1. ## Name of Theorem

Is there a standard name for the theorem proving that the expected value of the mean of a random sample without replacement equals the mean of the population from which the sample is drawn?

2. ## Re: Name of Theorem

isn't this just due to the linearity of the Expected value?

$E[\bar{x}] = E\left[\dfrac 1 N \displaystyle{\sum_{k=1}^N}~x_k \right ] = \dfrac 1 N \displaystyle{\sum_{k=1}^N}~E[x_k] = \dfrac 1 N \cdot N \mu = \mu$

3. ## Re: Name of Theorem

Originally Posted by romsek
isn't this just due to the linearity of the Expected value?

$E[\bar{x}] = E\left[\dfrac 1 N \displaystyle{\sum_{k=1}^N}~x_k \right ] = \dfrac 1 N \displaystyle{\sum_{k=1}^N}~E[x_k] = \dfrac 1 N \cdot N \mu = \mu$
I don't think so. We are dealing with a mean of means from samples of size 1 through (n - 1).

I can prove it, but I would just as soon not encumber my text with a proof that is well known to statisticians. I'd like to be able to say "According to the X Theorem ...." rather than spend a page on a proof.

4. ## Re: Name of Theorem

Originally Posted by JeffM
I don't think so. We are dealing with a mean of means from samples of size 1 through (n - 1).

I can prove it, but I would just as soon not encumber my text with a proof that is well known to statisticians. I'd like to be able to say "According to the X Theorem ...." rather than spend a page on a proof.
this is what appendices are for

I'm not entirely sure what you are doing. But if you repeat what I wrote for $N=1, \text{ population size}$

And then take all those resulting means and average them you will just be taking the average of a constant stream of the value $\mu$ which clearly has an average value of $\mu$

What am I missing?

5. ## Re: Name of Theorem

What you are missing is that not all samples will have a mean equal to the population mean.

Population: 1, 3, 4, 5, 12. Mean is 5. If I do samples of 3 without replacement, I can have sample means ranging from a low (8 / 3) through a high of (21 / 3).

So it is not a constant stream of 5. With a population size of 5 and a sample size of 3, there are 10 possible distinct samples, and quite a few have means differing from 5.

It is not a difficult proof that the expected value of the mean of a sample without replacement (regardless of sample size) equals the population mean, but I already have numerous appendices. If this is a standard, well known theorem, as I believe is true, it would certainly make sense simply to reference it by name.

But thank you for responding.

6. ## Re: Name of Theorem

none of the groups of samples are likely to have the same mean as the population mean

nevertheless the expected value of the mean of any group of samples will be the population mean for the reason I showed.

7. ## Re: Name of Theorem

Hey JeffM?

It seems that what you are describing is a conditional expectation.

Just note though that you will have to conditional expectation with respect to choices of subsets. If you evaluate the conditional expectation across all subsets then you should get the normal value of the population mean.

If you don't do that though then you will need a specific filtration [the term used in graduate probability] and your conditional expectation will be based on that.