#### JK423

Gold Member
Suppose that you have a probability distribution P of a parameter which, for simplicity, it is two-valued. For example, it could be a coin where we denote heads with "+" and tails with "-". Suppose that we throw the coin N times, and the results tend to follow the probability ditribution P for large enough N (which naturally could be 50%-50%).
We then calculate the mean value of all these results, which are a series of "+" and "-" (N in total), and find a number <S>.
Suppose now, that, we erase M of these results (with M<N) from the total N, and we are left with the rest N-M results. BUT, we do the erasure with a completely random way. We calculate, again, the mean value with the N-M results, <S'>.

QUESTION
Will the mean values <S> and <S'> be equal?

The answer will surely depend on the numbers N, M. Is there any known theorem about this, that guarantees the equality for appropriate N, M?

Edit: For finite N, M ofcourse the equality is impossible. What i mean is whether <S'> can approach <S> very very close, for appropriate N, M.

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#### mathman

Is your question about the sample mean or the theoretical mean? For the sample means, they could be different. For the theoretical means, they are equal.

#### JK423

Gold Member
Hmm, what is the difference? Perhaps, you mean that the sample mean includes a finite N while the theoretical an infinite N?

I am looking for a theorem on sample mean with finite N, so that it's applicable in real applications..
I am glad that the theoretical means are equal though :). Is there any proof of this to your knowledge?

#### mathman

The theoretical mean is determined from the probability distribution. It has nothing to do with sample size. There is a theorem (law of large numbers) which states (under the proper conditions) that the sample mean -> theoretical mean as N becomes infinite.

I don't know what kind of theorem you are looking for. Perhaps the following may help.
http://en.wikipedia.org/wiki/Sampling_(statistics [Broken])

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