Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was going through this wikipedia article on standard error. I could not understand the crucial step here. It goes like this:

This formula may be derived from what we know about the variance of a sum of independent random variables.

If $X_1, X_2 , \ldots, X_n$ are n independent observations from a population that has a mean $\mu$ and standard deviation $\sigma$ , then the variance of the total

$T = (X_1 + X_2 + \cdots + X_n)$ is $n\sigma^2$. Understood.

The variance of T/n must be $\frac{1}{n^2}n\sigma^2=\frac{\sigma^2}{n}$. Not understood.

And the standard deviation of T/n must be $\sigma/{\sqrt{n}}$ . Of course, T/n is the sample mean $\bar{x}$ .

I went to some basics:

$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}({x_i-\mu})^2$

$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}({x_i^2+\mu^2-2x_i\mu})$

$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}x_i^2+\mu^2-\frac{2}{n}\sum_{i=1}^{n}x_i\mu$

As Sample mean is an unbiased estimate of population mean, we get

$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}x_i^2-\mu^2$

$\displaystyle Var(X)=E(X^2)-(E(X))^2$

Nothing useful found from this.

Why is there a $1/n^2$ in that step to get variance ?

share|improve this question

3 Answers 3

up vote 1 down vote accepted

Let $Y$ be any random variable. Let $Z = Y/n$. Then $$Z^2 = \frac1{n^2} Y^2,$$ $$E(Z^2) = E\left(\frac1{n^2} Y^2\right) = \frac1{n^2} E(Y^2)$$ and therefore $$E\left(\left(\frac Yn\right)^2\right) = \frac1{n^2} E(Y^2).$$ Also, $$E(Z) = E\left(\frac1n Y\right) = \frac1n E(Y).$$ So from $Var(Y)=E(Y^2)-(E(Y))^2$ and $Var(Z)=E(Z^2)-(E(Z))^2,$ we find $$\begin{eqnarray} Var\left(\frac Yn\right) = Var(Z) &=& E(Z^2)-(E(Z))^2\\ &=& \frac1{n^2} E(Y^2) - \left(\frac1n E(Y)\right)^2 \\ &=& \frac1{n^2} (E(Y^2) - \left( E(Y)\right)^2 \\ &=& \frac1{n^2} Var(Y). \end{eqnarray}$$ Now consider the case where $Y = T$.

share|improve this answer

The only thing we need to prove here is that for any scalar constant $c$, and for a random variable $X$, $$\mathrm{Var}[cX] = c^2 \mathrm{Var}[X].$$ This follows from the property of expectation $$\mathrm{E}[cX] = c\mathrm{E}[X]$$ as follows: $$\begin{align*} \mathrm{Var}[cX] &= \mathrm{E}[(cX - \mathrm{E}[cX])^2] \\ &= \mathrm{E}[(cX - c\mathrm{E}[X])^2] \\ &= \mathrm{E}[c^2(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{E}[(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{Var}[X]. \end{align*}$$

share|improve this answer

Recall the definition of (population) variance: $$var\xi := E(\xi - E\xi)^{2}$$ for $\xi$ a random variable. Then we have $var\xi = E\xi^{2} - 2(E\xi)^{2} + (E\xi)^{2} = E\xi^{2} - (E\xi)^{2}$ so that $var(\xi/n) = E(\xi^{2})/n^{2} - (E\xi)^{2}/n^{2}.$

Thus we have

$$var(T/n) := var(X_{1}/n) + \cdots + var(X_{n}/n) = n\sigma^{2}/n^{2} = \sigma^{2}/n.$$

share|improve this answer
What if $E(\xi) \neq 0$. Does this mean that an underlying assumption that population mean is zero is required for this formula to hold true ?I am not sure if I am missing something obvious here..but can't wrap my head around this –  square_one Aug 23 at 14:47
Let me revise the answer. Letting $E\xi := 0$ is to simply calculation. In your question, since $varX_{i}\ (i = 1, \dots, n)$ are given, there is no need to do these derivations from the great beginning. –  Chou Aug 23 at 14:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.