Show that $E(S)=\sqrt{\frac{1}{n-1}}\frac{\Gamma(n/2)}{\Gamma[(n-1)/2]}\sigma$

Question

Let $X_1,...,X_n$ be a random sample of size $n$ from the normal distribution with mean $\mu$ and variance $\sigma^2$ and let $S^2=\frac{1}{n-1}\sum^n_{i=1}(X_i-\bar{X})^2$ be the sample variance. Show that $$E(S)=\sqrt{\frac{1}{n-1}}\frac{\Gamma(n/2)}{\Gamma[(n-1)/2]}\sigma$$

Edited: Below is my trial $$X_i~N(\mu,\sigma^2)$$

Let $Y_i=(X_i-\bar{X})^2$ $$Y_i=X_i-2X_i E(X_i )+[E(X_i )]^2$$ $$Y_i=X_i^2-2X_i μ+μ^2$$ $$∑_{i=1}^n(X_i^2-2X_i μ+μ^2)=∑_{i=1}^n(X_i^2)-2\mu∑_{i=1}^n(X_i)+n\mu^2$$

Then I not able to continue, because it seem can't simplify further. Any hint? Because I don't think my steps have the right approach.

Answer 1

To get you started, define a random variable $X$ by

$$X = \frac{(n-1)S^{2}}{\sigma^{2}} \sim \chi_{n-1}^{2}$$

Then

$$S = \sqrt{\frac{\sigma^{2}X}{n-1}}$$

and

$$\mathbb{E}[S] = \mathbb{E}\left[\sqrt{\frac{\sigma^{2}X}{n-1}}\right]= \sqrt{\frac{\sigma^{2}}{n-1}}\mathbb{E}[\sqrt{X}]$$