# A simple question about sampling variance [duplicate]

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We defind a variance of a random variable $X$ as $E(X-X_{\text{mean}})^2$=$E(X^2)-(E(X))^2$ and the mean = $E(X)$. However, for the case of sampling , Why would find out the sample mean as $\frac{X_1+\cdots+X_n}{n}$ for the case $X_1, \dots,X_n$ are iid and the sample variance are $\frac{\sum_{i=1}^{n} (X_i-X_{\text{mean}})^2}{n-1}$.

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## marked as duplicate by Did, Henry, Sasha, cardinal, Mike SpiveyJan 20 '12 at 3:34

This question was marked as an exact duplicate of an existing question.

I have inserted a power of 2 in your sample variance formula – Henry Jan 19 '12 at 8:24
what i mean is like why the sample mean = $\frac{X_1+\cdots+X_n}{n}$ – Mathematics Jan 19 '12 at 8:46

In your comment you concentrated your question on the why the sample mean is defined as $\dfrac{X_1+\cdots+X_n}{n}$. There are various reasons, some better than others:

1. It is easy to calculate, in a single pass
2. It is the mean (arithmetic average) of the sample
3. If you had sampled the whole (finite) population once each then the sample mean defined this way is the population mean.
4. The expectation of the sample mean is the population mean, so it is an unbiased estimator of the population mean
5. By the law of large numbers, the sample mean converges in some sense to the population mean as the sample size increases
6. If the population has a finite variance then by the central limit theorem the distribution of the sample mean converges towards a normal distribution with the same mean as the population mean as the sample size increases
7. For some families of distributions (but not others) the sample mean is a sufficient statistic, holding the useful information the sample provides about the parameters of the distribution
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>"If the population has a finite variance then by the central limit theorem the distribution of the sample mean converges towards a normal distribution with the same mean as the population mean as the sample size increases." Actually, this does not give the full story, The variance of the sample mean is converging to $0$, and so the distribution of the sample mean is converging to a constant: the population mean. In other words, the normal distribution to which the distribution of the sample mean is converging is degenerate (standard dev. $0$) in the limit as the sample size increases. – Dilip Sarwate Jan 20 '12 at 1:57

If the question is about the $n-1$ in the denominator, then this is a tradeoff between bias and accuracy of the estimator.