Following are the results obtained for two independent random variable x and y. $E(x)=4,E(y)=6,v(x)=5$ and $V(y)=4$, so find the value of
$1)\ $ $E(2x-y)^2$
$2)\ $ $E(7-2x-5y)$
That's the way I am trying :
$$V(x)=E(x^2)-E(x)^2 $$
$$5=E(x^2)-16$$ $$E(x^2)=21$$ In the same way $$E(y^2)=40$$ $$E(xy)=6\times 4=24$$
\begin{align}E(2x-y)^2 &=4E(x^2)-4E(xy)+E(y^2) \\ &=4(21)-4(24)+40\\ &=84-96+40\\ &=28 \end{align} But actual answer is $60$ which doesn't match.
For the first part,\begin{align} E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \\ &= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2 \end{align}