Why is the sample variance divided by (n-1)

introduction

There was a problem in probability theory and mathematical statistics ： Why is the sample variance divided by (n-1), At that time, I didn't understand very well when I was studying , However, if you ask the teacher, the teacher can't tell why （ I feel that the teacher is good at water …）, So find your own materials to learn . It is arranged as follows .

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1. Pre knowledge

The little friend who has studied probability theory and mathematical statistics must also know the following formula ：

1. If the mean （ expect ）$\mathbf{E}(x)=\mu$, variance $\mathbf{D}(x)={\sigma }^2$, that $\mathbf{E}(\overline{x})=\mu$,$\mathbf{D}(\overline{x})={\sigma }^2/n$

2. Note the population variance ${\sigma }^2$ And sample variance $S^2$ The formula is different , The first denominator is divided by n, One is divided by (n-1), Second, the sum of squares minus the inner one is the overall mean $\mu$, One subtracts the sample mean $\overline{x}$, That is to say .
$${\sigma }^2=\frac{{\sum }_{i=1}^n(x_i-\mu {)}^2}{n},S^2=\frac{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}{n-1}$$

3.${\sum }_{1=1}^k(x_i-\overline{x})=0$

2. Proving ideas

Actually, the sample variance $S^2$ It is essentially the overall mean $\mu$ Or total variance ${\sigma }^2$ A point estimation of , It's a random variable , Good point estimation has two most important properties ：

（1） The point estimate is unbiased , The expected value of the point estimate should be the estimated parameter , But this is not enough , Because there may be many forms of point estimation , So there's a third 2 strip .

（2） The unbiased estimator has a minimum variance , The variance of the minimum variance point estimate is smaller than that of any other estimator of the parameter .

The following proof $S^2$ yes ${\sigma }^2$ An unbiased estimator of . It is proved that the expected value of the point estimate should be the estimated population parameter .

3. Proof process

There are two ways to prove it ： The first one is commonly given in books , The second is better understood .

Method of proof 1：$$\begin{array}{rl}\mathbf{E}(S^2)& =\mathbf{E}(\frac{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}{n-1})\\ & =\frac{1}{n-1}\mathbf{E}[\sum _{i=1}^n(x_i-\overline{x}{)}^2]\\ & =\frac{1}{n-1}\mathbf{E}[\sum _{i=1}^n{x}_i^2-n{\overline{x}}^2]\\ & =\frac{1}{n-1}[\sum _{i=1}^n({\mu }^2+{\sigma }^2)-n({\mu }^2+\frac{{\sigma }^2}{n})]\\ & =\frac{1}{n-1}(n-1){\sigma }^2\\ & ={\sigma }^2\end{array}$$
Method of proof 2：
hypothesis $t$ It's a constant ：$$\begin{array}{rl}\sum _{i=1}^n(x_i-t{)}^2& =\sum _{i=1}^n(x_i-\overline{x}+\overline{x}-t{)}^2\\ & =\sum _{i=1}^n(x_i-\overline{x}{)}^2+2\sum _{i=1}^n(x_i-\overline{x})(\overline{x}-t)+\sum _{i=1}^n(\overline{x}-t{)}^2\\ & =\sum _{i=1}^n(x_i-\overline{x}{)}^2+2(\overline{x}-t)\sum _{i=1}^n(x_i-\overline{x})+\sum _{i=1}^n(\overline{x}-t{)}^2\\ & =\sum _{i=1}^n(x_i-\overline{x}{)}^2+\sum _{i=1}^n(\overline{x}-t{)}^2\\ & =\sum _{i=1}^n(x_i-\overline{x}{)}^2+n(\overline{x}-t{)}^2\end{array}$$
In the order $t$ Is the overall mean $\mu$, Then there are $$\begin{array}{r}\sum _{i=1}^n(x_i-\overline{x}{)}^2=\sum _{i=1}^n(\overline{x}-\mu {)}^2-n(\overline{x}-\mu {)}^2\end{array}$$
You can see ${\sum }_{i=1}^n(x_i-\overline{x}{)}^2$ and ${\sum }_{i=1}^n(\overline{x}-\mu {)}^2$ Are not strictly equal , There is still a difference $n(\overline{x}-\mu {)}^2$. be $$\begin{array}{rl}\mathbf{E}(S^2)& =\mathbf{E}(\frac{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}{n-1})\\ & =\frac{1}{n-1}\mathbf{E}[\sum _{i=1}^n(x_i-\overline{x}{)}^2]\\ & =\frac{1}{n-1}\mathbf{E}[\sum _{i=1}^n(\left(\overline{x}-\mu {)}^2-n(\overline{x}-\mu {)}^2\right)]\\ & =\frac{1}{n-1}[\mathbf{E}\left(\sum _{i=1}^n(\overline{x}-\mu {)}^2\right)-\mathbf{E}\left(\sum _{i=1}^{n}n(\overline{x}-\mu {)}^2\right)]\\ & =\frac{1}{n-1}[\mathbf{E}\left(\sum _{i=1}^n(\overline{x}-\mu {)}^2\right)-n\mathbf{E}\left(\sum _{i=1}^n(\overline{x}-\mu {)}^2\right)]\\ & =\frac{1}{n-1}(n{\sigma }^2-n\cdot \frac{{\sigma }^2}{n})\\ & ={\sigma }^2\end{array}$$

Both methods of proof can help understand . I hope it can help you .