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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 $X^2$?

foobar
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    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. – Chris Haug May 02 '17 at 01:05
  • 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. – Michael R. Chernick May 02 '17 at 01:16
  • Ok, so let's say the distribution of X is Normal, then what is the distribution of X^2 (its second moment)? From the notation, it looks like it would be Chi-Squared - but it can't be, since if it were Chi-Squared, the mean would be 2, but clearly the expected value doesn't have to be 2. – foobar May 02 '17 at 01:39
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    @esjd The square of a *standard* normal is indeed $\chi^2$, and its mean is $1$, not $2$. – Glen_b May 02 '17 at 05:16
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    My answer was mostly wrong, so I just deleted it. But see this question https://stats.stackexchange.com/questions/192807/pdf-of-the-square-of-a-standard-normal-random-variable for the details of the square of a standard normal r.v. – Flounderer May 02 '17 at 06:12
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    @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. – whuber May 02 '17 at 20:51

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