Properties of Variance for Discrete Random Variables

Use the computational formula Var(X)=E[X^2]-(E[X])^2, the linear transformation rule Var(aX+b)=a^2 Var(X), and the additivity of variance for independent random variables Var(X±Y)=Var(X)+Var(Y).

Step 1 of 157%

Tutorial

The Computational Formula for Variance

The variance of a discrete random variable XX is defined as Var(X)=E[(Xμ)2],\operatorname{Var}(X) = E[(X-\mu)^2], where μ=E[X].\mu = E[X]. Expanding the square and applying linearity of expectation gives an equivalent — and usually faster — computational formula:

Var(X)=E[X2](E[X])2.\operatorname{Var}(X) = E[X^2] - (E[X])^2.

The derivation is short:

Var(X)=E[(Xμ)2]=E[X22μX+μ2]=E[X2]2μE[X]+μ2=E[X2]2μ2+μ2=E[X2]μ2.\begin{align*} \operatorname{Var}(X) &= E[(X-\mu)^2] \\ &= E[X^2 - 2\mu X + \mu^2] \\ &= E[X^2] - 2\mu E[X] + \mu^2 \\ &= E[X^2] - 2\mu^2 + \mu^2 \\ &= E[X^2] - \mu^2. \end{align*}

To illustrate, suppose XX has the PMF

x012P(X=x)0.50.30.2\begin{array}{c|ccc} x & 0 & 1 & 2 \\ \hline P(X=x) & 0.5 & 0.3 & 0.2 \end{array}

Then

E[X]=0(0.5)+1(0.3)+2(0.2)=0.7,E[X2]=02(0.5)+12(0.3)+22(0.2)=1.1,Var(X)=1.10.72=1.10.49=0.61.\begin{align*} E[X] &= 0(0.5) + 1(0.3) + 2(0.2) = 0.7, \\ E[X^2] &= 0^2(0.5) + 1^2(0.3) + 2^2(0.2) = 1.1, \\ \operatorname{Var}(X) &= 1.1 - 0.7^2 = 1.1 - 0.49 = 0.61. \end{align*}
navigate · Enter open · Esc close · ⌘K/Ctrl K toggle