# Why is the sampling distribution of variance a chi-squared distribution?

The statement

The sampling distribution of the sample variance is a chi-squared distribution with degree of freedom equals to $n-1$, where $n$ is the sample size (given that the random variable of interest is normally distributed).

Source

My intuition

It kinda makes intuitive sense to me 1) because a chi-square test looks like a sum of square and 2) because a Chi-squared distribution is just a sum of squared normal distribution. But still, I don't have a good understanding of it.

Question

Is the statement true? Why?

• The initial statement is false in general (it's false for two separate reasons). What is your source (your link is missing), and what does it actually say? Oct 27, 2014 at 23:31
• My question also comes to reaction to a question-answer in a introductory stats class for which the access is protected. The question is "What distribution is the sampling distribution of variance in wing length in flies?" and the answer is "Chi-squared distribution" Oct 27, 2014 at 23:42
• The quoted statement in your first comment is still false in general. The comment at the end of the source is true (with the necessary assumptions): "when samples of size n are taken from a normal distribution with variance $\sigma^2$, the sampling distribution of the $(n-1)s^2/\sigma^2$ has a chi-square distribution with n-1 degrees of freedom." ... The answer to the question in your second comment will also be false -- unless, I suppose, someone has shown that wing length is normally distributed. (What basis could there be for asserting this to be true?) Oct 28, 2014 at 0:07
• So let's assume the wings are normally distributed, then the sampling distribution of $(n-1)s^2/\sigma^2$ would be chi-squared distributed. Why is it so? Oct 28, 2014 at 0:27
• Are you aware that a sum of squares of $k$ iid N(0,1) random variables is chi-squared with $k$ d.f.? Or is that the part you seek proof of? Oct 28, 2014 at 1:05

[I'll assume from the discussion in your question that you're happy to accept as fact that if $$Z_i, i=1,2,\ldots,k$$ are independent identically distributed $$N(0,1)$$ random variables then $$\sum_{i=1}^{k}Z_i^2\sim \chi^2_k$$.]

Formally, the result you need follows from Cochran's theorem. (Though it can be shown in other ways)

Less formally, consider that if we knew the population mean, and estimated the variance about it (rather than about the sample mean): $$s_0^2 = \frac{1}{n} \sum_{i=1}^{n}(X_i-\mu)^2$$, then $$s_0^2/\sigma^2 = \frac{1}{n} \sum_{i=1}^{n}\left(\frac{X_i-\mu}{\sigma}\right)^2=\frac{1}{n} \sum_{i=1}^{n}Z_i^2$$, ($$Z_i=(X_i-\mu)/\sigma$$) which will be $$\frac{1}{n}$$ times a $$\chi^2_n$$ random variable.

The fact that the sample mean is used, instead of the population mean ($$Z_i^*=(X_i-\bar{X})/\sigma$$) makes the sum of squares of deviations smaller, but in just such a way that $$\sum_{i=1}^{n}(Z_i^*)^2\,\sim\chi^2_{n-1}$$ (about which, see Cochran's theorem). That is, rather than $$ns_0^2/\sigma^2\sim \chi^2_n$$ we now have $$(n-1)s^2/\sigma^2\sim\chi^2_{n-1}$$.

• @Glen_b Can you give a reference for other proofs on this fact? I really want to know it. Sep 17, 2015 at 1:09
• Which of several facts are you after proof of? Sep 17, 2015 at 2:44
• @Glen_b The only two methods besides Cochran-Madow Theorem of proving this fact that the sample variance and the sample mean is statistiacally independent with a chi-square distribution are:(1)Scheffe's canonical basis(Scheffe,1959)(2)Cumulant methods(Or mgfs, which is equivalent to it). If you know more methods, I really want to know them. Nov 1, 2015 at 12:27
• One more comment I want to add is that altough sample mean is used, but sometimes we want a fixed power independent of the fixed variance, this method is replaced by Stein's two-stage method(1949). Nov 1, 2015 at 12:30
• $\bar{X}$ is independent of $X_i-\bar{X}$ in this circumstance Nov 12, 2017 at 20:25