Which is the better estimator for the mean?

Question

measurements x1, x2, x3 from three independent runs of an experiment with variance sigma-squared.

(x1+x2+x3)/3 or (x1+2x2+x3)/4

I am having trouble with this homework problem. I think the first one would be the better estimator because it is an actual mean. I have learned that the best point estimator for a sample population mean would be µ.

Now I need to justify this answer and I am stuck.

Please help!

score 2 · Accepted Answer · answered Sep 25 '13 at 15:25

2

Both estimators are unbiased.

The variance of the first estimator is equal to $$\frac{1}{9}(\sigma^2+\sigma^2+\sigma^2)=\frac{\sigma^2}{3}.$$

The variance of the second estimator is $$\frac{1}{16}(\sigma^2+4\sigma^2+\sigma^2)=\frac{6\sigma^2}{16}.$$

The second estimator has larger variance than the first. Thus the first estimator is "better."

answered Sep 25 '13 at 15:25

André Nicolas

507,029

Thank you! This makes sense but why did you get 1/9 and 1/16? I understand you squared the original fraction but I don't understand why you would have to do that. – SamHaim Sep 25 '13 at 15:56
In general, the variance of $kY$ is $k^2$ times the variance of $Y$. The other fact that we used is that the variance of a sum of independent random variables is the sum of the variances. – André Nicolas Sep 25 '13 at 16:15
Ah, yes. Thank you very much! This helped a lot! – SamHaim Sep 26 '13 at 00:43
You are welcome. – André Nicolas Sep 26 '13 at 00:43

Which is the better estimator for the mean?

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