Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$ given a random sample from a normal distribution ...

Question

Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.

GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.

I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.

I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$

Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom? Could anyone please tell me which one is the correct answer with explanation?

Thank you

Answer 1

With $x$ degrees of freedom, the chi-square distribution is $$ f(v)\,dv = \left.\left(\frac v 2\right)^{x/2 - 1} e^{-v/2}\,\frac{dv}{2}\right/\Gamma\left(\frac x 2\right) $$ \begin{align} \int_0^\infty v^{1/2} f(v)\,dv & = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \left.\left(\frac v 2\right)^{(n-1)/2 - 1} e^{-v/2}\,\frac{dv}{2}\right/\Gamma\left(\frac {n-1} 2\right) \\[10pt] & = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty 2^{1/2} \left.\left( \frac v 2 \right)^{(n-2)/2} e^{-v/2} \frac{dv}{2} \right/\Gamma\left(\frac{n-1}{2}\right) \\[10pt] & = \sigma\sqrt{\frac{2}{n-1}} \left. \int_0^\infty u^{n/2-1} e^{-u}\,du \right/\Gamma\left(\frac{n-1}{2}\right) \\[10pt] & = \sigma\sqrt{\frac{2}{n-1}} \cdot \frac{\Gamma\left(\frac n 2\right)}{\Gamma\left(\frac{n-1}{2}\right)}. \end{align}

It remains to show that the factor by which $\sigma$ is multiplied is not $1$. (Maybe I'll say more about that later.)

Answer 2

You might think about this general situation Let $X,Y$ be two positive random variables such that $\mathbb{E}(X^2)=\mathbb{E}(Y^2)$. Then, by Jensen inequality, $$ \mathbb{E}(Y)\leq \sqrt{\mathbb{E}(Y^2)}=\sqrt{\mathbb{E}(X^2)} $$ with equality if and only if $Y$ is deterministic. Thus, if $Y$ isn't deterministic, then $$ \boxed{\mathbb{E}(Y^2)=\mathbb{E}(X^2)\Rightarrow \mathbb{E}(Y) < \sqrt{\mathbb{E}(X^2)}.} $$

Now in your case, take $X=Z-\mathbb{E}(Z)$ with $Z\sim {\cal N}(\mu,\sigma)$ and take $Y$ equal to $S$ (which is not deterministic). Applying the above inequality, we get $$ \mathbb{E}(S)<\sigma. $$ In particular, $S$ is a biased estimator of $\sigma$.

Edit: I just realized that I answered the GOAL but not the actual question... Still, it might help you for your homework!