suppose
$f(y_{1}, y_{2}) = \begin{cases} 1 & 0\leq y_{1}\leq 2, \, 0 \leq y_{2} \leq 1, \, 2y_{2} \leq y_{1} \\ 0 & otherwise \end{cases}$
The goal is to find $Var(y_{1} - y_{2})$
Without using Covariance, I have decided to use the property of expectation, namely, $ Var\,(y_{1} - y_{2}) = E\,[(y_{1} - y_{2})^2] - (E\,[y_{1} - y_{2}])^2$
$E\,[y_{1} - y_{2}]= \int_{0}^{2}\int_{0}^{1} (y_{1}-y_{2})* 1 \,\,dy_{2}\,dy_{1} = 1$
$E\,[y_{1} - y_{2}]= \int_{0}^{2}\int_{0}^{1} (y_{1}-y_{2})^2* 1 \,\,dy_{2}\,dy_{1} = \frac{4}{3}$ but my calculation is not correct.
The correct one should be:
$E\,[y_{1} - y_{2}]= \int_{0}^{2}\int_{0}^{\frac{y_{1}}{2}} (y_{1}-y_{2})^2* 1 \,\,dy_{2}\,dy_{1} = \frac{7}{6}$
Why do the bounds differ from E[A] to E[A^2]? Could anyone provide me with some explanation, please? Many thanks
$y_2$ is bounded as well at $E(A)$:
$$ \int_{0}^{2}\int_{0}^{\frac{y_1}{2}} (y_1-y_2)\cdot 1 \ dy_2 \ dy_1 =1$$
So in both cases the upper bound for $y_2$ is $\frac{y_1}{2}$.