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November 28th, 2016, 10:23 AM  #1 
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479  I Have Forgotten All My Statistics
I studied statistics about 50 years ago so I do not remember much. I am guessing that if the following is true, it is probably a basic theorem. I have a proof, but it may not be rigorous. Given an urn containing $n > 1$ marbles with UNKNOWN weights $r_i,\ for\ 1 \le i \le n$, and a KNOWN mean weight of $\bar r$ and a random drawing of m marbles without replacement, where $1 \le m < n$, the m marbles withdrawn will have an expected mean weight of $\bar r$ as will the $n  m$ marbles remaining in the urn. Consider the base case where $m = 1.$ The expected mean weight of the 1 marble withdrawn will be $\displaystyle \sum_{i=1}^n \left (r_i * \dfrac{1}{n} \right ) = \dfrac{1}{n} * \sum_{i=1}^n r_i = \bar r.$ The expected mean weight of the (n  1) marbles remaining in the urn is $\displaystyle \dfrac{1}{n  1} \left \{ n \bar r  \sum_{i=1}^n \left (r_i * \dfrac{1}{n} \right ) \right \} = \dfrac{1}{n  1} * (n \bar r  \bar r) = \dfrac{1}{\cancel {n  1}} * \cancel {(n  1)} \bar r = \bar r.$ This Theorem 1 proves the proposition if n = 2 or if n > 2 and m = 1. Now consider cases if n > 2. $Theorem\ 1 \implies \exists\ at\ least\ one\ integer\ in\ [1,\ (n  2)]\ for\ which\ the\ proposition\ is\ true.$ $Let\ k\ be\ an\ arbitrary\ such\ integer.$ $Withdraw\ k\ marbles\ at\ random\ from\ urn.$ $Expected\ mean\ weight\ of\ k\ marbles\ withdrawn = \bar r\ by\ Theorem\ 1.$ $Expected\ mean\ weight\ of\ (n  k)\ marbles\ remaining\ in\ urn = \bar r\ by\ Theorem\ 1.$ $Withdraw\ 1\ marble\ at\ random\ from\ urn.$ $Expected\ mean\ weight\ of\ last\ marble\ withdrawn = \bar r\ by\ Theorem\ 1.$ $Expected\ mean\ weight\ of marbles\ remaining\ in\ urn = \bar r\ by\ Theorem\ 1.$ $Expected\ weight\ of\ (k + 1)\ marbles\ withdrawn = k \bar r + \bar r = (k + 1) \bar r \implies$ $Expected\ mean\ weight\ of\ (k + 1)\ marbles\ withdrawn = \dfrac{1}{\cancel {k + 1}} * \cancel{(k + 1)} \bar r = \bar r.$ That is Theorem 2. Putting Theorems 1 and 2 together gives a proof by induction that the proposition is true for any integer m such that $1 \le m < n.$ I do not care about this proof if the proposition is a standard theorem. The name of that theorem will be enough. But if it is not a standard theorem, does the proof above have a hole in it? 
November 28th, 2016, 11:43 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1103 
This appears to be true only if $1) \text{ n is even} \\ \\ 2) \ m=\dfrac n 2$ Last edited by skipjack; November 28th, 2016 at 04:32 PM. 
November 28th, 2016, 02:18 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 
Romsek Thank you for answering. I appear not to have explained the problem correctly. Let's take an example with n = 5, m = 2, and population of 1, 3, 4, 5, 7. So the population mean is (1 + 3 + 4 + 5 + 7) / 5 = 20 / 5 = 4. Removal of 2 marbles at random entails that there are $\dbinom{5}{2} = \dfrac{5!}{2! * 3!} = \dfrac{5 * 4 * 3}{3 * 2} = 10$ equally likely removals. (1) 1, 3 mean 2 times 0.1 = 0.2 (2) 1, 4 mean 2.5 times 0.1 = 0.25 Cumulative 0.45 (3) 1, 5 mean 3 times 0.1 = 0.3 Cumulative 0.75 (4) 1, 7 mean 4 times 0.1 = 0.4 Cumulative 1.15 (5) 3, 4 mean 3.5 times 0.1 = 0.35 Cumulative 1.5 (6) 3, 5 mean 4 times 0.1 = 0.4 Cumulative 1.9 (7) 3, 7 mean 5 times 0.1 = 0.5 Cumulative 2.4 (8) 4, 5 mean 4.5 times 0.1 = 0.45 Cumulative 2.85 (9) 4, 7 mean 5.5 times 0.1 = 0.55 Cumulative 3.40 (10) 5, 7 mean 6 times 0.1 = 0.6 Cumulative 4.00 That is just an example and obviously proves nothing, but it works for a population of 5 and a random sample of 2. And 5 is not even, and 2 is not 5 / 2. My fault for not being clear. It really comes down to a proof that the EXPECTED mean of any random sample without replacement is equal to the population mean. I recollect that as being true, but I do not remember whether there is a difference between sampling with replacement and without replacement. And I have no recollection of what the name of any theorem proving that may be (if indeed it is true). Last edited by skipjack; November 28th, 2016 at 04:33 PM. 
November 28th, 2016, 02:52 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,613 Thanks: 617 
Mean with or without replacement is equal to the population mean. Therefore your original proposition is trivially true for both cases (drawn and left behind).

November 28th, 2016, 02:55 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 
Mathman Is there a name for the theorem proving that proposition, or do I have to derive it? 

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