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?
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$\begingroup$ 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 $\hat{\sigma}_k^2\sim N(0,\frac{\sigma^2}{2n})$. $\endgroup$– StijnDeVuystAug 24, 2015 at 14:44
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$\begingroup$ OK, this makes sense now! $\endgroup$– StijnDeVuystAug 24, 2015 at 16:13
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1$\begingroup$ Can you tell the source from where you arrived at this result that constant times chi squared is gamma RV ? $\endgroup$– Akhil SinghOct 7, 2019 at 19:43
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1$\begingroup$ @Akhil Wikipedia is a good online source of such information. $\endgroup$– whuber ♦Oct 7, 2019 at 19:57
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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 $\sigma^2$.
