Let's say $X \sim D$, ie. $X$ is a random variable following some distribution $D$.

Then the first moment of $X$ is defined as $E[X] = \int x\hspace{1mm}f(x)\hspace{2mm}dx$

And the first moment of $X$ is defined as $E[X^2] = \int x^2\hspace{1mm}f(x)\hspace{2mm}dx$

Looking at this, I'm interpreting the notation as "there exists a distribution $X^2$, we don't know what its cdf/pdf is... but it just so happens that we can calculate its expectation by that integral".

That is, let $S = X^2$, then $E[S] = \int s\hspace{1mm}f(s)\hspace{2mm}ds = \int x^2\hspace{1mm}f(x)\hspace{2mm}dx$

If I understood this correctly, then what exactly is the distribution of $X^2$?

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    $\begingroup$ If $X$ is a random variable, then you can always consider the distribution of the random variable $S=X^2$. Whether it has finite moments is a different question and I'm not sure it's the one you're asking here. $\endgroup$
    – Chris Haug
    May 2, 2017 at 1:05
  • $\begingroup$ Chris is right. It is the moments that need not exist. The Cauchy is the classic case. All I can say in general is that S$*2$ is non-negative. I don't think anything else can be done without some restriction to a family of distributions. $\endgroup$ May 2, 2017 at 1:16
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    $\begingroup$ @esjd The square of a standard normal is indeed $\chi^2$, and its mean is $1$, not $2$. $\endgroup$
    – Glen_b
    May 2, 2017 at 5:16
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    $\begingroup$ My answer was mostly wrong, so I just deleted it. But see this question stats.stackexchange.com/questions/192807/… for the details of the square of a standard normal r.v. $\endgroup$
    – Flounderer
    May 2, 2017 at 6:12
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    $\begingroup$ @Flounderer gave you a good reference. The result there, suitably written, is fully general. To see that, let $F_X$ be the distribution function of $X$, defined as $F_X(x)=\Pr(x\le X)$. Then the distribution of $X^2$ by definition is $$F_{X^2}(t)=\Pr(X^2\le t)=\Pr(|X|\le\sqrt{t})=F_X(\sqrt{t})-F_X(-\sqrt{t})$$for any $t\ge 0$. (Of course $F_{X^2}(t)=0$ for $t\lt 0$ since squares are positive.) It's that simple. $\endgroup$
    – whuber
    May 2, 2017 at 20:51


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