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How would you go about calculating the distribution of the sample standard deviation $T(X_1,...,X_n)=S$ where

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

Here $X_1,...,X_n$ are iid and distributed by $N(\mu, \mu^2)$.

Is there a way to do this using standard results about the sum and the transformation formula?

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Note that you can factor out the $\mu$ term from both the variance and the mean, you can represent $X_i - \bar{X}$ as $\mu(Z_i -\bar{Z})$ where $Z_i \sim N(0,1)$ and $\bar{Z} \sim N(0,\frac{1}{n})$.

(Dividing out the $\mu$ term still leaves both distributions nonstandard with mean 1, but because it's $X_i - \bar{X}$ you can add $0 = -1 + 1$ to remove the unit mean.)

Plugging this into the previous formula gives you:

\begin{equation} \begin{split} S^2 & = \frac{1}{n-1}\sum_i (X_i - \bar{X})^2\\ & = \frac{\mu^2}{n-1} \sum_i (Z_i - \bar{Z})^2 \end{split} \end{equation}

By Cochran's theorem ($\sum_i(Z_i - \bar{Z})^2 \sim \sigma_Z^2\chi^2_{n-1} \overset{D}{=} \chi^2_{n-1}$) we have:

\begin{equation} \frac{n-1}{\mu^2}S^2 \sim \chi^2_{n-1} \end{equation}

Where the $\frac{n-1}{\mu^2}$ factor is to correct for the scalar multiple in front of the Chi-square distribution. Re-arranging this as:

\begin{equation} S^2 \sim \frac{\mu^2}{n-1}\chi^2_{n-1} \end{equation}

We can use the identity that if $Y \sim \chi^2_v$ and $U = cY$ then $U \sim \textrm{Gamma}(\frac{v}{2},2c)$. This gives us that:

\begin{equation} S^2 \sim \textrm{Gamma}(\frac{n-1}{2},\frac{2\mu^2}{n-1}) \end{equation}

Edit Cochran's theorem is taught in a lot of earlier courses, they just might not call it that, or only teach a special case. I'm not sure how to do it without the theorem other than a lot of algebra.

If you want to pivot with the standard deviation, you can scale $S^2$ by $\frac{n-1}{\mu^2}$ to remove the unknown parameter and go back to the earlier parameterization. Going back to what we had before:

\begin{equation} \frac{S^2(n-1)}{\mu^2} \sim \chi^2_{n-1} \end{equation}

Then using the continuous mapping theorem to take the square root of both sides:

\begin{equation} \frac{S\sqrt{n-1}}{\mu} \sim \chi_{n-1} \end{equation}

Where $\chi_{n-1}$ is the Chi distribution with $n-1$ degrees of freedom. I believe this can be used to get confidence intervals for $s$:

\begin{equation} \begin{split} CI(\alpha) & = [\chi_{{n-1}_{\frac{\alpha}{2}}}< \frac{S\sqrt{n-1}}{\mu} < \chi_{{n-1}_{1-\frac{\alpha}{2}}}]\\ & = [\frac{1}{S\sqrt{n-1}}\chi_{{n-1}_{1-\frac{\alpha}{2}}}< \mu < \frac{1}{S\sqrt{n-1}}\chi_{{n-1}_{\frac{\alpha}{2}}}] \end{split} \end{equation}

It seems though that this confidence interval does pretty poorly for small $n$ because the sample standard deviation is biased. I'm not sure off the top of my head how to fix that.

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  • $\begingroup$ Thank you for this answer. Is there a way to do it without Cochran's theorem, as this isn't covered in the scope of my course? Essentially what I am trying to do, is create a confidence interval for the pivot $T/\mu$. I was under the impression that to do this I need to work out what the distribution of $T$ is first. Is this correct? Also, I was looking for the distribution of $S$ instead of $S^2$ $\endgroup$ Commented Dec 3, 2017 at 23:08
  • $\begingroup$ I edited the answer to take into account your comment. $\endgroup$ Commented Dec 3, 2017 at 23:38

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