TITLE: MSE (mean square error) of normal distribution estimator
QUESTION [0 upvotes]: suppose we have the sample variance $s^2 $ as our estimator
i know that $E(s^2)=\sigma^2$ , that imply $bias(s^2 ,\sigma^2)=0$
but how should i handle the MSE
$MSE=E((s^2-\sigma^2)^2)=Var(s^2 ,\sigma^2)+bias^2(s^2 ,\sigma^2)$
expanding it to $E(s^4)$ make it very complicated and i do not know how to handle that.

REPLY [1 votes]: Comment continued: In my main comment I showed you how to find $Var(S^2)$ analytically by using elementary methods. Here are simulation results for the case $\sigma=4$ and $n = 6.$
You can compare them with your answers when you get them. (Simulation results
should be accurate to at least a couple of significant digits.)
set.seed(509);  m=10^6;  n=6;  mu=0;  sg=4
x = rnorm(m*n, mu, sg)
DTA = matrix(x, nrow=m)  # each of m rows is sample of size 6
v = apply(DTA, 1, var)   # m sample variances
mean(v);  var(v);  mean((v-sg^2)^2)
# 16.00542  # aprx E(S^2) = 16
# 102.4153  # aprx Var(S^2) = 256*10/25 = 102.4
# 102.4152  # aprx MSE(S^2) = Var(S^2) because unbiased

Q = 5*v/sg^2
hist(Q, prob=T, br=50, col="skyblue2")         # histogram of simulated Q ~ CHISQ(5)
  curve(dchisq(x, 5), add=T, lwd=2, col="red") #  density of CHISQ(5)
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