Suppose $X_1,\ldots,X_n$ are i.i.d. normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2$. Let $\bar{X}_n$ be the sample mean. Consider estimating $\sigma$. For any given constant $c$, define the estimator\footnote{followed this video for both parts. In (ii) I use his calculation to shorten my shown work. https://www.youtube.com/watch?v=MqeS2NWB3I4}
\begin{equation*} \hat{\sigma}_{n,c} = \frac{c}{n} \sum_{i=1}^n |X_i - \bar{X}_n| \end{equation*}
\textbf{(i)} Compute $E(\hat{\sigma}_{n,c})$
\begin{equation*} \begin{aligned} & E[\hat{\sigma}_{n,c}] = E \left[ \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] \\ & = \frac{c}{n} E \left[ \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] \\ & = \frac{c}{n} E \left[ \sum_{i=1}^{n} \sqrt{ (X_i - \bar{X}_n)^2 } \right] \\ & = \frac{c}{n} E \left[ \sum_{i=1}^{n} \sqrt{\left[(X_i - \mu) (\bar{X}_n - \mu)\right]^2} \right] \\ & = \frac{c}{n} E \left[ \sqrt{\sum_{i=1}^{n}(X_i - \mu)^2 - 2n(\bar{X}_n - \mu)^2 + n(X_i - \mu)^2} \right] \\ & = \frac{c}{n} \left[ \sqrt{ n\sigma^2 - nE((X_i - \mu)^2)} \right] \\ & = \frac{c}{n} \left[\sqrt{ n\sigma^2 - n\frac{\sigma^2}{n}} \right] \\ & \frac{c\sigma\sqrt{n-1}}{n} \end{aligned} \end{equation*}
\textbf{(ii)} For what $c$ is $\hat{\sigma}_{n,c}$ a consistent estimator of $\sigma$. Explain.\ \ \indent We know that the estimator is consistent if $lim_{n \to \infty}Var(\hat{\sigma}_{n,c}) = 0$. So we show:
\begin{equation*} \begin{aligned} & \operatorname{Var}(\hat{\sigma}_{n,c}) = \operatorname{Var} \left( \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right) \\ & = E\left[\left( \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right)^2 \right] - E\left[ \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] ^2 \\ & = \frac{c}{n} E\left[ \left( \sqrt{ (X_i - \bar{X}_n)^2} \right)^2 \right] - \frac{c^2\sigma^2(n-1)}{n^2} = \\ & \text{saving space by citing the proof from the video here}\\ & = \frac{c\sigma^2}{n} - \frac{c^2\sigma^2(n-1)}{n^2} \\ & \sigma^2 \left[ \frac{c}{n} - \frac{c^2(n-1)}{n^2} \right] \end{aligned} \end{equation*}
Notice that $\operatorname E(X_i-\overline X) = 0$ and \begin{align} \operatorname{var}(X_i-\overline X) & = \operatorname{var}(X_i) + \operatorname{var}(\overline X) - 2\operatorname{cov}(X_i,\overline X) \\[8pt] & = \sigma^2 + \frac{\sigma^2} n - \frac{2\sigma^2} n \\[8pt] &= \frac{n-1} n \sigma^2. \end{align}
So \begin{align} \operatorname E\big( |X_i - \overline X| \big) & = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} \sqrt{\frac{n-1} n}\cdot\sigma|z| e^{-z^2/2} \, dz \\[12pt] & = \sigma\sqrt{\frac{n-1}{2\pi n}} \cdot 2 \int_0^{+\infty} e^{-z^2/2} (z \, dz) \\[10pt] & = 2\sigma \sqrt{\frac{n-1}{2\pi n}} \int_0^{+\infty} e^{-u} \, du \\[8pt] & = 2\sigma \sqrt{\frac{n-1}{2\pi n}}. \end{align}