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So I'm trying to solve this problem.

Random vector $(X,Y)$ has the joint density function given by $f(x,y)=6 x y^2$, when $0<x<1$, $0<y<1$, and zero otherwise.

What is $E(7Y)$? Answer with at least one decimal place.

I know that $$E[G(X,Y)]=\int\int{g(x,y)f(x,y)}dydx, $$ so I thought that I could do like follows; $E[7Y]=\int\int{7y \cdot 6 x y^2}dydx=\frac{21}{4}$, but it is wrong. Any help and insight would be appreciated!

mrrobot
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    It is not wrong. $\mathsf E(7Y) = \int_0^1\int_0^1 7y\cdot 6xy^2,\mathrm d y,\mathrm d x = 21/4$. Which equals $5.25$ to at least one decimal. – Graham Kemp Oct 15 '21 at 11:14
  • So my solution is right? I guess it's a misprint then. Thank you! – mrrobot Oct 15 '21 at 11:22

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