I'm reviewing probability and statistics.The textbooks said that

if the sampled population is infinite, then


I'm curious about how does this result come from. Wikipedia does not tell me much. Is there any provement? How does it come from?

  • 2
    $\begingroup$ Why does this question have 3 up-votes????? Correction - why does it have any upvotes? $\endgroup$
    – Epictetus
    Nov 14, 2012 at 9:52
  • $\begingroup$ What is $n$? What is $\sigma$? What is $\bar x$? $\endgroup$
    – martini
    Nov 14, 2012 at 9:56
  • $\begingroup$ $n$ is the number of data, $\sigma$ is the standard deviation of those data, and $\bar{x}$ is the mean. @martini $\endgroup$ Nov 14, 2012 at 11:26

2 Answers 2


If our population consists of $N$ individuals and $x_i$, $i=1,\ldots,N$, is the variable of interest, then the population mean and population variance are given by $$ \bar{X}=\frac{1}{N}\sum_{i=1}^N x_i,\qquad \sigma^2=\frac{1}{N-1}\sum_{i=1}^N(\,x_i-\bar{X})^2. $$ Suppose a simple random sample $x_1,\ldots,x_n$ of size $n$ is drawn from this population (i.e. every sample of size $n$ has equal probability of being drawn). Then the corresponding sample mean and sample variance is $$ \bar{x}=\frac{1}{n}\sum_{i=1}^n x_i,\qquad \mathrm{var}(\bar{x})=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2. $$ It can be shown that $$ \mathrm{var}(\bar{x})=\frac{N-n}{Nn}\sigma^2=\left(1-\frac{n}{N}\right)\frac{\sigma^2}{n}. $$ Now, let $N\to\infty$ and see what happens.


Assume we have a group of data, to get the SD of the mean, we can calculate the mean for many subgroups of our data and calculate it's SD. To do this, $\sigma_{\bar{y}}$ can be defined as follows: $$\sigma_{\bar{y}}^2=\frac{\sum\limits_{j=1}^{k}(\bar{y_j}-y_{tm})^2}{k}$$ Where $\bar{y_j}$ is the mean of a finite number of data, $y_{tm}$ is the mean for an infinite number of data, and k is the number of times you took the finite number of data. Since $$\bar{y_j}=\frac{\sum\limits_{i=1}^{n}y_{i}}{n}$$ Substituting that in the first equation we get: $$\sigma_{\bar{y}}^2=\frac{1}{k}\sum\limits_{j=1}^{k}\left(\frac{1}{n}\sum\limits_{i=1}^{n}y_i-y_{tm}\right)^2$$ $$\sigma_{\bar{y}}^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{n}y_i-ny_{tm}\right)^2$$ But $n=\sum_{i=1}^{n}1$, so substituting on the $n$ inside the parentesis: $$\sigma_{\bar{y}}^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{n}y_i-\sum\limits_{i=1}^{n}y_{tm}\right)^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\sum\limits_{i=1}^{n}\left(y_i-y_{tm}\right)^2$$Since $\frac{1}{n}\sum_{i=1}^{n}(y_i-y_{tm})^2=\sigma_{j}^2$ because it is the variance of each group of data: $$\sigma_{\bar{y}}^2=\frac{1}{nk}\sum\limits_{j=1}^{k}\sigma_{j}^2$$ Since $\sigma_j^2$ is the variance of a subgroup of the same date, it can be considered the same for all $j$, and at last we get: $$\sigma_{\bar{y}}^2=\frac{\sigma^2}{n}$$ $$\sigma_{\bar{y}}=\frac{\sigma^2}{\sqrt{n}}$$


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