Description
Solve the exercises by hand.
1. Let K, T, σ, and r be positive constants and let
g(x) = 1
√
2π
Z b(x)
0
e
−
y
2
2 dy
where b(x) = 1
σ
√
T
h
log
x
K
+
r +
σ
2
2
T
i
. Compute g
0
(x).
2. Let φ(u) = 1
√
2π
e
−u
2/2
so that Φ(x) = Z x
−∞
φ(u) du (i.e., the Φ(x) in Black-Scholes).
(a) For x > 0, show that φ(−x) = φ(x).
(b) Given that limx→∞
Φ(x) = 1, use the properties of the integral as well as a substitution
to show that Φ(−x) = 1 − Φ(x) (again, assuming x > 0).
3. (a) Under what condition does the following hold?
Z Z
D
f(x, y) dA =
Z Z
D
f(x, y) dy dx =
Z Z
D
f(x, y) dx dy
(b) Evaluate the double integral
Z Z
D
e
y
2
dA
where D = {(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y}
