1
$\begingroup$

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 $X^2$?

$\endgroup$
6
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.