Vtyjkyuk

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.

$$T_c = sum_j=1^n frac(X_j - bar X)^2c$$

I found the expectation of $T_c$ to be $fracsigma^2(n-1)c$ 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:

$operatornameVar(T_c) = mathbb E (T_c^2)+ mathbb E (T_c)^2$

$operatornameVar(T_c) = mathbb E ((sum_j=1^n frac(X_j - bar X)^2c)^2) + fracsigma^2(n-1)c$

Any tips/solutions?

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

Observe that $X_i-barX$ is a Gaussian of mean $0$, let us compute the covariance of two of those
beginalign*
operatornameCov[X_i-barX, X_j-barX] &= mathbbE[(X_i-barX)(X_j-barX)]\
&= mathbbE[((X_i-mu)-(barX-mu))((X_j-mu)-(barX-mu))]\
&=mathbbE[(X_i-mu) (X_j-mu)] - 2 mathbbE[(X_i-mu) (barX-mu)] + mathbbE[(barX-mu)(barX-mu)]\
&= sigma_ij - 2 fracsigma^2n + fracsigma^2n\
&= sigma_ij-fracsigma^2n
endalign*
Where $sigma_ii=sigma^2$ and for $ineq j$, $sigma_ij=0$.

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 $Sigma_ij=operatornameCov[X_i-barX, X_j-barX]$. Observe that $Sigma = sigma^2 (pmbI-1/n cdot pmb1)$, we are looking at the eigen values of this. First note that $Sigma-sigma^2 pmbI=1/n cdot pmb1$ (where $pmb1$ 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 $Sigma pmb1 = sigma^2 (pmbI-1/n cdot pmb1) pmb1 = sigma^2 (pmb1 - pmb1) = 0 cdot pmb1$, 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
$$2(n-1) fracsigma^4c^2$$

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 ?