I am trying to solve the following question:

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