Exercise 4, Ref. [7]
Let c(t, x) be the Black-Scholes price function of European calls. Prove that
lim c(t, x) = (x − Ke−rτ )+ ,
σ→0+
lim c(t, x) = x.
σ→∞
Compute also the following limits:
lim c(t, x),
lim c(t, x),
K→0+
lim c(t, x),
T →+∞
K→+∞
lim c(t, x).
x→0+
Repeat all the above for put options.
Solution
Recall that
c(t, x) = xΦ(d1 ) − Ke−rτ Φ(d2 ),
where
d2 =
and where Φ(x) =
d1 → d2 and
√1
2π
Rx
log
x
K
+ r − 12 σ 2 τ
√
,
σ τ
√
d1 = d2 + σ τ ,
1 2
−∞
(1)
(2)
e− 2 y dy is the standard normal distribution. As σ → 0+ we have
x
1
+ rτ )σ −1 .
d2 ∼ √ (log
K
τ
Hence
d2 → +∞, if x > Ke−rτ ,
d2 → −∞, if x < Ke−rτ ,
d2 → 0, if x = Ke−rτ ,
Thus
lim Φ(d1 ) = lim+ Φ(d2 ) = 1,
if x > Ke−rτ ,
lim+ Φ(d1 ) = lim+ Φ(d2 ) = 0,
if x < Ke−rτ ,
σ→0+
σ→0
σ→0
σ→0
lim+ Φ(d1 ) = lim+ Φ(d2 ) = Φ(0),
σ→0
if x = Ke−rτ .
σ→0
It follows that
lim c(t, x) = x − Ke−rτ
σ→0+
lim+ c(t, x) = 0,
if x > Ke−rτ ,
if x ≤ Ke−rτ ,
σ→0
1
i.e., limσ→0+ c(t, x) = (x − Ke−rτ )+ . For σ → +∞ we have d2 → −∞ and d1 → +∞, hence
Φ(d1 ) → 1 and Φ(d2 ) → 0. Thus c(t, x) → x as σ → +∞. As K → 0+ , both d1 and d2
diverge to +∞, hence
lim+ c(t, x) = x.
K→0
For K → +∞, d1 , d2 diverge to −∞. Hence the first term in c(t, x) converges to zero. As
the first term in c(t, x) always dominates the second term (since c(t, x) > 0), then the second
term also goes to zero and thus
lim c(t, x) = 0.
K→+∞
For T → +∞ we have d2 → −∞ and d1 → +∞, hence
lim c(t, x) = x.
T →+∞
Finally, for x → 0+ , both d1 , d2 diverge to −∞ and thus
lim c(t, x) = 0.
x→0+
To compute the limits for put options we use the put-call parity:
c(t, x) − p(t, x) = x − Ke−rτ ,
by which it follows that
lim p(t, x) = (Ke−rτ − x)+ ,
σ→0+
lim p(t, x) = 0,
K→0+
lim p(t, x) = 0,
T →+∞
lim p(t, x) = Ke−rτ
σ→+∞
lim p(t, x) = +∞,
K→+∞
lim p(t, x) = Ke−rτ .
x→0+
Exercise 6, Ref. [7]
Consider a European derivative with maturity T and pay-off Y given by
Y = k + S(T ) log S(T ),
where k > 0 is a constant. Find the Black-Scholes price of the derivative at time t < T and
the hedging self-financing portfolio. Find the probability that the derivative expires in the
money.
2
Solution
The pay-off function is g(z) = k + z log z. Hence the Black-Scholes price of the derivative is
ΠY (t) = v(t, S(t)), where
Z 2 √ x2 dx
r− σ2 τ −σ τ x
−rτ
g se
v(t, s) = e
e− 2 √
2π
ZR 2
√
√
x2 dx
σ2
(r− σ2 )τ −σ τ x
−rτ
k + se
=e
(log s + (r − )τ − σ τ x) e− 2 √
2
2π
R
Z
√
1
2 dx
= ke−rτ + s log s e− 2 (x+σ τ ) √
2π
Z R
Z
2
√
√ 2 dx
√
1
σ
− 12 (x+σ τ )2 dx
√ − sσ τ
+ s(r − )τ
e
xe− 2 (x+σ τ ) √
2
2π
2π
R
R
Using that
Z
e
R
√
− 12 (x+σ τ )2
dx
√ = 1,
2π
Z
1
xe− 2 (x+σ
√
τ )2
R
we obtain
v(t, s) = ke−rτ + s log s + s(r +
√
dx
√ = −σ τ ,
2π
σ2
)τ.
2
Hence
σ2
)τ.
2
This completes the first part of the exercise. The number of shares of the stock in the hedging
portfolio is given by
hS (t) = ∆(t, S(t)),
ΠY (t) = ke−rτ + S(t) log S(t) + S(t)(r +
where ∆(t, s) =
∂v
∂s
= log s + 1 + (r +
σ2
)τ .
2
Hence
hS (t) = 1 + (r +
σ2
)τ + log S(t).
2
The number of shares of the bond is obtained by using that
ΠY (t) = hS (t)S(t) + B(t)hB (t),
hence
hB (t) =
1
(ΠY (t) − hS (t)S(t))
B(t)
σ2
σ2
= e (ke
+ S(t) log S(t) + S(t)(r + )τ − S(t) − S(t)(r + )τ − S(t) log S(t))
2
2
= ke−rT − S(t)e−rt .
−rt
−rτ
This completes the second part of the exercise. To compute the probability that Y > 0,
we first observe that the pay-off function g(z) has a minimum at z = e−1 and we have
3
g(e−1 ) = k − e−1 . Hence if k ≥ e−1 , the derivative has probability 1 to expire in the money.
If k < e−1 , there exist a < b such that
g(z) > 0 if and only if 0 < z < a or z > b.
Hence for k < e−1 we have
P(Y > 0) = P(S(T ) < a) + P(S(T ) > b).
Since S(T ) = S(0)eαT −σ
√
TG
, with G ∈ N (0, 1), then
log S(0)
+ αT
a
√
S(T ) < a ⇔ G >
:= A,
σ T
+ αT
log S(0)
b
√
S(t) > b ⇔ G <
:= B.
σ T
Thus
Z
+∞
2
− x2
e
P(Y > 0) = P(G > A) + P(G < B) =
A
dx
√ +
2π
= 1 − Φ(A) + Φ(B).
This completes the solution of the third part of the exercise.
4
Z
B
x2 dx
e− 2 √
2π
−∞