The question I tried to solve, but failed, goes like this:

Find the expected value of $(\bar{X}_n)^2$ and find an ubiased estimator for $\mu^2$.

This is the solution given by the TA:

$$E[(\frac{1}{n}\sum^n_{i=1} X_i)^2]$$ $$= E[\frac{1}{n^2}\sum^n_{i=1} X_i \sum^n_{k=1} X_k]$$ $$= \frac{1}{n^2}\sum^n_{i=1} E[X^2_{i}] + \frac{1}{n^2}\sum^n_{i=1} \sum_{k \neq i} E[X_i]E[X_k]$$ $$= \frac{1}{n}(\sigma^2 + \mu^2) + \frac{n-1}{n}\mu^2$$ $$= \mu^2 + \frac{\sigma^2}{n}$$

Hence, the unbiased estimator is: $$(\bar{X}_n)^2 - \frac{\hat{\sigma}^2}{n}$$

I have a few questions about this. The TA will take days to answer and I can't wait that long:

  • Where did this $X_k$ come from? why didn't he just use $\frac{1}{n}\sum^n_{i=1} X_i$ twice?
  • I don't understand the second line at all. Why is there $\frac{1}{n^2}$ two times? and how did he separate the sum into two parts like that?
  • The transition from the second line to the third line is also unclear.

Can anyone please assist me with this question? thank you.


1 Answer 1


Here is a more expanded version of the derivation: \begin{align} E\left[\bar X_n^2\right] &= E\left[\left(\frac{1}{n}\sum^n_{i=1} X_i\right)^2\right] \\ &= E\left[\left(\frac{1}{n}\sum^n_{i=1} X_i\right)\left(\frac{1}{n}\sum^n_{i=1} X_i\right)\right] \\ &= E\left[\left(\frac{1}{n}\sum^n_{i=1} X_i\right)\left(\frac{1}{n}\sum^n_{k=1} X_k\right)\right] \\ &= E\left[\frac{1}{n^2}\left(\sum^n_{i=1} X_i\right)\left(\sum^n_{k=1} X_k\right)\right] \\ &= E\left[\frac{1}{n^2}\left(\sum^n_{i=1} X_i\right)a\right] \\ &= E\left[\frac{1}{n^2}\sum^n_{i=1} a X_i\right] \\ &= E\left[\frac{1}{n^2}\sum^n_{i=1} \left(\sum^n_{k=1} X_k\right) X_i\right] \\ &= E\left[\frac{1}{n^2}\sum^n_{i=1} \left(\sum^n_{k=1} X_kX_i\right)\right] \\ &= \frac{1}{n^2}\sum^n_{i=1} \sum^n_{k=1} E\left[X_kX_i\right] \\ \end{align} Notice that, in the inner summation, as we are iterating over $k$, and because $k$ and $i$ both range from $1$ to $n$, then there will come a point when $k = i$. This means that the inner summation becomes $$\sum^n_{k=1} E\left[X_kX_i\right] = E[X_iX_i] + \sum_{k \neq i} E[X_kX_i] = E[X_i^2] + \sum_{k \neq i} E[X_kX_i]$$ and so \begin{align} E\left[\bar X_n^2\right] &= \frac{1}{n^2}\sum^n_{i=1} \sum^n_{k=1} E\left[X_kX_i\right] \\ &= \frac{1}{n^2}\sum^n_{i=1} \left(E[X_i^2] + \sum_{k \neq i} E[X_kX_i]\right) \\ &= \frac{1}{n^2}\sum^n_{i=1} E[X_i^2] + \frac{1}{n^2}\sum^n_{i=1}\sum_{k \neq i} E[X_kX_i] \end{align} Because $X_k$ and $X_i$ are assumed to be independent and identically distributed for $k \neq i$, then $E\left[X_kX_i\right] = E[X_k]E[X_i]$ and $E[X_k] = E[X_i]$ and so \begin{align} E\left[\bar X_n^2\right] &= \frac{1}{n^2}\sum^n_{i=1} E[X_i^2] + \frac{1}{n^2}\sum^n_{i=1}\sum_{k \neq i} E[X_k]E[X_i] \end{align} I'll leave the rest up to you.


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