Probability Question on Conditional Distributions?

Question

We have been through this problem many times and can't quite get it; any help is appreciated.

Let T be triangle with verts (0,0) (1,0) and (1,2).
Suppose X,Y have joint pdf (5/2)(x^2)*y for x,y in T.

Find E[Y|X], E[Y^2|X] and (we can probably handle) Var[Y|X].

Again, thanks for any help!

Answer 1

The first step is to find the marginal distribution of X

f(x) = ∫ f(x, y) dy =

2x
∫ (5/2 x² y) dy = 5x^4
0

now find the conditional distribution of y given x

f(y | x) = f(x, y) / f(x) = y / (2x²)

E(Y | X) = ∫ y * f(y | x) dy (limits are 0 to 2x)
E(Y | X) = 4 x / 3


E(Y² | X) = ∫ y² * f(y | x) dy (limits are 0 to 2x)
E(Y² | X) = 2x²


Var(Y | X) = E(Y² | X) - E(Y|X)²
= 4x/3 - 2x²

Answer 2

To compute the expectations, you need the conditional pdf f_Y|X(y|x), which is the given joint pdf divided by the marginal pdf f_X(x) for X. You get the latter by integrating the joint pdf with respect to y.

Note that the region over which the joint pdf is nonzero may be described by

0 < x < 1, 0 < y < 2x

(If we were looking for f_X|Y(x|y), we would instead describe it by y/2 < x < 1, 0 < y < 2)

So

f_X(x) = (integral from 0 to 2x) (5/2) x²y dy = 5x^4 for 0 < x < 1
and 0 otherwise

And thus

f_Y|X(y|x) = (5/2)x²y/(5x^4) = y/(2x²), 0 < x < 1, 0 < y < 2x
and 0 otherwise

Then

E[Y|X] = (integral from 0 to 2x) y f_Y|X(y|x) dy

I leave the details to you.