I am working on an exercise asking which of the two following estimators; $X1, X2 $for the population mean of a normal population with parameters $ \mu, \sigma$ is best and why

$X1:=\frac {X_1+X_2+....X_{2n}}{2n}$,

$X2:=\frac {X_1+X_2+....X_n}{n}$

I computed the expectation and both are unbiased estimators. I tried to find the variance of the estimators, but the computations became unwieldy:

$Var(X1)= E[(X1- \mu)^2]=$

$\frac {1}{4n^2}E[( \Sigma _{i=1}^{2n}X_i-\mu)^2]=$

$E[\frac{1}{4n^2} (X_1+X_2+...+ X_{2n})^2 - 2\mu X1+ \mu^2]$

and in order to clear this I need to compute $E[X_i^2] $ ,

Which I don't see how to do. I tried standardizing but did not see it helping. Without this computation, I cant decide the best estimator for the pop mean based on neither variance nor bias-variance tradeoff.

Any ideas?



1 Answer 1


It's possible to derive the Expectation of $X_i^2$, however there's a much simpler way to calculate your variance:

$$ Var(X1) = Var(\frac{1}{2n} \sum_{i=1}^{2n} X_i) \\ = \frac{1}{4n^2} \sum_{i=1}^{2n} Var(X_i) \\ = \frac{1}{4n^2} 2 n \sigma^2 \\ = \frac{\sigma^2}{2n} $$ This comes from the fact that the variance is linear.

  • $\begingroup$ Thank you Erin. I guess we need indepence to use this formula, otherwise we need to use covariance/correlation, right? $\endgroup$
    – MSIS
    Commented Apr 8, 2020 at 23:43
  • $\begingroup$ That's true. But the exercise you are working on is of typical nature where draws of the population are supposedly stochastically independent. If you are interested in getting the expectation of X_i^2 I suggest to look at the theoretical variance by integration and reformulate that. Then you'll see that E[X_i^2] is Var(X) + mu^2. $\endgroup$ Commented Apr 8, 2020 at 23:47
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    $\begingroup$ Erin, I selected your answer. One last question, please : If we have two estimators E1, E2 of the same parameter and they are both unbiased and they have the same variance/s.e, what criteria would we use to select one over the other. Or maybe that should be a separate question? $\endgroup$
    – MSIS
    Commented Apr 11, 2020 at 4:29
  • 1
    $\begingroup$ I'd recommend opening another question for this since other people may have a different view or more experience. However, If they're both unbiased and equal variances, then I'd choose the one which is easier to compute (or faster to compute on a PC). Although, what I teach to my students, is that unbiasedness is not the non-plus-ultra. It is really nice to have and sometimes important. However, there are cases where a biased estimator has a lower MSE. Just as an additional information. $\endgroup$ Commented Apr 11, 2020 at 14:29

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