Suppose you have $X_i$ be i.i.d, such that $E(X_i) = \mu$ and $Var(X_i) = \sigma^2$. I know $\overline{X}_n = \sum\limits_{i=1}^n X_i$ and, as is standard $E(\overline{X_n}) = \mu$ and $Var(\overline{X}_n) = \frac{\sigma^2}{n}$.

I know you can derive that $E\left(\overline{X}_n^2\right) = \frac{\sigma^2}{n} + \mu^2$ using the identity $E(\overline{X}^2_n) = Var(\overline{X}_n) + E(\overline{X_n})^2$.

However, when I try to prove it directly from the definition of the sample mean, I get: $$ E(\overline{X}^2_n) = E\left( \left( \frac{1}{n} \sum\limits_{i=1}^n X_i\right)^2\right) = \frac{1}{n^2} \sum\limits_{i=1}^n E\left(X_i^2 \right) = \frac{1}{n^2} \sum\limits_{i=1}^n (\sigma^2 + \mu^2) = \frac{1}{n^2} n(\sigma^2 + \mu^2)= \frac{\sigma^2 + \mu^2}{n} $$.

I am trying to figure out what I'm doing wrong

  • 2
    $\begingroup$ HI: When you square the sum of the $X_{i}$ in the last equation, the sum is squared. So, all of the $X_i$ multiply all of the other $X_i$. So, you can't just say that it's equal to $\sum_{i=1}^{n} E(X_{i}^2)$. I didn't go through the algebra but, if you fix that, you'll probably get the answer above that you expect to get. $\endgroup$
    – mlofton
    Commented Oct 19, 2020 at 1:27

1 Answer 1


\begin{equation*} \text{E}(\bar{X}^{2}_{n})=\text{E}\left[\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)^{2}\right] \neq \frac{1}{n^2}\sum_{i=1}^{n}\text{E}\left(X_{i}^2\right). \end{equation*} For example, suppose that we have the following data: $X=1,3,2$. So, \begin{equation*} \text{E}(\bar{X}^{2}_{n})=\text{E}\left[\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)^{2}\right]=\text{E}\left[\left(\frac{1}{3}\sum_{i=1}^{3}X_{i}\right)^{2}\right]=\text{E}\left[\left(\frac{1}{3}[1+3+2]\right)^{2}\right]=\text{E}(2^{2})=4 \end{equation*} on the other hand, \begin{equation*} \frac{1}{n^2}\sum_{i=1}^{n}\text{E}\left(X_{i}^2\right)=\frac{1}{3^2}\text{E}\left[\sum_{i=1}^{3}\left(X_{i}^2\right)\right]=\frac{1}{9}\text{E}\left(1^{2}+3^{2}+2^{2}\right)=\frac{14}{9} \end{equation*} therefore, we can conclude that $\text{E}(\bar{X}^{2}_{n}) \neq \frac{1}{n^2}\sum_{i=1}^{n}\text{E}\left(X_{i}^2\right)$ as we have 4 $\neq$ 14/9.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.