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I'm given a certain $f_{X,Y}$ joint density function and I'm asked to find $E(X)$. I know you can find the marginal distribution $f_X$ and then easily compute $E(X)$. However:

Question: I want to know if there's a smart way to go directly from the joint density to the expectation. Is there a specific theorem that could be applied in this case?

Thanks for helping! :D

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    $\begingroup$ $$E(X)=\iint xf_{X,Y}(x,y)dxdy$$ $\endgroup$ – Did Sep 16 '15 at 22:47
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    $\begingroup$ See the first comment on this question. Your $Z$ would just be $g(X,Y)=X$. $\endgroup$ – Ben Whitney Sep 16 '15 at 22:51
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    $\begingroup$ Is there a special theorem? Just the definition of expectation. $$\begin{align} \mathsf E(X) & = \int x f_X(x)\operatorname d x \\ ~ & = \int x \int f_{X,Y}(x,y)\operatorname d y \operatorname d x \\ ~ & = \iint x f_{X,Y}(x,y)\operatorname d y\operatorname d x \end{align}$$ $\endgroup$ – Graham Kemp Sep 16 '15 at 23:07
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$$ E(X)=\int xf_X(x)dx\\=\int x\int f_{X,Y}(x,y)dydx\\=\int\int xf_{X,Y}(x,y)dydx $$

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