Finding an unbiased estimator for $\mu^2$

Question

I'm currently studying my first course in statistics. I am going through some question and have come across one that has stumped me. The question is as follows:

Suppose $\bar{X}_1, \ldots, \bar{X}_n$ are $n$ identically distributed independent random variables each with mean $\mu$ and variance $1$. Find an unbiased estimator for $\mu^2$.

To my understanding an unbiased estimator for $\mu^2$ is one such that $E[\text{?}] = \ldots = \mu^2$. However, I don't know how to go about solving for $?$

Any help or hints will be greatly appreciated. Thank you for your time.

score 2 · Answer 1 · answered Feb 12 '15 at 02:34

2

Hint: For any $k\in\{1,\dots,n\}$, $$E[\bar X_k^2]=\mathrm{Var}(\bar X_k)+E[\bar X_k]^2=1+\mu^2.$$

Can you use this to find an unbiased estimator for $\mu^2$?

answered Feb 12 '15 at 02:34

Ian

2,773

Great hint, I was actually on that path and forgot to calculate $E[\bar{X}^2]$ – wonggr Feb 12 '15 at 02:42

score 2 · Accepted Answer · answered Feb 12 '15 at 02:37

Just try the first thing that comes to mind. What about $S_n := \frac1n\sum_1^n X_i$? If we do $E[S_n^2]$ we get $$ \frac{1}{n^2}\left(\sum_{i\neq j} E[X_iX_j] + \sum_{i=1}^n E[X_i^2]\right) =\frac{1}{n^2}\left(\sum_{i\neq j} E[X_i]E[X_j] + \sum_{i=1}^n [\mathrm{Var}(X_i)+E[X_i]^2]\right) $$

$$ =\frac{1}{n^2}\left(\sum_{i\neq j} \mu^2 + \sum_{i=1}^n [1+\mu^2]\right) =\frac{1}{n^2}\left(\sum_{i,j} \mu^2 + \sum_{i=1}^n1\right) = \frac{n^2\mu^2+n}{n^2} = \mu^2 + \frac{1}{n}. $$

So $S_n$ doesn't work. But we can modify it a little to see $E[S_n^2 -\frac{1}{n}] = \mu^2$. Thus $S_n^2-\frac{1}{n}$ is an unbiased estimator.

Thanks for the clear justification! – wonggr Feb 12 '15 at 02:39 — wonggr, Feb 12 '15 at 02:39
This is so neat, thank you – ImOnlyListening Mar 14 '21 at 19:36 — ImOnlyListening, Mar 14 '21 at 19:36

Finding an unbiased estimator for $\mu^2$

2 Answers2