# Sample variance of a random sample from a normal distribution with mean and variance

I know that if the sample variance of a random sample from a normal distribution $$(\mu,\sigma^2)$$ is

$$S_1^2 = \frac{1}{n-1}\sum{(X_i-\bar{X})}^2$$

then,

$$U =\frac{(n-1)S_1^2}{\sigma^2}$$ has a $$\chi^2$$ distribution with $$n-1$$ degrees of freedom.

Does this mean that if my sample variance of a random sample from a normal distribution $$(\mu,\sigma^2)$$ is

$$S_2^2=\frac{1}{n}\sum{(X_i-\bar{X})}^2$$

then,

$$V = \dfrac{nS_2^2}{\sigma^2}$$, has a $$\chi^2$$ distribution with $$n$$ degrees of freedom?

• If you had asked about $\dfrac{n S_1^2}{\sigma^2}$ then it would have a $\chi^2$ distribution with $n-1$ degrees of freedom, as it would be equal to your original $U$. If you had defined $S_2^2= \dfrac1n \sum (X-\mu)^2$ then $\dfrac{n S_2^2}{\sigma^2}$ would have a $\chi^2$ distribution with $n$ degrees of freedom Commented Feb 20, 2023 at 17:11
• Thank you, I did made a mistake on that.
– user1141374
Commented Feb 21, 2023 at 0:23

No. Not quite. In general, the sample variance $$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2,$$ given $$X_i\overset{\mathrm{iid}}{\sim}\mathcal N(\mu,\sigma^2)$$, is distributed according to $$S^2\sim\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1})$$ (shape-scale parameterization). Through a slight abuse of notation we write $$S^2\sim \operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1})$$ $$\tfrac{n}{\sigma^2}S^2\sim \tfrac{n}{\sigma^2}\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1})$$ $$\tag{1} \tfrac{n}{\sigma^2}S^2\sim\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2n}{n-1}).$$ The right hand side of $$(1)$$ cannot be written as a simple $$\chi^2$$-distribution as it is equivalent to $$\tfrac{n}{\sigma^2}S^2\sim\tfrac{n}{n-1}\underbrace{\operatorname{Gamma}(\tfrac{n-1}{2},2)}_{\chi^2(n-1)},$$ which is to say $$\tfrac{n}{\sigma^2}S^2$$ has the same distribution as a $$\chi^2(n-1)$$ random variable multiplied by $$n/(n-1)$$.
• Does this mean that $\tfrac{n}{\sigma^2}S^2$ has a ${\chi^2}$-distribution with n-1 degrees of freedom?
• No, the last line means $\tfrac{n}{\sigma^2}S^2$ has the same distribution as a $\chi^2(n-1)$ random variable multiplied by $n/(n-1)$. Commented Feb 20, 2023 at 13:55
• @cronky $\tfrac{n}{\sigma^2}S_1^2$ - note the subscript - would be equal to $\tfrac{n-1}{\sigma^2}S^2$ and so would have a $\chi^2$ -distribution with $n-1$ degrees of freedom Commented Feb 20, 2023 at 17:14