y, z) ∈ L2 (dP × dt), for all y, z.
(C) P -a.s, g(t, y, 0) = 0 for all t and for all y.
(D) g is independent of y.
(E) |g(t, z)| ≤ µ|z|.
such that:
ρ0 (X) = E[−X] = Eg [−X]
and
ρt (X) = Eg [−X|Ft ],
∀t, ∀X ∈ L2 (Ft ).
If ρt , for 0 ≤ t ≤ T is continuous in t, then g is also sublinear (resp. convex).
The End
81