proof confidence interval is unbiased for normal sample

Question

I am trying to solve the following question:

A $(1−\alpha)100\%$ confidence interval for a parameter, $\theta$, is called unbiased if the expectation of the midpoint of the two endpoints is equal to $\theta$. Consider a sample $\mathbf{Y} = (Y_1,\ldots,Y_n)$, where $Y_i$ is iid $N(\mu,\sigma^2)$ and consider the confidence intervals for $\mu$ and $\sigma^s$ presented in Section 1.3.1.

1. Show that the confidence interval for $\mu$ is unbiased.
2. Show that the confidence interval for $\sigma^2$ is biased

Now the confidence interval that it is talking about are the following:

Confidence Interval for $\sigma^2$

$1−\alpha = P\left(\frac{(n−1)\hat{\sigma}^2}{ c_2 }≤ σ^2 ≤\frac{ (n−1)\hat{\sigma}^2}{ c_1 }\right)$ and when we observe $\underline{y} = (y_1,\ldots,y_n)$ it becomes $\left(\frac{(n−1)s^2}{ c_2 },\frac{ (n−1)s^2}{ c_1 }\right)$

Confidence Interval for $\mu$

$1-\alpha = P\left(\hat{\mu}-c\frac{\hat{\sigma}}{\sqrt{n}}\leq \mu \leq \hat{\mu}+c\frac{\hat{\sigma}}{\sqrt{n}}\right)$ and when we observe $\underline{y} = (y_1,\ldots,y_n)$ it becomes $\left(\hat{\mu}-c\frac{s}{\sqrt{n}},\hat{\mu}+c\frac{s}{\sqrt{n}}\right)$

My attempt

I tried with both but really I got nowhere. We never done a problem like this, so I'm improvising. For example for $\mu$ I would calculate the following $$E\left[\frac{1}{2}\left(\hat{\mu}+c\frac{\hat{\sigma}}{\sqrt{n}}-\hat{\mu}+c\frac{\hat{\sigma}}{\sqrt{n}}\right)\right] = E\left[c\frac{\hat{\sigma}}{\sqrt{n}}\right] = \frac{c}{\sqrt{n}}E[\hat{\sigma}] $$ but then I don't know how to go further as I know that $E[\hat{\sigma}^2] = \sigma^2$ and that $Var[\hat{\sigma}^2]=\frac{2\sigma^4}{n-1}$ but I know nothing about $E[\hat{\sigma}]$.

Similarly for $\sigma$ I get $$E\left[\frac{1}{2}\left(\frac{ (n−1)\hat{\sigma}^2}{ c_1 }-\frac{ (n−1)\hat{\sigma}^2}{ c_2 }\right)\right] = E\left[\frac{(c_2-c_1)(n-1)\hat{\sigma}^2}{2c_1c_2}\right] = \frac{(c_2-c_1)(n-1)}{2c_1c_2}E[\hat{\sigma}^2] =\frac{(c_2-c_1)(n-1)\sigma^2}{2c_1c_2} $$

But as you see I don't get nowhere.. Can you help me?

Edit after the answer

So following the answer I continued as follows:

$$E\left[\frac{1}{2}\left(\hat{\mu}+\frac{c\hat{\sigma}}{\sqrt{n}}+\hat{\mu}-\frac{c\hat{\sigma}}{\sqrt{n}}\right)\right] = E[\hat{\mu}] = \mu$$ hence it is unbiased. And $$E\left[\frac{1}{2}\left(\frac{(n-1)\hat{\sigma}^2}{c_1}+\frac{(n-1)\hat{\sigma}^2}{c_2}+\right)\right] = \frac{(n-1)(c_1+c_2)}{2c_1c_2}E[\hat{\sigma}^2] = \frac{(n-1)(c_1+c_2)}{2c_1c_2}\sigma^2 \neq \sigma^2$$ and so it is biased.

Answer 1

The endpoints of the confidence interval for $\mu$ are $\hat \mu \pm\text{something},$ and that's enough to conclude that the midpoint of the interval is $\hat \mu,$ without knowing what the "something" is. Thus you need only prove that $\operatorname{E}(\hat\mu)=\mu.$ Do you know how to do that?

The endpoints of the confidence interval for $\sigma^2$ are not given as $\hat\sigma^2\pm\text{something},$ where $\hat\sigma^2$ is some unbiased estimator of $\sigma^2;$ therefore this must be approached differently.

You have not told us what $c_1$ and $c_2$ are. I will surmise that they are so chosen that $$ \Pr\left(\sigma^2 < \frac{(n-1)s^2}{c_2} \right) = \frac \alpha 2 = \Pr\left( \frac{(n-1) s^2}{c_1} < \sigma^2 \right) $$ The midpoint is $$ (n-1)s^2 \left( \frac 1 {2c_2} + \frac 1 {2c_1} \right). $$ The expected value (if we take $s^2$ to be the usual unbiased estimator of $\sigma^2$) is $$ (n-1)\sigma^2 \left( \frac 1 {2c_2} + \frac 1 {2c_1} \right). $$ Showing this is not $\sigma^2$ is the same as showing that $$ (n-1) \left( \frac 1 {2c_2} + \frac 1 {2c_1} \right) \ne 1. $$ Here the asymmetry of the chi-square distribution is used. We have $$ \Pr(\chi^2_{n-1} < c_2) = \frac \alpha 2 = \Pr(\chi^2_{n-1} > c_1) $$ You need the fact that $c_1$ is farther from the expected value of $\chi^2_{n-1}$ (which is $n-1$) than $c_2$ is, so that their average is above the expected value rather then equal to the expected value.

Answer 2

The midpoint of the CI for $\mu$ is simply $\hat \mu$. What you've calculated appears to be the half-width of the CI. Specifically, $$L = \hat \mu - c \frac{\hat \sigma}{\sqrt{n}} , \quad U = \hat \mu + c \frac{\hat \sigma}{\sqrt{n}},$$ implies the midpoint is $$\frac{L + U}{2} = \frac{2 \hat \mu}{2} = \hat \mu.$$