How to calculate the bias of the statistic A given statistic : $T_c = \sum_{j=1}^n \frac{{(X_j - \bar X)}^2}{c}$, where $c$ is a constant, as an estimator of variance $\sigma^2.$
$X_1,\ldots, X_n$ denote a random sample from a population which has normal distribution with unknown mean $\mu$ and unknown variance $\sigma^2$.
The statistic is distributed as $x^2_{n-1}$ (a chi-squared variate with $n-1$ degrees of freedom).
I am tasked to find the bias of $T_c$.
I know the formula for bias is $\mathbb E \hat \theta - \theta$.
I found $\theta$ as $\mu = n - 1$ for a chi-squared distribution of $n-1$ degrees of freedom.
However, I am confused as to how to calculate $\mathbb E \hat \theta$.
What i thought of doing is to calculate $\mathbb E T_c$, but i got stuck halfway through, so I am not sure if I am doing the right thing. Any help please. thanks in advance.
 A: $$
\frac 1 {\sigma^2} \sum_{i=1}^n (X_i-\overline X)^2 \sim \chi^2_{n-1}.
$$
Therefore
$$
\operatorname E\left( \frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \overline X)^2 \right) = n-1.
$$
So
$$
\operatorname E\left( \frac 1 c \sum_{i=1}^n (X_i-\overline X)^2 \right) = \frac{\sigma^2} c (n-1).
$$
Subtract $\sigma^2$ from that to get the bias.
A: Since $T_c$ is an estimator of $\sigma^2$. $T_c$ will be unbiased for $\theta = \sigma^2$ if $\mathbb{E}_{\mu,\sigma^2}(T_c) = \sigma^2 \quad \forall \mu, \sigma^2 \in \mathbb{R} \times \mathbb{R}_0^+$
\begin{align*}
\mathbb{E}(T_c)&= \frac{1}{c}\mathbb{E} \Big( \sum_{j=1}^n X_j^2 + \bar{X}^2 -2X_j \bar{X} \Big) \\  
&= \frac{1}{c} \sum_{j=1}^n \mathbb{E}(X_j^2) + \mathbb{E}(\bar{X}^2) - 2  \mathbb{E} (X_j \bar{X})
\end{align*}
Now calculate $\mathbb{E}(X_j^2),\mathbb{E}(\bar{X}^2)$ & $ \mathbb{E} (X_j \bar{X})$.
1.
 Using Var$(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$ we find $\mathbb{E}(X_j^2) = \sigma^2 + \mu^2$.
2.
\begin{align*}
\mathbb{E}(\bar{X}^2) &= \mathbb{E}\Big( \big(\frac{1}{n} \sum_i^n X_i\big) \big(\frac{1}{n} \sum_i^n X_i\big)\Big) \\
&= \frac{1}{n^2} \mathbb{E}\Big( \big(\sum_i^n X_i\big) \big(\sum_i^n X_i\big)\Big)\\
&= \frac{1}{n^2} \mathbb{E}\Big( \sum_{i \neq j} X_iX_j  + \sum_i X_i^2\Big) \\
&=\frac{1}{n^2} \Big(\sum_{i \neq j} \mathbb{E}(X_i)\mathbb{E}(X_j)  + \sum_i \mathbb{E}(X_i^2)\Big)\\
&= \frac{1}{n^2} \Big( n(n-1)\mu^2 + n(\mu^2 + \sigma^2) \Big)\\
&= \frac{(n-1)\mu^2 + \mu^2 + \sigma^2}{n} =\mu^2 + \frac{\sigma^2}{n}
\end{align*} 
3.
\begin{align*}
\mathbb{E} (X_j \bar{X})&=  \frac{1}{n} \sum_i^n \mathbb{E}(X_jX_i) \\
&=\frac{1}{n} \big( \sum_{i\neq j}\mathbb{E}(X_j)\mathbb{E}(X_i) \big) + \frac{1}{n} \mathbb{E}(X_j^2)\\
&= \frac{1}{n} (n-1) \mu^2 + \frac{1}{n}(\mu^2 + \sigma^2)\\
&= \mu^2 + \frac{\sigma^2}{n}
\end{align*}
By using the values we have found we obtain
\begin{align*}
\mathbb{E}(T_c) 
&= \frac{1}{c} \sum_{j=1}^n \mathbb{E}(X_j^2) + \mathbb{E}(\bar{X}^2) - 2  \mathbb{E} (X_j \bar{X})\\
&= \frac{1}{c} \sum_{j=1}^n  (\sigma^2 + \mu^2) + ( \mu^2 + \frac{\sigma^2}{n} ) - 2 (\mu^2 + \frac{\sigma^2}{n}) \\
&= \frac{(n-1)\sigma^2}{c}
\end{align*}
So $T_c$ is biased for $\sigma^2$ for all values of $c \neq n-1$
