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The expected value or mean of a random variable $g(X)$, denoted by $Eg(X)$. If $E|g(X)|=\infty$, we say that $Eg(X)$ does not exist. (Casella, G., & Berger, R. L. Statistical inference. Definition 2.2.1)

I am curious why we need to discuss $E|g(X)|$, rather than $Eg(X)$ directly, if expectation does not exist.

Mariana
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  • Because $\mathbb E|g(X)|$ (in contrast with $\mathbb Eg(X)$) is always defined (it can take "value" $+\infty$) so can always be used to inspect whether $\mathbb Eg(X)$ exists. If $\mathbb Eg(X)$ would be the thing in our toolbox (instead of $\mathbb E|g(X)|$) then - if it does not exist - our toolbox is emtpy, so we cannot check. – drhab May 05 '21 at 13:56

2 Answers2

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Define $X^+=\max \{ X,0 \},X^-=\max \{ -X,0 \}$. There are really three cases:

  • $E[X^+]$ and $E[X^-]$ are both finite, this is the normal situation. In this case you can just define $E[X]=E[X^+]-E[X^-]$ and there's no problem.
  • Exactly one of $E[X^+]$ and $E[X^-]$ is infinite. In this situation it makes sense to interpret $E[X]=E[X^+]-E[X^-]=\pm \infty$. However many theorems actually need finite expectation anyway.
  • Both $E[X^+]$ and $E[X^-]$ are infinite. In this situation it turns out that there is no real sense in defining the expectation. You might intuitively expect that some symmetric truncation procedure along the lines of the Cauchy principal value would help...but it really doesn't. See Why do we ask for *absolute* convergence of a series to define the mean of a discrete random variable? for more.
Ian
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Usually, one first defines the expectation of a non-negative random variable $X$ by considering simple variables and taking a supremum. Now, any r.v. $X$ can be written as $X^+ - X^-$ for non-negative r.v.s $X^+$ and $X^-$. Then, one would like to define $$\Bbb E[X] := \Bbb E[X^+] - \Bbb E[X^-].$$ For things to go nicely, we restrict ourselves to the case where both the expectations on the right are finite. This is equivalent to wanting the expectation of $|X| = X^+ + X^-$ being finite.

Aryaman Maithani
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