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Define $Z = \min(X, Y)$ and the joint pdf of $X$ and $Y$ as $f_{XY}(x,y)$.

I saw an approach that said

$$ E[Z] = \int \int \min(x,y) f_{XY}(x,y) \,dy\,dx $$

Is this readily obvious, or do you need to convert the following: $$ E[Z] = \int \min(x,y)f_Z(z) \,dz $$

to the above?

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  • $\begingroup$ Obvious. For any function $g(x,y)$, the expected value of $g$ is $\int \int g f_{XY}dxdy$$. $\endgroup$
    – YJT
    Apr 19, 2021 at 13:32
  • $\begingroup$ @YJT Is this the law of unconscious statistician? If so, I've only seen that for the one variable case, but it seems it applies to any number of variables? $\endgroup$
    – David
    Apr 19, 2021 at 13:34
  • $\begingroup$ @David Yes. Consider the (generically 2-to-1) change of variables $(Z,W)=(\min(X,Y),\max(X,Y))$. $\endgroup$
    – user10354138
    Apr 19, 2021 at 13:42
  • $\begingroup$ Yes.it applies to more than one variable. $\endgroup$
    – YJT
    Apr 19, 2021 at 13:42

2 Answers 2

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You do not need to do that. In general you have $$ \operatorname E(g(W)) = \int_{\mathcal W} u\cdot f_{g(W)}(u)\, du = \int_{\mathcal X} g(x)f_W(x)\, dx. $$ Using the second integral is often simpler because you are given $f_W$ and there is no need to find $f_{g(W)}.$

Some people call this the "law of the unconscious statistician," and if you google that term you will find it.

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The two definitions are the same, but I think that in a real problem you have to use the second one....

For example, let $X,Y$ be i.i.d. rv's following a $Exp(1)$ distribution.

to find

$$E(Z)=\int_0^{\infty}2t e^{-2t}dt=\frac{1}{2}\int_0^{\infty}w e^{-w}dw=\frac{1}{2}$$

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  • $\begingroup$ This is quite incorrect. There is not in fact a need to find the density function of $Z. \qquad$ $\endgroup$
    – Michael Hardy
    Apr 19, 2021 at 15:39

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