Derivation of standard error of mean

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 $\fracn\sigma^2=\frac$. Not understood. And the standard deviation of T/n must be $\sigma/<\sqrt>$ . Of course, T/n is the sample mean $\bar$ . I went to some basics: $\displaystyle Var(X)=\frac\sum_^()^2$ $\displaystyle Var(X)=\frac\sum_^()$ $\displaystyle Var(X)=\frac\sum_^x_i^2+\mu^2-\frac\sum_^x_i\mu$ As Sample mean is an unbiased estimate of population mean, we get $\displaystyle Var(X)=\frac\sum_^x_i^2-\mu^2$ $\displaystyle Var(X)=E(X^2)-(E(X))^2$ Nothing useful found from this. Why is there a $1/n^2$ in that step to get variance ?

• statistics
• statistical-inference

3 Answers 3

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

$\begingroup$ Why do we take Z = Y/n in the first place? Just to introduce the constant 1/n? I do not understand why we would want to introduce that constant in the first place. $\endgroup$

$\begingroup$ @Oleksandra The question that was asked above was essentially, why is it that when I divide a variable by $n$, the variance is divided by $n^2$ instead of just $n$? So this answer divides a variable by $n$ to show what happens. The reason why to divide by $n$ in the first place was evidently already understood. If you do not understand, you may ask a new question about that. $\endgroup$