Question

Suppose (X,Y ) is chosen according to the continuous uniform distribution on the triangle with vertices (0,0), (0,1) and (2,0), that is, the joint pdf of (X,Y ) is fX,Y (x,y) =c, for 0 ≤ x ≤ 2,0 ≤ y ≤ 1, 1/ 2x + y ≤ 1,    0 , else.

(a) Find the value of c.

(b) Calculate the pdf, the mean and variance of X.

(c) Calculate the pdf and the mean of Y .

(d) Calculate the conditional mean and variance of X given Y .

(e) Derive E[X] using E[X] = E[E(X|Y )] and compare the answer in (b). Comment on your findings

. (f) Compute V[X] = E[V (X|Y )]+V[E(X|Y )] and compare the answer in (b). Comment on your findings.

Answer 1

(X,Y ) is chosen according to the continuous uniform distribution on the triangle with vertices (0,0), (0,1) and (2,0).

The joint PDF is $f_{X,Y}\left ( x,y \right )=c;0\leqslant x\leqslant 2,y<1-\frac{x}{2}$

a) The condition for PDF is

2 1-x/2 fx,y(x,y) dydx = 1 o Jo 2 1-x/2 Jo Jo r 2 c | (1-x/2) dx = 1 0 c (x-x74]:-1 0

b) The marginal PDF of   is

1-x/2 x,y (X 1-x/2 ду fy(x) = 0

The mean is

| x(1-x/2) dx E(X) = 0 2 E(X) = ริ

The variance is found as

E(x2)-x*(1 - x/2)dx 0 2 E(X-) Var(X) = E(X2)-E(X)2 Var(X) = 3-9 Var(X) 2/9

c) The marginal PDF of is

2-2y 0 2-2y dx fy(y) = 0

The mean is

E(Y)y (2 -2y) dy E(Y) = (y2-2y3/3] E(Y) = 1,3 0

d) The conditional distribution,

fy (y) 2 - 2y

The conditional expectation is

E(X|Y) = 2

2-2y 0 4

The conditional variance is

Var(X Y) = E(X2 Y)-E(X Y)2 4 = (1-y)2-(1-y)2 Var(XY) Var(XY) 2