My Math Forum I Have Forgotten All My Statistics

 Probability and Statistics Basic Probability and Statistics Math Forum

 November 28th, 2016, 11:23 AM #1 Senior Member   Joined: May 2016 From: USA Posts: 1,253 Thanks: 519 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 n$, the m marbles withdrawn will have an expected mean weight of $\bar r$ as will the $n - m$ marbles remaining in the urn. Consider the base case where $m = 1.$ The expected mean weight of the 1 marble withdrawn will be $\displaystyle \sum_{i=1}^n \left (r_i * \dfrac{1}{n} \right ) = \dfrac{1}{n} * \sum_{i=1}^n r_i = \bar r.$ The expected mean weight of the (n - 1)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an\ arbitrary\ such\ integer.$ $Withdraw\ k\ marbles\ at\ random\ from\ urn.$ $Expected\ mean\ weight\ of\ k\ marbles\ withdrawn = \bar r\ by\ Theorem\ 1.$ $Expected\ mean\ weight\ of\ (n - k)\ marbles\ remaining\ in\ urn = \bar r\ by\ Theorem\ 1.$ $Withdraw\ 1\ marble\ at\ random\ from\ urn.$ $Expected\ mean\ weight\ of\ last\ marble\ withdrawn = \bar r\ by\ Theorem\ 1.$ $Expected\ mean\ weight\ of marbles\ remaining\ in\ urn = \bar r\ by\ Theorem\ 1.$ $Expected\ weight\ of\ (k + 1)\ marbles\ withdrawn = k \bar r + \bar r = (k + 1) \bar r \implies$ $Expected\ mean\ weight\ of\ (k + 1)\ marbles\ withdrawn = \dfrac{1}{\cancel {k + 1}} * \cancel{(k + 1)} \bar r = \bar r.$ That is Theorem 2.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, 12:43 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 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 05:32 PM.
 November 28th, 2016, 03:18 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,253 Thanks: 519 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 05:33 PM.
 November 28th, 2016, 03:52 PM #4 Global Moderator   Joined: May 2007 Posts: 6,665 Thanks: 651 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). Thanks from JeffM1
 November 28th, 2016, 03:55 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,253 Thanks: 519 Mathman Is there a name for the theorem proving that proposition, or do I have to derive it?

 Tags forgotten, statistics

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Jounetsu Calculus 1 March 6th, 2015 02:21 AM ElMarsh Advanced Statistics 2 October 23rd, 2009 06:10 PM wontonsoup Calculus 2 May 5th, 2009 07:18 AM Wiglets Algebra 2 August 17th, 2008 07:54 PM Wiglets Abstract Algebra 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top