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.
Best Answer
$$ \frac 1 {\sigma^2} \sum_{i=1}^n (X_i-\overline X)^2 \sim \chi^2_{n-1}. $$ 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.