Variance of Sums of Independent Random Variables

When two random variables are independent, the variance of their sum equals the sum of their variances.

For independent discrete random variables $X$ and $Y\!:$

$$\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y).$$

In general (without independence), there is an additional covariance term, but independence forces that term to vanish.

A direct consequence is that the same formula holds for the difference. Since $\text{Var}(-Y) = (-1)^2\text{Var}(Y) = \text{Var}(Y),$ we have

$$\text{Var}(X-Y) = \text{Var}(X) + \text{Var}(Y).$$

Subtraction does not reduce variability — independent fluctuations always accumulate.

For example, if $\text{Var}(X)=4$ and $\text{Var}(Y)=9$ with $X,Y$ independent, then

$$\text{Var}(X+Y) = 4+9 = 13, \qquad \text{Var}(X-Y) = 4+9 = 13.$$