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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?

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    $\begingroup$ 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 $\endgroup$
    – Henry
    Commented Feb 20, 2023 at 17:11
  • $\begingroup$ Thank you, I did made a mistake on that. $\endgroup$
    – user1141374
    Commented Feb 21, 2023 at 0:23

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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 $n/(n-1)$.

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  • $\begingroup$ Does this mean that $\tfrac{n}{\sigma^2}S^2$ has a ${\chi^2}$-distribution with n-1 degrees of freedom? $\endgroup$
    – user1141374
    Commented Feb 20, 2023 at 13:53
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    $\begingroup$ 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 $n/(n-1)$. $\endgroup$ Commented Feb 20, 2023 at 13:55
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    $\begingroup$ I see, thank you so much. $\endgroup$
    – user1141374
    Commented Feb 20, 2023 at 13:56
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    $\begingroup$ @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 $\endgroup$
    – Henry
    Commented Feb 20, 2023 at 17:14

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