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The output of a process is $XY^3$, where $X$ and $Y$ are mutually dependent. $Y$ is $\sim U(0,2)$ and if $Y=y$, $X$ is exponential with mean $y$. What is the expected output of the process?

I realize that the $E[XY^3]$ is the $\int_{-\infty}^\infty \int_{-\infty}^\infty XY^3 f(x,y) \, dx \, dy$, but I'm not sure what my joint distribution function $P[XY^3]$ is?

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    $\operatorname{E}(XY^3)$ is not $\displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty XY^3 f(x,y)\,dx\,dy;$ rather, it is $\displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty xy^3 f(x,y)\,dx\,dy.$ Capital $X$ is the random variable; lower-case $x$ is what goes from $-\infty$ to $\infty$ in this integral. Without being careful about which is which, how would you understand something like $\Pr(X\le x)\text{?} \qquad$ – 2017-01-20

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Given the event $Y=y$, you have $$ \operatorname{E}(XY^3 \mid Y=y) = \operatorname{E}(Xy^3\mid Y=y) = y^3 \operatorname{E}(X\mid Y=y) = y^3 \cdot y = y^4. $$ So then you need the expected value of $Y^4$: $$ \operatorname{E}(Y^4) = \int_0^2 y^4 \left( \frac 1 2 \, dy \right) = \cdots. $$

This is an instance of the law of total expectation: $$ \operatorname{E}(XY^3) = \operatorname{E}(\operatorname{E}(XY^3\mid Y)) = \operatorname{E}(Y^4). $$