My Math Forum proof

 November 10th, 2017, 04:53 AM #1 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 Thanks: 60 Math Focus: सामान्य गणित proof How I can prove the sample mean of "n" samples is the unbiased estimator of the population mean. SHAZAM Unbiased Estimation Is this link helpful? Last edited by MATHEMATICIAN; November 10th, 2017 at 05:03 AM.
 November 10th, 2017, 05:05 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198 \begin{align*} &E[S_n] = \\ &E\left[\dfrac 1 n \sum \limits_{k=1}^n~X_k\right] = \\ &\dfrac 1 n \sum \limits_{k=1}^n E[X_k] \end{align*} Assuming your samples all come from the same distribution $E[X_k] = \mu,~\forall k \in \{1,2, \dots n\}$ so $E[S_n] = \dfrac 1 n (n \mu) = \mu$
November 10th, 2017, 07:38 AM   #3
Math Team

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Math Focus: सामान्य गणित
Quote:
 Originally Posted by romsek \begin{align*} &E[S_n] = \\ &E\left[\dfrac 1 n \sum \limits_{k=1}^n~X_k\right] = \\ &\dfrac 1 n \sum \limits_{k=1}^n E[X_k] \end{align*} Assuming your samples all come from the same distribution $E[X_k] = \mu,~\forall k \in \{1,2, \dots n\}$ so $E[S_n] = \dfrac 1 n (n \mu) = \mu$

can you please explain it equation by equation in simple words?

 November 10th, 2017, 09:50 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,808 Thanks: 971 "SHAZAM" sounds like a secret word in Ali-Baba-like old stories !!
November 11th, 2017, 07:29 AM   #5
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From: काठमाडौं, नेपाल

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Math Focus: सामान्य गणित
Quote:
 Originally Posted by Denis "SHAZAM" sounds like a secret word in Ali-Baba-like old stories !!
DENIS could you please explain me those equations?

 November 11th, 2017, 09:51 AM #6 Senior Member     Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198 I don't know how you can use language like "sample mean" and not understand the notation below. You know the sample mean is the sum of say $n$ samples divided by $n$. $S_n$ represents this sample mean, and we want to find the expectation of it. That is what $E[S_n] = E\left[\dfrac 1 n \sum \limits_{k=1}^n~X_k\right]$ says. Then due to properties of expectation we can push the expectation into the sum to obtain $E[S_n] = \dfrac 1 n \sum \limits_{k=1}^n E[X_k]$ You are told you are sampling from a single population with a given mean $E[X_k] = \mu,~\forall k$ so substituting this in we get $E[S_n] = \dfrac 1 n \sum \limits_{k=1}^n \mu = \dfrac 1 n (n \mu) = \mu$ I don't mean to be rude but if you can't understand the above you need to do some more reading.

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