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
3 X 4 9 5 37 EP (X) = i P(X = i) = +2· +3· = . 18 18 18 18 i=1
On the other hand, we get
3 X 10 5 3 12 5 37 EP EP (X| Y ) = EP (X| Y = j)P(Y = j) = 2 · + · + · = . 18 3 18 5 18 18 j=1
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
4 5 P(X = 1, Y = 3) = 0 6= · = P(X = 1) P(Y = 3). 18 18 Exercise 2 (a) We need to show that Z ∞Z ∞ Z ∞Z ∞ 1 − xy −y ? f(X,Y ) (x, y) dxdy = e e dxdy = 1. −∞ −∞ 0 0 y
1 We first compute the marginal density of Y . For y ≥ 0, we obtain Z ∞ Z ∞ 1 − xy −y 1 −x 1 −x ∞ h i fY (y) = e e dx = e−y e y dx = e−y −ye y = e−y . 0 y y 0 y 0
Of course, we have fY (y) = 0 for all y < 0. Therefore,
Z ∞Z ∞ Z ∞ Z ∞ 1 − xy −y e e dxdy = fY (y) dy = e−y dy = 1. 0 0 y 0 0
(b) For any fixed y ≥ 0, the conditional density of X given Y = y equals
f(X,Y ) (x, y) 1 −x fX|Y (x| y) = = e y, ∀ x ≥ 0, fY (y) y
and fX|Y (x| y) = 0 for x < 0. Consequently, for every y ≥ 0
Z ∞ Z ∞ Z ∞ x − xy EP (X| Y = y) = xfX|Y (x| y) dx = e dx = y ze−z dz = y. 0 0 y 0
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 0, if x ≤ 0, P(X ≤ x, X < 0.5) FX|X< 0.5 (x) = = 2 P(X ≤ x) = 2x, if x ∈ (0, 0.5), P(X < 0.5) 1, if x ≥ 0.5.
so that the conditional density of X given the event {X < 0.5} equals 0, if x ≤ 0, fX|X< 0.5 (x) = 2, if x ∈ (0, 0.5), 0, if x ≥ 0.5.
Therefore, Z ∞ Z 0.5 EP (X| X < 0.5) = xfX|X< 0.5 (x) dx = 2x dx = 0.25. −∞ 0
Exercise 4 We have 1 −x fX (x) = e λ , ∀ x > 0, λ and thus Z ∞ x P(X > x) = 1 − FX (x) = fX (u) du = e− λ , ∀ x > 0. x Consequently, x−1 P(X>x) − P(X > x, X > 1) P(X>1) = e λ , if x ≥ 1, P(X > x| X > 1) = = P(X > 1) 1, if x < 1.
2 Hence the conditional density equals ( x−1 1 λ e− λ , if x ≥ 1, fX|X>1 (x) = 0, if x < 1,
and thus Z ∞ x − x−1 EP (X| X > 1) = e λ dx = 1 + λ = 1 + EP (X). 1 λ
Exercise 5 Since
Cov(X, Y ) = EP (XY ) − EP (X) EP (Y ) = EP (X 3 ) − EP (X) EP (X 2 ) = 0
since EP (X) = EP (X 3 ) = 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 U 2 − V 2 − EP (U + V ) EP (U ) − EP (V ) = 0.
since EP (U ) = EP (V ) and EP (U 2 ) = 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.
