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MAT4701: Voluntary Assignment 1
Sindre Froyn
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Problem 1
Let ti = ni , i = 0, 1, . . . , n, i.e, ∆ti = n1 .
a) For n ∈ N compute
h n−1
X
i
Bti Bti+1 − Bti Btj Btj+1 − Btj .
F (n) = E
i,j=0
——————————————–
For convenience we write Btj = Bj . We will consider three separate cases. If
i < j, then Bi , Bj and (Bi+1 − Bi ) are all Ftj -measurable, since i + 1 ≤ j,
leaving (Bj+1 − Bj ) as an independent factor in the expression.
n−1
X
i,j=0
E Bi Bi+1 − Bi Bj · E Bj+1 − Bj
Using that E[Bj+1 − Bj ] = E[Bj+1 ] − E[Bj ] = 0, it follows that
n−1
X
i,j=0
E Bi Bi+1 − Bi Bj · 0 = 0.
The second case is when j < i, but that is completely analogous to the first
case and we get zero in the same way. The third case is when we have i = j.
We can rewrite the expression with one sum, since we get zero whenever
i 6= j and use the fact that we have independent increments.
n−1
n−1
X X
2 2 E Bi2 E Bi+1 − Bi
=
E Bi2 Bi+1 − Bi
i=0
i=0
We can find the expectations.
E[Bi2 ] = E[Bt2i ] = ti =
i
n
2
1
E[ Bti+1 − Bti ] = ti+1 − ti = ∆ti =
n
Substituting this back into the expression, we get (since all the other terms
cancel to zero):
n−1
X i
F (n) =
n2
i=0
1
——————————————–
PN
b) Simplify the answer in a) using
k=1 k =
limn→∞ F (n).
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Applying the formula for a geometric series:
N (N +1)
2
and compute
N
N
X
N (N + 1)
1
1 X
N +1
N2 + N
i
i
=
=
0+
=
=
F (N ) =
2
2
2
N
N
N
2
2N 2
2N
i=1
i=0
Passing to the limit.
1+
n→∞
2
lim F (n) = lim
n→∞
1
n
1
2
=
——————————————–
c) Explain the result in b) in light of the Ito isometry.
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Like we showed in a), we only have to count the terms where i = j, which
resulted in the expression (this time we include both sums):
E
n−1
X
i=0
Bi Bi+1 − Bi
2 = E
n−1
X
Bi ∆Bi
i=0
2 We observe that the partition is from 0 to 1.
0 = t0 =
1
n−1
n
0
< < ... <
< = tn = 1
n
n
n
n
so if we let ∆ti → 0, we get the stochastic integral from 0 to 1.
n−1
2 X
Bi ∆Bi
E
i=0
−−−→
∆tj →0
Z
E
1
Bs dBs
0
2 We can now apply the Ito isometry, find the expectation and calculate the
Lebesgue integral.
Z 1
Z 1
1 1 1
E[Bs2 ]ds =
sds = s2 =
2 0 2
0
0
The calculations we performed in a) are very closely related to the methods
used in proving the Ito isometry. In part b), instead of using the Ito isometry,
i.e. writing the sum as an integral like we did in part c), we calculated the
sum of an infinite series. As we can see, we arrived at the same result in the
end.
2
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Problem 2
t 2
a) Use the formula E[eαBt ] = e 2 α , that is valid for all α in C, and standard
trigonometric formulas to compute
i) E[cos(αBt )]
ii) E[sin(αBt )]
iii) E[cos2 (αBt )]
iv) E[sin2 (αBt )]
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a)
(i) With the trigonometric identity
cos(x) =
1 ix
e + e−ix
2
where x = αBt , we can write the expectation as
1 iαBt
E[cos(αBt )] =
E[e
] + E[e−iαBt ] =
2
We can define the complex numbers γ1 = iα and γ2 = −iα and use the given
formula with γ12 = γ22 = −α2 .
1
α2
α2
α2 1 − α2 t
e 2 + e− 2 t = 2e− 2 t = e− 2 t
2
2
(ii) This time we use
1 ix
e − e−ix .
2i
α2
1 − α2 t
1
E[eiαBt ] − E[e−iαBt ] =
e 2 − e− 2 t = 0
E[sin(αBt )] =
2i
2i
(iii) Multiplying cos x · cos x with the formula above, we get
sin(x) =
2
cos x =
eix + e−ix
2
2
=
i
e2ix + 2e0 + e−2ix
1h
= e2ix + e−2ix + 2 .
4
4
With this we can rewrite the initial expectation.
E[cos2 (αBt )] =
1 2iαBt
] + E[e−2iαBt ] + E[2] =
E[e
4
Defining γ1 = 2iα and γ2 = −2iα we use the given formula with γ12 = γ22 =
−4α2 .
i e−2α2 t + 1
1 h −2α2 t
−2α2 t
e
+e
+2 =
4
2
3
(iv)
2
sin x =
eix − e−ix
2
2
=
i
e2ix − 2e0 + e−2ix
1h
= e2ix + e−2ix − 2 ⇒
4
4
i
1 h 2iαBt
E[e
] + E[e−2iαBt ] − E[2] =
4
h
i e−2α2 t − 1
1 −2α2 t
−2α2 t
e
−e
−2 =
4
2
E[sin2 (αBt )] =
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b) Assume that t ≥ s. Use a) to compute E[cos(Bt ) cos(Bs )].
Hint: Use a trigonometric formula and independence of increments.
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Since Bt = Bt − Bs + Bs ;
cos(Bt ) cos(Bs ) = cos(Bt − Bs + Bs ) cos(Bs ).
We can now use the trigonometric identity
cos(θ + ϕ) = cos(θ) cos(ϕ) − sin(θ) sin(ϕ)
with θ = Bt − Bs , ϕ = Bs and get
cos(Bt − Bs + Bs ) = cos(Bt − Bs ) cos(Bs ) − sin(Bt − Bs ) sin(Bs ) =⇒
cos(Bt ) cos(Bs ) = cos(Bt − Bs ) cos2 (Bs ) − sin(Bt − Bs ) sin(Bs ) cos(Bs ).
We can make the substitution sin(ϕ) cos(ϕ) =
= cos(Bt − Bs ) cos2 (Bs ) −
1
2
sin(2ϕ).
1
sin(Bt − Bs ) sin(2Bs )
2
Taking the expectation and using linearity on the right side.
1 E cos(Bt ) cos(Bs ) = E cos(Bt −Bs ) cos2 (Bs ) − E sin(Bt −Bs ) sin(2Bs )
2
Independent increments.
1 E cos(Bt − Bs ) E cos2 (Bs ) − E sin(Bt − Bs ) E sin(2Bs )
2
|
{z
}
We have now established that
|
{z
=0 By 2a)
=0
E[cos(Bt ) cos(Bs )] = E[cos(Bt − Bs )]E[cos2 (Bs )]
4
}
We define a new Brownian motion (as in exercise 2.12 in the text book):
eu = Bt − Bs where u = t − s ≥ 0. Now we can use the results from 2a) to
B
get our expectation.
−2t
+1
2a) − 1 u e
2
e
2
E[cos(Bt ) cos(Bs )] = E[cos(Bu )]E[cos (Bs )] == e
2
−2t
1 1
1
1
5
1 +1
1 1 s− 5 t
u=t−s − 1 (t−s) e
=== e 2
e 2 2 + e 2 s− 2 t =
e 2 s e− 2 t + e− 2 t
=
2
2
2
——————————————–
Rt
c) Use b) to compute E[( 0 cos(Bs )ds)2 ].
Rt
RtRu
Hint: For any reasonable function f , ( 0 f (s)ds)2 = 2 0 0 f (u)f (v)dvdu.
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Applying the hint we want to find the expectation
Z tZ s h
h Z tZ s
i
i
E 2
cos(Bs ) cos(Bu )duds = 2
E cos(Bs ) cos(Bu ) duds
0
0
0
0
Using the result from 2b) we now have the double integral
Z tZ s 1
1
1 1u −5s
I =2
e 2 e 2 + e 2 u e− 2 s duds
0
0 2
Z tZ s
Z t
1
5
1
1
u
−
s
u
−
s
=
e 2 e 2 + e 2 e 2 duds =
I1 ds
0
0
0
First we calculate the inner integral I1 .
Z s
is
h 1
1
1
5
1
5
1
1
e 2 u e− 2 s + e 2 u e− 2 s du = 2e 2 u e− 2 s + 2e 2 u e− 2 s =
I1 =
0
0
1
1
5
1
1
5
1
1
2e 2 s e− 2 s + 2e 2 s e− 2 s − 2e 2 (0) e− 2 s + 2e 2 (0) e− 2 s =
1
5
I1 = 2 e−2s − e− 2 s − e− 2 s + 1
We can now find the outer integral I.
Z t
Z t
5
1
e−2s − e− 2 s − e− 2 s + 1 ds =
I1 ds = 2
I=
0
0
it
h 1
1
2 5
2 − e−2s + e− 2 s + 2e− 2 s + s =
2
5
0
1
1
1
1
5
5
2
2
2 − e−2t + e− 2 t + 2e− 2 t + t − 2 − e−2(0) + e− 2 (0) + 2e− 2 (0) + 0
2
5
2
5
5
1
4
4
I = −e−2t + e− 2 t + 4e− 2 t + 2t − (−1 + + 4)
5
5
The integral is done. We have shown that
i
hZ t
1
4 5
19
E ( cos(Bs )ds)2 = −e−2t + e− 2 t + 4e− 2 t + 2t − .
5
5
0
5
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Problem 3 R
Rt
t
Compute E[ 0 Bs dBs · 0 Bs2 dBs ].
——————————————–
Rt
Rt
Setting X = 0 Bs dBs and Y = 0 Bs2 dBs we can use that
E[XY ] =
We have
1
1
E[(X + Y )2 ] − E[(X − Y )2 ].
4
4
Z
2 i 1 h Z t
2 i
1 h t
E
− E
Bs + Bs2 dBs
Bs − Bs2 dBs
4
4
0
0
Ito isometry.
1
4
Z
t
0
Z
h
i
2 i
1 t h
2 2
E Bs + Bs ds −
E Bs − Bs2 ds
4 0
Some intermediate calculations.
2 E Bs + Bs2
= E Bs2 + 2Bs3 + Bs4 = s + 0 + 3s2 = s + 3s2
2 E Bs − Bs2
= E Bs2 − 2Bs3 + Bs4 = s − 0 + 3s2 = s + 3s2
This means we have the same integrals, which cancel to zero:
Z
Z
1 t
1 t
2
s + 3s ds −
s + 3s2 ds = 0.
4 0
4 0
So, to summarize.
Z t
i
hZ t
E
Bs2 dBs = 0
Bs dBs ·
0
0
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Problem 4
Let t ≥ s and compute E[(Bs + sin(Bs ))e2Bt | Fs ].
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We begin by observing that the factor (Bs +sin(Bs )) is Fs -measurable, which
means we can factor it out of the conditional expectation.
Bs + sin(Bs ) E e2Bt | Fs
Using that Bt = Bt − Bs + Bs we get
Bs + sin(Bs ) E e2(Bt −Bs +Bs ) | Fs =
Bs + sin(Bs ) E e2(Bt −Bs ) e2Bs | Fs
Using that exp(2Bs ) is Fs -measurable, and that exp(2(Bt − Bs )) is
independent of Fs , we get
Bs + sin(Bs ) e2Bs E e2(Bt −Bs )
This expectation is known, by exercise 2.20 in the text.
Bs + sin(Bs ) e2Bs e2(t−s)
6
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Problem R5t
R
1 t 2
Let Zt = e 0 Bs dBs − 2 0 Bs ds .
a) Prove that dZt = Zt · Bt dBt and use this to compute E[Zt ].
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To simplify the notation we write
Z
Z t
1 t 2
B ds
Bs dBs −
ζ(t) =
2 0 s
0
so Zt = eζ(t) . By the Ito formula we get
2
2 1
1
dZt = eζ(t) dζ(t) + eζ(t) dζ(t) = eζ(t) dζ(t) + dζ(t)
2
2
(⋆)
where
1
dζ(t) = Bt dBt − Bt2 dt
2
and when we use that dtdt = dBt dt = dtdBt = 0 and dBt dBt = dt;
2
2
1
dζ(t) = (Bt dBt )2 − Bt3 dBt dt + Bt4 dt = Bt2 dt
4
Substituting these in to the parenthesis in (⋆), we get
2 1
1 1
= Bt dBt − Bt2 + Bt2 = Bt dBt
dζ(t) + dζ(t)
2
2
2
Using this, eζ(t) = Zt and (⋆) we have shown that
dZt = Zt · Bt dBt .
In the general form dYt = uds + vdBt we note that we don’t have a uds
part. This means that Zt is a martingale, and we can use the fact that
E[Zt ] = E[Z0 ], and since Z0 = e0 = 1, then E[Zt ] = E[Z0 ] = E[1] = 1.
——————————————–
2
b) Prove that if t ≥ 0, then |Zt |2 ≤ eBt −t for all ω.
——————————————–
Since the exponential function is always non-negative, we have
2
2
|Zt |2 = eζ(t) = eζ(t) = e2ζ(t) .
2
To prove that e2ζ(t) ≤ eBt −t we only need to show that 2ζ(t) ≤ Bt2 − t, since
the exponential funtion is strictly increasing. Writing out the full exponent.
Z
Z t
1 t 2 B ds ≤ Bt2 − t
2
Bs dBs −
2 0 s
0
7
2
Z
t
0
Bs dBs −
Z
t
0
Bs2 ds ≤ Bt2 − t.
With the Ito formula, we get
Z t
1
1 Bs dBs = 2 Bt2 − t = Bt2 − t
2
2
2
0
which results in
Bt2
−t−
Z
t
0
Bs2 ds ≤ Bt2 − t
2
which must be true. This means |Zt |2 ≤ eBt −t for t ≥ 0.
——————————————–
c) Use the result in b) to prove that if 0 ≤ t < 12 , then E[Zt2 Bt2 ] < ∞.
Hint: Bt is Gaussian.
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2
Since |Zt |2 = Zt2 ≤ eBt −t , it follows that
2
2
2
Bt2 Zt2 ≤ Bt2 eBt −t =⇒ E[Zt2 Bt2 ] ≤ E[Bt2 eBt −t ] ≤ E[Bt2 eBt ]
(⋆⋆)
where the last inequality holds since t ≥ 0. So, to prove this result we only
2
need to show that E[Bt2 eBt ] is finite. Since Bt is Gaussian, we can consider
2
2
the expectation as E[Bt2 eBt ] = E[f (Bt )] for f (x) = x2 ex . This yields the
integral
Z
x2
1
2
2
x2 ex e− 2t dx.
I = E[Bt2 eBt ] = √
2πt R
We have the condition that t < 21 , which means that
1
constant γ = 2t
− 1 > 0 which we can use:
2
x2
ex e− 2t = ex
2 (1− 1 )
2t
1
2t
> 1. We define the
2
= e−γx .
We now have the form of a Gaussian integral.
r
Z
π
2
e−γx dx =
γ
R
Differentiating with regard to γ and multiplying with −1 on both sides.
Z
r
d
π
d
−γx2
e
dx = (−1)
(−1)
dγ R
dγ γ
Z
r
2
1
π
x2 e−γx dx =
2γ
γ
R
Setting this back in the integral, we have shown it is finite.
r Z
1
1
1
1
π
2 −γx2
x e
dx = √
< ∞.
I = √
= p
2πt R
2πt 2γ γ
2 2tγ 3
By the inequalities in (⋆⋆) the desired result follows.
8
——————————————–
d) Why do we need to prove an inequality like the one in c), and what can
we conclude from that?
——————————————–
Proving that expectations are finite are important when we want to show that
a process is a martingale. This inequality is used to prove that the stochastic
process Zt Bt is a martingale with regard to some filtration Mt . There are
three properties which must be veried and the first property, measurability,
follows by definition. The third property, E[Zt Bt | Ms ] = E[Zs Bs ] for some
t > s, follows since Bt is a martingale, and as we saw in (a), so is Zt . In this
instance it is the second property, finite expectation, that needs some work
to verify.
1
E[Zt Bt ] ≤ E[|Zt Bt |] ≤ E[|Zt Bt |2 ] 2 ≤ E[Zt2 Bt2 ] < ∞.
We see that all the three properties of martingales are met, and we have
formally proved that Zt Bt is a martingale.
9