Variance of sums of independent random variables

Question

I have the following formula -

$Var(\overline{X}) = Var(\frac{1}{n}\sum_{i=1}^n X_i) = \frac{1}{n^2}\sum_{i=1}^n Var(X_i)$

I know that the variance of the sum of independent random variables is equal to the sum of the variances of the random variables but I don't see where the $\frac{1}{n^2}$ is coming from? Why isn't it $\frac{1}{n}$?

Nevermind, I see why by looking at properties of variance...I'm going to try and derive it now. — csss, Mar 03 '14 at 15:09
Try calculating $\operatorname{var}(aY)$ in terms of $\operatorname{var}(Y)$ to see why the constant gets squared. Do not rely on formulas that you find in your book; do the calculation for yourself. — Dilip Sarwate, Mar 03 '14 at 15:10

score 2 · Answer 1 · answered Mar 03 '14 at 15:13

$\text{Var}\left(cX\right)=\mathbb{E}\left[\left(cX\right)^{2}\right]-\left(\mathbb{E}\left[cX\right]\right)^{2}=\mathbb{E}\left[c^{2}X^{2}\right]-\left(c\mathbb{E}\left[X\right]\right)^{2}=c^{2}\mathbb{E}\left[X^{2}\right]-c^{2}\left(\mathbb{E}\left[X\right]\right)^{2}=c^{2}\left[\mathbb{E}\left[X^{2}\right]-\left(\mathbb{E}\left[X\right]\right)^{2}\right]=c^{2}\text{Var}\left(X\right)$

score 1 · Answer 2 · answered Mar 03 '14 at 15:13

In general,

$$ \begin{aligned} \ \\ Var(aX) &= E[(aX)^2] - (E[aX])^2 \ \\ &= a^2E[X^2] - (aE[X])^2 \ \\ &= a^2E[X^2] - a^2 E[X]^2 \ \\ &= a^2 \left(E[X^2] - E[X]^2\right) \ \\ &= a^2 Var(X) \end{aligned} $$

Use this to work out the variance of the mean of random variables; let me know if you get stuck.

Variance of sums of independent random variables

2 Answers2