# Thread: 'Alternate' proof that the expected value of the sample mean is the population mean

1. ## 'Alternate' proof that the expected value of the sample mean is the population mean

It would be appreciated if someone could verify that this makes sense.
By definition

So taking its expectation we get

Now, as we have a population of size and a sample size of size , we have different samples and of those, contain each of the values .

Then clearly,

So, the expectation of will be given by,

Therefore

2. ## Re: 'Alternate' proof that the expected value of the sample mean is the population me

You aren't being very clear about what the summations are over and what terms are involved. I also don't understand what you're trying to say when you claim that
Originally Posted by sid9221
contain each of the values .
What? You have from X1 up to Xn but the count implies you're looking at the number of combinations of size k-1 from a set of size N-1.

Overall I think you need to be more explicit in what you're referring to which might clear up some of the ambiguity that I see in the proof.

3. ## Re: 'Alternate' proof that the expected value of the sample mean is the population me

Originally Posted by Dason
You aren't being very clear about what the summations are over and what terms are involved. I also don't understand what you're trying to say when you claim that

What? You have from X1 up to Xn but the count implies you're looking at the number of combinations of size k-1 from a set of size N-1.

Overall I think you need to be more explicit in what you're referring to which might clear up some of the ambiguity that I see in the proof.
Apologies, the k's were all typos, then were meant to be n's. Does it make sense now.

I'm working on the 'Latex' error bit. Its meant to be [[N-1 Choose n-1 ]*sum X_j]/[N Choose n]

The summations for x_i go from 1..n and for X_j go from 1..N. Trying to update the latex versions, doesn't seem to be working..

4. ## Re: 'Alternate' proof that the expected value of the sample mean is the population me

I guess I understand your question a little bit more now.

In your question, you have a population with distinct elements:

Now sampling elements out of without replacement, forming a set of random sample

with the assumption that each element is equally likely to be sampled.

Now, the population mean is defined by

and the sample mean is defined by

OP may want to calculate the expected value of the sample mean without invoking the linearity of expectation, but just rely on the first principle with combinatorics argument.

As mentioned by OP, the number of ways to form such a random sample is and by the equally likely assumption, each way has a equal probability of

So the expected value may be expressed like this:

where are the number of times that the elements appear in the above summation.

The number of ways of forming a random set of elements, with the condition that a particular elements must be included in the set, is

and thus we have

Finally, putting these things together,

and this should be what OP trying to argue for.

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