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November 10th, 2017, 04:53 AM  #1 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 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: Southern California, USA Posts: 1,602 Thanks: 816 
$\begin{align*} &E[S_n] = \\ &E\left[\dfrac 1 n \sum \limits_{k=1}^n~X_k\right] = \\ &\dfrac 1 n \sum \limits_{k=1}^n E[X_k] \end{align*}$ 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 Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित  Quote:
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: 10,898 Thanks: 716 
"SHAZAM" sounds like a secret word in AliBabalike old stories !!

November 11th, 2017, 07:29 AM  #5 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित  
November 11th, 2017, 09:51 AM  #6 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816 
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. 