Determine the variance of estimator T

Question

I'm having trouble figuring out how to find the variance of the following estimator.

Let $X_1,X_2,...,X_n$ denote random sample from a population which has a normal distribution with unknown mean $\mu$ and unknown variance $\sigma^2$. The statistic below is an estimator for $\sigma^2$, where $c$ is a constant.

$$T_c = \sum_{j=1}^n \frac{(X_j - \bar X)^2}{c}$$

I found the expectation of $T_c$ to be $\frac{\sigma^2(n-1)}{c}$ using the definition of $T_c$, however I am stumped over how to determine the $Var(T_c)$.

I started to try to determine it like so but got stuck:

$\operatorname{Var}(T_c) = \mathbb E (T_c^2)+ \mathbb E (T_c)^2$

$\operatorname{Var}(T_c) = \mathbb E ((\sum_{j=1}^n \frac{(X_j - \bar X)^2}{c})^2) + \frac{\sigma^2(n-1)}{c}$

Any tips/solutions?

Answer 1

I suppose that the samples are independents, so they have $0$ covariance.

Observe that $X_i-\bar{X}$ is a Gaussian of mean $0$, let us compute the covariance of two of those \begin{align*} \operatorname{Cov}[X_i-\bar{X}, X_j-\bar{X}] &= \mathbb{E}[(X_i-\bar{X})(X_j-\bar{X})]\\ &= \mathbb{E}[((X_i-\mu)-(\bar{X}-\mu))((X_j-\mu)-(\bar{X}-\mu))]\\ &=\mathbb{E}[(X_i-\mu) (X_j-\mu)] - 2 \mathbb{E}[(X_i-\mu) (\bar{X}-\mu)] + \mathbb{E}[(\bar{X}-\mu)(\bar{X}-\mu)]\\ &= \sigma_{ij} - 2 \frac{\sigma^2}{n} + \frac{\sigma^2}{n}\\ &= \sigma_{ij}-\frac{\sigma^2}{n} \end{align*} Where $\sigma_{ii}=\sigma^2$ and for $i\neq j$, $\sigma_{ij}=0$.

Now let's apply the trick in this post sum of squares of dependent gaussian random variables

the goal is to determine the coefficient $\lambda_i$ in front of the chi-squared variables and then use the variance of variance of $\chi_1^2$. So we must find the eigen decomposition of $\Sigma$ where $\Sigma_{ij}=\operatorname{Cov}[X_i-\bar{X}, X_j-\bar{X}]$. Observe that $\Sigma = \sigma^2 (\pmb{I}-1/n \cdot \pmb{1})$, we are looking at the eigen values of this. First note that $\Sigma-\sigma^2 \pmb{I}=1/n \cdot \pmb{1}$ (where $\pmb{1}$ denote the all $1$ matrix or all $1$ vector depending on context) have all same rows and so $\Sigma$ have eigen value $\sigma^2$ with multiplicity $n-1$ (If a symmetric matrix $A$ has $m$ identical rows show that $0$ is an eigen value of $A$ whose geometric multiplicity is atleast $m-1$.). Then observe that $\Sigma \pmb{1} = \sigma^2 (\pmb{I}-1/n \cdot \pmb{1}) \pmb{1} = \sigma^2 (\pmb{1} - \pmb{1}) = 0 \cdot \pmb{1}$, so that $0$ is also an eigen value.

Applying the forum trick we get that $T$ have the distribution of a sum of $n-1$ independant $\chi^2_1$ multiplied by $\sigma^2/c$. The variance of $\chi_1^2$ is $2$ and so the variance you seek (modulo all the errors I made) is $$2(n-1) \frac{\sigma^4}{c^2}$$

Comment : I am not satisfied by not being able to distinguish in my notation the matrix of all ones and the vector of all one, someone have a suggestion ?