I am learning about the idea of estimators in statistics and understand that the role of an estimator is to estimate a parameter of the population distribution. In my book it says that the estimation function for the population mean is the sample mean, and there is a proof of why this is, but I don't understand this.

Here is a link to something showing the same proof. This starts with the expectation of the sample mean and substituting our formula for the sample mean into there and then moving 1/n out of the summation and then saying this is the same as the sum over the expectation values of the individual data points. They then say this is the same as n time the population mean.

I thought the point of this was to show that the expectation of the sample mean is the population mean, but it looks like we are just asserting this rather than showing it. We are saying the expectation of our data point is the population mean.

Can someone explain this to me?

  • $\begingroup$ Your question is about the properties of the Expectation operator. The proof uses the linearity of the operator and the identical distribution of the summands, which is assumed at the beginning. $\endgroup$ – JohnK Dec 31 '15 at 13:24
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    $\begingroup$ Does this help stats.stackexchange.com/questions/184189/… ? $\endgroup$ – Tim Dec 31 '15 at 13:34
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    $\begingroup$ It's worth noting that the link asserts nothing about the mean being "best": it only demonstrates some useful properties of the sample mean qua estimator. Note, too, that there are many "estimation functions for the population mean": the sample mean is merely one of the more popular ones and is not universally "best" in all circumstances. $\endgroup$ – whuber Dec 31 '15 at 14:13
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    $\begingroup$ To add to Whuber's comment, the "best" estimator is often a function of the relative costs and variances. For different scenarios it is useful to find the cost function (e.g. $C=C_0 + \sum^n_{i=1} C_n n^2_i$) and the variance of each estimator (e.g. variance of linear regression estimator of $\mu$ versus a ratio estimator of $\mu$). You can then combine these to find the option with the smallest variance for the least cost. The proof you have linked is just showing that $\bar{y}$ is an unbiased estimator of the population mean $\bar{Y}$, not that it is the best one. $\endgroup$ – Chris C Dec 31 '15 at 16:34

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