# Help to understand this. Expected value of $S^\alpha$ in Gaussian distribution [closed]

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? • The box is too small to see clear – Deep North Dec 14 '15 at 23:21
• The original graphic has disappeared from the Web, rendering this question practically incomprehensible. – whuber Aug 22 '19 at 13:28

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.