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Lets $X_1,\cdots,X_n$ be simple random sample from $\mathcal{N}(\mu,\sigma)$. $\overline{x}$ is sample mean. Let $$S^2=\begin{cases}\sum_{i=1}^n (x_i-\mu)^2, \mathrm{ where\ } \mu \mathrm{\ is\ known} \\ \sum_{i=1}^n (x_i-\overline{x})^2, \mathrm{ where\ } \mu \mathrm{\ is\ not\ known}\end{cases}$$

$\frac{S^2}{\sigma^2}\sim \chi^2_{\nu}$ where $\nu=n \mathrm{\ or\ } n+1 \mathrm{\ respectively}$. Density of $\chi^2_{\nu}$ is given by $$g_{\nu}(x)=\frac{1}{2^\frac{\nu}{2} \Gamma(\frac{\nu}{2})}x^{\frac{\nu}{2}-1}\mathrm{e}^\frac{x}{2},$$ for $x>0$.

And now I don't understand things I have put into red boxes. Empty red box stands for missing $-1$. The $(\ast\ast)$ stands for saying that $g$ is density function for all $\nu>0$.

Why in 1st red box there is $x$ instead of $\frac{S^\alpha}{\sigma^\alpha}$? Why in 2nd red box there is no $-1$? Why in 3rd red box there is $\Gamma(\frac{\nu}{2})$ and not $\Gamma(\frac{\nu+\alpha}{2})$? From where the 4th red box comes? And 5th?

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    $\begingroup$ The box is too small to see clear $\endgroup$ – Deep North Dec 14 '15 at 23:21
  • $\begingroup$ The original graphic has disappeared from the Web, rendering this question practically incomprehensible. $\endgroup$ – whuber Aug 22 '19 at 13:28
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The derivation seems to be full of typos, though the final answer seems OK.

  1. The first box is correct, since the distribution $g_\nu$ applies $x=S^2/\sigma^2$, but it would have been clearer if the first two expressions would be flipped as $E((S/\sigma)^\alpha) = E((S^2/\sigma^2)^{\alpha/2})$.
  2. In your second box the $-1$ in the exponent of $x$ is indeed missing.
  3. There are lots of problems with the double-starred part. I think it would make more sense if the $=$ sign between boxes 4 and 5 would be replaced with a multiplication sign. Then I see no problem with your box 3, which is just the term from the previous denominator, but the term to the left of it should be $2^{\nu/2}$. The new terms in the numerator just cancel out the new denominator within the integral. And note that the $-1$ you were missing has reappeared! The value of the integral when written in this form is just $1$, so the result follows.
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