Calculating the expectation of a multivariate probability function

Question

I was given the excercise with solution:

And I am not finding the correct value.

My approach

$E[Y_1-Y_2]=\int^1_0\int^1_0(y_1-y_2)3y_1dy_1dy_2$

$=\int^1_0\int^1_0(3y_1^2-3y_1y_2)dy_1dy_2$

$=\int^1_0[y_1^3-\frac{3}{2}y_1^2y_2]^1_0dy_2$

$=\int^1_0[1-\frac{3}{2}y_2]dy_2$

$=[y_2-\frac{3}{4}y^2]^1_0$

$=1-3/4=1/4$

Obviously my approach is wrong, so not only do I need to understand why the solution' s method is correct, but also why my method doesn't work.

Answer 1

Your steps are correct but you started off wrong. Integration of $y_2$ should be from $0$ till $y_1$, i.e. $$E[Y_1-Y_2]=\int^1_0\int^{y_1}_0(y_1-y_2)3y_1dy_2dy_1$$ This is so because $0 < y_2 < y_1$. Hence you get:

$E[Y_1-Y_2]=\int^1_0\int^{y_1}_0(y_1-y_2)3y_1dy_2dy_1$

$=\int^1_0\int^{y_1}_0(3y_1^2-3y_1y_2)dy_2dy_1$

$=\int^1_0[3y_1^2y_2-\frac{3}{2}y_1y_2^2]^{y_1}_0dy_1$

$=\int^1_0[3y_1^3 - \frac{3}{2}y_1^3]dy_1$

$=\frac{3}{2}\int^1_0y_1^3 dy_1$

$=\frac{3}{2} \frac{y_1^4}{4}]_0^1 = \frac{3}{8}$