0
$\begingroup$

I know that if $U\sim\chi^2(k)$ then $aU\sim \Gamma(k/2,2a)$ for $a>0$. But i read about the estimator and its distribution $$\hat{\sigma}_k^2=\frac{1}{2k}\sum_{i=1}^k (X_{2i}-X_{2i-1})^2=\frac{2\sigma^2}{2k}\sum_{i=1}^k \frac{(X_{2i}-X_{2i-1})^2}{2\sigma^2}\sim \frac{\sigma^2}{k}\chi^2(k)$$ where $X_i\sim\mathcal{N}(\mu,\sigma^2)$ iid distributed. Here, it can be shown that $$ \sum_{i=1}^k \frac{(X_{2i}-X_{2i-1})^2}{2\sigma^2}\sim \chi^2(k).$$ Is it correct to write $\frac{\sigma^2}{k}\chi^2(k)$ instead of $aU\sim \Gamma(k/2,2a)$ where $a=\frac{\sigma^2}{k}$. If this is ok when can i use this type notation?

$\endgroup$
4
  • $\begingroup$ Are you sure your expression for $\hat{\sigma}_k^2$ is correct? Don't you need to take the square of the terms in the sum? As it is now, it is a very bad estimator for the population variance $\sigma^2$ since $\hat{\sigma}_k^2\sim N(0,\frac{\sigma^2}{2n})$. $\endgroup$ Aug 24, 2015 at 14:44
  • $\begingroup$ OK, this makes sense now! $\endgroup$ Aug 24, 2015 at 16:13
  • 1
    $\begingroup$ Can you tell the source from where you arrived at this result that constant times chi squared is gamma RV ? $\endgroup$ Oct 7, 2019 at 19:43
  • 1
    $\begingroup$ @Akhil Wikipedia is a good online source of such information. $\endgroup$
    – whuber
    Oct 7, 2019 at 19:57

1 Answer 1

2
$\begingroup$

Everything in the post is correct. The question is really about notation, and I think it comes down to: can $\chi^2(k)$ be used as a notation for both the type of distribution that is intended and as a generic random variable having that distribution. For the normal distribution (and many others), we clearly make a distinction: the distribution is denoted as $N(\mu,\sigma^2)$ and a random variable having that distribution is denoted as $X$. For the chi-squared distribution, both are often written as $\chi^2(k)$ or $\chi^2_k$ which may lead to confusion.

If you are aware of this issue, there is no problem in writing $\frac{\sigma^2}{k}\chi^2(k)$ instead of $aU$. Both have the same distribution. Depending on how you define them they can be the same, $aU=\frac{\sigma^2}{k}\chi^2(k)$, or not.

A remark on the quality of the estimator $\hat{\sigma}_k$: its mean is $\sigma^2$ (so it is unbiased) but its variance is $\frac{2\sigma^4}{k}$. Note that the sample variance $S^2$ of the $2k$ observations is also unbiased but has a smaller variance $\frac{2\sigma^4}{2k-1}$. So unless you have good reasons not to, it is better to use the sample variance as an estimate for $\sigma^2$.

$\endgroup$

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.