# Thread: Homework help with conditional probability distribution

1. ## Homework help with conditional probability distribution

Hi, I was wondering if I could get some help with this homework problem.

We are given that X~U(0,1) and conditional distribution of Y|X=x is U(x,1). I have to find E(Y|X=x), E(X+Y), and Var(X+Y).

This is what I have so far in working with the problem:

For I assumed that E(Y|X=x)=mX+b (where m and b are constants) so then E(Y)=E(E(Y|X=x))=E(mX+b) since in the condition X=x so then I get E(Y)=mE(X)+b. Is that correct?

Then for E(X+Y), I know that E(X+Y)=E(X)+E(Y)

So then E(X)=1/2, but then I don't know how to find E(Y). Would E(Y|X=x)=E(Y)?

And then for Var(X+Y), I know that Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)

I know:

Var(X)=1/12

Var(Y)=E(Y^2)-E(Y)^2
Var(Y|X=x)=E(Y^2|X=x)-E(Y|X=x)^2
E(Y^2)=E(E(Y^2|X=x)
So then does Var(Y)=E(Var(Y|X=x)+E(Y|X=x)^2)

and then Cov(X,Y)=E(XY)-E(X)E(Y)
E(XY)=E(E(XY|X=x))=E(XE(Y|X=x))
Cov(X,Y)=E(XE(Y|X=x))-(1/2)(E(Y|X=x))

So then I have: Var(X+Y)= (1/12) + [E(Var(Y|X=x)+E(Y|X=x)^2)] + 2[E(XE(Y|X=x))-(1/2)(E(Y|X=x))]

Is that right or am I on the right path?

2. ## Re: Homework help with conditional probability distribution

hi,
I you do not need to make any assumptions about the conditional distribution of Y given x, all the necessary information is already given with U(x,1),

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