1. ExpectedValue

Can anyone tell me how to calculate the expected value of A^2 when A=x*y and of B^2 when B=2x+2y.

2. I assume that X and Y are rvs?

Then $E(A^2)=E(X^2Y^2)$

and if they are continuous, then.....

$E(A^2)=E(X^2Y^2)=\int\int x^2y^2f(x,y)dxdy$

If they are indep....

$E(A^2)=E(X^2)E(Y^2)=\int x^2f_X(x)dx\int y^2f_Y(y)dy$

Likewise $E(B^2)=4E[(X+Y)^2]=4\int\int (x+y)^2f(x,y)dxdy$

3. Originally Posted by matheagle
I assume that X and Y are rvs?

Then $E(A^2)=E(X^2Y^2)$

and if they are continuous, then.....

$E(A^2)=E(X^2Y^2)=\int\int x^2y^2f(x,y)dxdy$

If they are indep....

$E(A^2)=E(X^2)E(Y^2)=\int x^2f_X(x)dx\int y^2f_Y(y)dy$

Likewise $E(B^2)=4E[(X+Y)^2]=4\int\int (x+y)^2f(x,y)dxdy$
Yes, they are both random v., independent and uniform on [0, 1].

4. It does make a slight difference knowing the distribution........

$E(A^2)=E(X^2)E(Y^2)=\int_0^1 x^2dx\int_0^1 y^2dy={1\over 9}$

and $E(B^2)=4E[(X+Y)^2]=4\int_0^1\int_0^1 (x+y)^2dxdy$

or $4E[(X+Y)^2]=4\left[E(X^2)+2E(X)E(Y)+E(Y^2)\right]$

$=4\left[ {1\over 3} +2\left({1\over 2}\right)\left( {1\over 2}\right) +{1\over 3} \right]$

5. matheagle, I appreciate your help.

Thank you