6
$\begingroup$

Let $X_i$ be independent, normally distributed random variables, for $1\leq i\leq N$. What is the distribution of $Y_m=\frac1 N \sum_{i=1}^N X_i^m$?

Every high school student knows part of the answer. The mean of $Y^m$ is the $m$'th moment of the normal distribution, and the variance of $Y^1$ is $1/N$. I'm interested in the width of the distribution of $Y^m$. How does its variance scale with the sample size $N$? (Is variance a useful measure?) I'm especially interested in the large sample limit, with m a small integer. Knowing where to look it up would be helpful.

$\endgroup$
6
  • 1
    $\begingroup$ Did you consider using Central Limit Theorem? $X_i^m$ will be iid., so $\sqrt{N}(Y^{(m)}-EX_1^m)\to N(0,var(X_1^m))$. You can find $EX_1^m$ and $Var(X_1^m)=EX_1^{2m}-(EX_1^m)^2$ by differencing the characteristic function of $X_1$ and evaluating it at zero, which for normal variables is not that hard to do. As a bonus, all the higher moments will be expressed as the functions of the first and second moment. $\endgroup$ – mpiktas Oct 31 '11 at 13:01
  • 1
    $\begingroup$ Why is the variance of Y $1/(n-1)$ ? Is X standard normal? And shouldn't it be $n$ in the denominator? $\endgroup$ – JohnRos Oct 31 '11 at 17:04
  • $\begingroup$ Rodney, to post a comment you need to log in with the same id used to create the post. $\endgroup$ – whuber Nov 1 '11 at 1:10
  • $\begingroup$ John: you're right. mpiktas: I didn't, but I should have. Thank you. $\endgroup$ – Rodney Polkinghorne Nov 1 '11 at 6:47
  • $\begingroup$ @mpiktas You write "As a bonus, all the higher moments will be expressed as the functions of the first and second moment" Does "bonus" mean that it is possible to derive formulas for the higher moments that do not involve the first or second moments, while, in contrast, the characteristic function approach has the advantage of producing formulas that express the higher moments in terms of the first and second moments? $\endgroup$ – Dilip Sarwate Nov 1 '11 at 11:05
6
$\begingroup$

I'm interested in the width of the distribution of $Y_m$. How does its variance scale with the sample size $N$?

$\text{var}(X_i^m) = E[X_i^{2m}] - (E[X_i^m])^2$ is easily evaluated from the moment-generating function $\exp(\sigma^2t^2/2 + \mu t)$ of the $N(\mu, \sigma^2)$ random variable $X_i$ (or from the characteristic function as suggested in the comment by mpiktas). Since this variance does not depend on $i$, let us denote it by $\text{var}(X^m)$. Then, since the $X_i^m$'s are independent random variables (they are functions of independent random variables), we have $$\text{var}(Y_m) = \text{var}\left(\frac{1}{N}\sum_{i=1}^N X_i^m\right) = \frac{1}{N^2}\left(\sum_{i=1}^N \text{var}(X_i^m)\right) = \frac{1}{N}\text{var}(X^m)$$ Note that the displayed equation above does not require that the $X_i$ be normal random variables. For i.i.d. random variables $Z_i$ (with finite second moment), the variance of $N^{-1}\sum_i Z_i$, the average of $N$ variables, is always $N^{-1}\text{var}(Z)$, that is the variance always scales as $1/N$. For general distributions, the Chebyshev inequality can be used to obtain a weak bound on the width of the distribution. See here for a related discussion.

$\endgroup$
1
  • $\begingroup$ Recently, the TeX formatting in many posts seems to have gone a bit weird. I tried to edit this post to make $\text into $ \text which would apparently have fixed the problem but singe-space edits are not allowed. - at least I'm assuming that this isn't just an error on my end but happens everywhere. $\endgroup$ – kram1032 Apr 15 '14 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.