MATH3075/3975 Financial Derivatives
Tutorial 1: Solutions
Exercise 1 (a) The conditional distributions of X given Y = j are:
for j = 1: (1/5, 3/5, 1/5),
for j = 2: (2/3, 0, 1/3),
for j = 3: (0, 3/5, 2/5),
and thus the conditional expectations EP(X|Y = j) are computed as follows:
EP(X|Y = 1) = 1/5 + 6/5 + 3/5 = 2,
EP(X|Y = 2) = 2/3 + 0 + 3/3 = 5/3,
EP(X|Y = 3) = 0 + 6/5 + 6/5 = 12/5.
(b) The marginal distributions of X and Y are:
for X : (4/18, 9/18, 5/18),
for Y : (10/18, 3/18, 5/18).
On the one hand, we obtain
EP(X) =
3∑
i=1
iP(X = i) =
4
18
+ 2 ·
9
18
+ 3 ·
5
18
=
37
18
.
On the other hand, we get
EP
(
EP(X|Y )
)
=
3∑
j=1
EP(X|Y = j)P(Y = j) = 2 ·
10
18
+
5
3
·
3
18
+
12
5
·
5
18
=
37
18
.
Hence the equality EP(X) = EP
(
EP(X|Y )
)
holds, as was expected.
(c) Random variables X and Y are not independent since the condition
P(X = i, Y = j) = P(X = i)P(Y = j), ∀ i, j = 1, 2, 3,
is not satisfied. For instance, if we take i = 1 and j = 2, then
P(X = 1, Y = 3) = 0 6=
4
18
·
5
18
= P(X = 1)P(Y = 3).
Exercise 2 (a) We need to show that∫ ∞
−∞
∫ ∞
−∞
f(X,Y )(x, y) dxdy =
∫ ∞
0
∫ ∞
0
1
y
e
−x
y e−y dxdy
?
= 1.
1
We first compute the marginal density of Y . For y ≥ 0, we obtain
fY (y) =
∫ ∞
0
1
y
e
−x
y e−y dx =
1
y
e−y
∫ ∞
0
e
−x
y dx =
1
y
e−y
[
−ye−
x
y
]∞
0
= e−y.
Of course, we have fY (y) = 0 for all y < 0. Therefore,∫ ∞ 0 ∫ ∞ 0 1 y e −x y e−y dxdy = ∫ ∞ 0 fY (y) dy = ∫ ∞ 0 e−y dy = 1. (b) For any fixed y ≥ 0, the conditional density of X given Y = y equals fX|Y (x| y) = f(X,Y )(x, y) fY (y) = 1 y e −x y , ∀x ≥ 0, and fX|Y (x| y) = 0 for x < 0. Consequently, for every y ≥ 0 EP(X|Y = y) = ∫ ∞ 0 xfX|Y (x| y) dx = ∫ ∞ 0 x y e −x y dx = y ∫ ∞ 0 ze−z dz = y. Exercise 3 We first compute the conditional cumulative distribution function of X given the event {X < 0.5} FX|X< 0.5(x) := P(X ≤ x|X < 0.5), ∀x ∈ R. We obtain FX|X< 0.5(x) = P(X ≤ x, X < 0.5) P(X < 0.5) = 0, if x ≤ 0, 2P(X ≤ x) = 2x, if x ∈ (0, 0.5), 1, if x ≥ 0.5. so that the conditional density of X given the event {X < 0.5} equals fX|X< 0.5(x) = 0, if x ≤ 0, 2, if x ∈ (0, 0.5), 0, if x ≥ 0.5. Therefore, EP(X|X < 0.5) = ∫ ∞ −∞ xfX|X< 0.5(x) dx = ∫ 0.5 0 2x dx = 0.25. Exercise 4 We have fX(x) = 1 λ e− x λ , ∀x > 0,
and thus
P(X > x) = 1− FX(x) =
∫ ∞
x
fX(u) du = e
− x
λ , ∀x > 0.
Consequently,
P(X > x|X > 1) =
P(X > x,X > 1)
P(X > 1)
=
P(X>x)
P(X>1) = e
−x−1
λ , if x ≥ 1,
1, if x < 1. 2 Hence the conditional density equals fX|X>1(x) =
{
1
λ
e−
x−1
λ , if x ≥ 1,
0, if x < 1,
and thus
EP(X|X > 1) =
∫ ∞
1
x
λ
e−
x−1
λ dx = 1 + λ = 1 + EP(X).
Exercise 5 Since
Cov(X,Y ) = EP(XY )− EP(X)EP(Y ) = EP(X3)− EP(X)EP(X2) = 0
since EP(X) = EP(X3) = 0. Therefore, the random variables X and Y are uncorrelated. They are
not independent, however, since, for instance
EP(Y |X = 1) = 1 6= 4 = EP(Y |X = 2).
Recall that under independence of X and Y we have EP(Y |X) = EP(Y ) and EP(X|Y ) = EP(X).
Exercise 6 (MATH3975) We have
Cov(X,Y ) = EP(XY )− EP(X)EP(Y )
= EP
(
(U + V )(U − V )
)
− EP(U + V )EP(U − V )
= EP
(
U2 − V 2
)
− EP(U + V )
(
EP(U)− EP(V )
)
= 0.
since EP(U) = EP(V ) and EP(U2) = EP(V 2). Hence the random variables X and Y are uncorrelated.
To check whether X and Y are independent, we need first to specify the joint distribution of X
and Y . For instance, if we take U = V , then X = 2U and Y = 0 so that X and Y are independent.
However, if we take U and V independent (but not deterministic), then X and Y are not necessarily
independent. For instance, we may take as U and V the i.i.d. Bernoulli random variables with
P(U = 1) = p = 1− P(U = 0). It is then easy to check that X and Y are not independent.
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