# Probability- expected value of Z, where z= X/(1+y)^2

1. Sep 15, 2011

### Roni1985

1. The problem statement, all variables and given/known data

X & Y are independent r.v.s with uniform distribution between 0 and 1.

Z= X/(1+Y)^2

find E[Z].
2. Relevant equations

3. The attempt at a solution

Here is what I did.

E[Z]= E[X]*E[1/(1+Y)^2]
E[X]=1/2
E[1/(1+Y)^2]=?
I think that once I know the distribution of (1+Y)^(-2), I'll be able to find the answer. Is it 1/(1+Y)^2~ U(1,2) ?

2. Sep 15, 2011

### micromass

Staff Emeritus
You can also use this theorem:

$$E[g(X,Y)]=\iint{g(x,y)f_{X,Y}(x,y)dxdy}$$

So, in your case, this comes down to

$$\int_0^1\int_0^1\frac{x}{1+y^2}dxdy$$

3. Sep 15, 2011

### Roni1985

Hey,

so adding a constant to a uniform r.v. doesn't change it's distribution?

4. Sep 15, 2011

### micromass

Staff Emeritus
Yes

Obviously it does change the distribution. But I don't see how that is important here.

5. Sep 15, 2011

### Ray Vickson

If W = 1/(1+Y)^2, then for any w > 0 we have P{W <= w} = P{1/(1+Y)^2 <= w} = P{-sqrt(w) <= 1/(1+Y) <= sqrt(w)}. The left-hand inequality -sqrt(w) <= 1/(1+Y) holds automatically because Y >= 0, so P{W <= w} = P{1/(1+Y) <= sqrt(w)} = P{Y >= -1 + 1/sqrt(w)}. Since Y ~ U(0,1), we can easily get P{W <= w} and hence can get the density function of W. Alternatively: just apply the standard formula for the density of a transformed random variable.

RGV

6. Sep 15, 2011

### Roni1985

Well I was thinking that now its uniformly distributed from 1 to 2... isn't it?

EDIT:
Oh right its still uniform from 0 to 1...
omg... I'm so rusty :\

Thanks.

Last edited: Sep 15, 2011
7. Sep 15, 2011

### Ray Vickson

You have all the information you need to work out the answer for yourself.

RGV