# Prove var(s^2) = 2sigma^4/(n-1)

#### gh33

##### New Member
I am trying to prove var(s^2) = 2sigma^4/(n-1) for N(u, sigma^2)
where S^2= sumof(xi-xbar)^2/(n-1)
I tried

var(s^2) = E[(s^2)^2] - E[(s^2)]^2
I have already proved S^2 is unbiased estimator of sigma^2
so
E[(s^2)]=sigma^2
hence E[(s^2)]^2= sigma^4

Can anyone give me hint/tips about how to find
E[(s^2)^2] for N(u, sigma^2)

if i do
var( sumof(xi-xbar)^2 / (n-1)) then I expand
= (1/(n-1))^2 var(sumof(xi-xbar)^2)
I have a dump question here,
does var of sum = sum of var here?

Thanks

#### BGM

##### TS Contributor
Your proof maybe much simpler if you are allowed to use the fact that
$$\frac {(n - 1)S^2} {\sigma^2} \sim \chi^2(n-1)$$

By the way, $$Var\left[\sum_{i=1}^n X\right] = \sum_{i=1}^n Var[X_i]$$

if and only if they are uncorrelated, i.e.

$$Cov[X_i, X_j] = 0 ~~ \forall i \neq j$$

#### gh33

##### New Member
Your proof maybe much simpler if you are allowed to use the fact that
$$\frac {(n - 1)S^2} {\sigma^2} \sim \chi^2(n-1)$$

By the way, $$Var\left[\sum_{i=1}^n X\right] = \sum_{i=1}^n Var[X_i]$$

if and only if they are uncorrelated, i.e.

$$Cov[X_i, X_j] = 0 ~~ \forall i \neq j$$
Cant use the chi-square distribution
That's why I stuck here.

#### Dragan

##### Super Moderator
Cant use the chi-square distribution
That's why I stuck here.
You should have said this explicitly in your original post.