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Proof of Sample Mean

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Submitted By gangan
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Why is?
Consider a population of size N with the following elements: .
Now suppose we perform a random sampling of size n of the population. The total number of unique samples that can be drawn is, where

To simplify the notation, we’ll letbe equal to.
Now sincerepresents the total number of unique samples that can be drawn, we know that there will bedifferent sample means, .
Therefore the mean of the sample mean will simply be

We don’t know the exact shape of eachbut we know that it will contain n random elements from the population

Here’s the ABSOLUTELY CRUCIAL PART of the proof. We know for a fact that the a’s from the population will appear on and off in each, the important thing is to ask how many times will each a appear? This result requires a proof even beyond my own capability, and I’m simply going to quote the result from my own observation:
“If the population size is N and the sample size is n, each element in the population will appeartimes in the sampling distribution.”
But we know that

Again we will adopt a simpler notation temporarily. Let

Reminder: represents the number of times each element in the population will appear in the sampling distribution.
We know that

where

Here’s the clincher. Since each element in the population will appeartimes, I can perform the following tour de force

Which simplifies to

Before I restore the original expressions forand, I need to perform a sleight of hand maneuver for

Convince yourself that the above is consistent with the factorial notation. The reason for me doing this will become very apparent soon.
With that done,now becomes

You should realise that many terms will cancel off, leaving us with just

But this is just the expression for, the population mean. Hence

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