unbiased estimator for $\mu^2$ Let $(X1, . . . , Xn)$ be $n$ random variables with $E(Xi) = \mu$ and $Var(Xi) = \sigma^2$
Is the following estimator an unbiased estimator for $\mu^2$ ?
$T = \frac{n}{n-1}$ $\sum_{j=1}^n {(Xj-\bar X)}^2$ - $\sum_{j=1}^n {Xj}^2$ + $X1\sum_{j=2}^n {Xj}$
What I have done is the following
I need to find the expectation of this estimator to check if it is unbiased for $\mu^2$
the first piece of the estimator is $ n VAR(X)$ the second one is $n E{(X^2)}$ and the last one is $n E{(X)}^2$  since the variance can be rewritten as $[E(X^2) - E{(X)}^2]$ I would get $0$ for this estimator because $ n [E(X^2) - E{(X)}^2]$ - $n E{(X^2)}$ + $n E{(X)}^2$ and and consequently it will be bias for $\mu^2$
My doubt is about the last estimator summation namely $X1\sum_{j=2}^n {Xj}$ because it starts with $J$ equals $2$ instead of $1$
does this affect the result of the estimator or would it be the same if it started with $1$? if not is my attempt correct? Thanks in advance
 A: Why not just use the expression
$$
\begin{align}
\mathsf{Var}X=\mathsf EX^2-(\mathsf EX)^2
&\implies(\mathsf EX)^2=\mathsf EX^2-\mathsf{Var}X\\
&\implies\mu^2=\mathsf EX^2-\mathsf{Var}X.
\end{align}
$$
Now replace the r.h.s. by unbiased estimators for each term, namely
$$
T=\frac{1}{n}\sum_{k=1}^nX_k^2-\frac{1}{n-1}\sum_{k=1}^n(X_k-\bar X)^2.
$$
Its easy to verify that
$$
\begin{align}
\mathsf ET
&=\mathsf E\left(\frac{1}{n}\sum_{k=1}^nX_k^2-\frac{1}{n-1}\sum_{k=1}^n(X_k-\bar X)^2\right)\\
&=\frac{1}{n}\sum_{k=1}^n\mathsf EX_k^2-\mathsf E\left(\frac{1}{n-1}\sum_{k=1}^n(X_k-\bar X)^2\right)\\
&=\mathsf EX^2-\mathsf{Var}X\\
&=\mu^2.
\end{align}
$$
You can compare this estimator to what you have in your post and check for equality.

Edit:
The expected value of the estimator $T$ in the OPs post can be verified to be
$$
\begin{align}
\mathsf ET
&=\mathsf E\left(\frac{n}{n-1}\sum_{j=1}^n (X_j-\bar X)^2 - \sum_{j=1}^n X_j^2 + X_1\sum_{j=2}^n X_j\right)\\
&=\mathsf E\left(\frac{n}{n-1}\sum_{j=1}^n (X_j-\bar X)^2\right) - \mathsf E\left(\sum_{j=1}^n X_j^2\right) + \mathsf E\left(X_1\sum_{j=2}^n X_j\right)\\
&=n\mathsf{Var}X - n\mathsf EX^2 + \sum_{j=2}^n \mathsf E(X_1X_j)\\
&=-n\mu^2 + \sum_{j=2}^n \mathsf E(X_1X_j)\\
\end{align}
$$
If $X_1$ is independent of $X_j$, $j\in \{2,\dots n\}$ then we further have
$$
\begin{align}
\mathsf ET
&=-n\mu^2 + (n-1)\mu^2=-\mu^2,
\end{align}
$$
which shows that $-T$ is an unbiased estimator of $\mu^2$.
