Suppose $_1, _2, ... , _ ∼ _{\mu,\sigma^2}$ are independent and identically distributed samples from some distribution $_{\mu,\sigma^2}$ with mean $\mu$ and variance $\sigma^2$. $$\hat{S_n^2}=\frac{1}{n-1}\sum(X_i-\bar{X}_n)^2$$ $$\hat{T}^2_n=\frac{1}{n}\sum(X_i-\bar{X}_n)^2$$ are both point estimators for the parameter $\sigma2$.
I calculated their expectations to be $$\mathbb{E}[\hat{S}_n^2]=\frac{1}{n-1}\mathbb{E}[\sum(X_i-\bar{X}_n)^2]\\=\\=\frac{1}{n-1}(\sum(\mu^2+\sigma^2)-n\mu-n\frac{\sigma^2}{n})\\=\frac{1}{n-1}(n(\mu^2+\sigma^2)-n\mu^2-\sigma^2)\\=\frac{1}{n-1}((n-1)\sigma^2)\\=\sigma^2$$
$$\mathbb{E}[\hat{T_n}]=\frac{1}{n}(\sum(X_i-\bar{X}_n)^2)\\=\frac{1}{n}\mathbb{E}[\sum\{X_i\}^2-n(\bar{X}_n)^2]\\=\frac{1}{n}(\sum(\mathbb{E}[\{X_i\}^2]-n\mathbb{E}[\bar{X}_n]^2)\\=\frac{1}{n}(\sum(\mu^2+\sigma^2)-n\mu-n\frac{\sigma^2}{n}))\\=\frac{1}{n}(n(\mu^2+\sigma^2)-n\mu^2-\sigma^2)\\=\frac{1}{n}((n-1)\sigma^2)\\=\frac{n-1}{n}\sigma^2$$
but now I have to compare the standard errors of the two estimators and I am confused of how to proceed with the variance:
$$\mathbb{V}[\hat{S}^2_n]=\mathbb{V}[\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X}_n)^2]\\=\frac{1}{n-1}\mathbb{V}[\sum_{i=1}^n(X_i-\bar{X}_n)^2]\\=\frac{1}{n-1}\sum_{i=1}^n\mathbb{V}[(X_i-\bar{X}_n)^2]\\=\frac{n}{n-1}(\mathbb{E}[(X_i-\bar{X}_n)^4]-\sigma^2)\\=\frac{n}{n-1}(\mathbb{E}[X_i^4-n(\bar{X}_n)^4]-\sigma^2)\\=\frac{n}{n-1}(\mathbb{E}[X_i^4]-n\mathbb{E}[\bar{X}_n^4]-\sigma^2)\\$$
