unbiased estimator for $\mu^2$

Question

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

Answer 1

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.

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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 $\mu^2$.