Assume that $X_1,...X_4$ are idd $N(0,\sigma^2)$ where $\sigma$ is unknown. My task is to show that $T=\frac{1}{3}\sum_{i=1}^4(X_i-\bar X)^2$ is unbiased. I am aware that this question has been asked before, but I am curious as to where I messed up. Here is my work:
$E[T]=\frac{1}{3}E[\sum_{i=1}^4X_i^2-2\bar X\sum_{i=1}^4X_i+4\bar X]$
$=\frac{4}{3}E[X_i^2]-\frac{2}{3}E[\bar X\sum_{i=1}^4X_i]+\frac{4}{3}E[\bar X^2]=\frac{4}{3}\sigma^2-\frac{2}{3}E[\bar X\sum_{i=1}^4X_i]+\frac{4}{3}\frac{\sigma^2}{4}$
$=\frac{5}{3}\sigma^2-\frac{2}{3}E[\bar X\sum_{i=1}^4X_i]=\frac{5}{3}\sigma^2-\frac{2}{3}\frac{1}{4}E[ (\sum_{i=1}^4X_i)^2]$ (using definition of $\bar X$).
At this point, I must have made some sort of mistake because the second term will not equal $(2/3)\sigma^2$. Does anyone see where I made a mistake?
In the first line $4\bar{X}$ should be $4\bar{X}^2$ but you correct this later.
In the last line you have $E [(\sum_{i=1}^4 X_i)^2] = E[\sum_{i=1}^4 X_i^2] = 4 \sigma^2$ as desired (the cross terms have zero expectation).
Alternatively, in the last line you can do $E[\bar{X} \sum_{i=1}^4 X_i] = 4 E[\bar{X}^2] = \sigma^2$, and leads to the same result.