Why Independent Variances Add

Why Study the Variance of a Sum?

Expected values add even when random variables are dependent: the expected value of a sum is the sum of the expected values. Variance measures something different: how far a random outcome tends to lie from its mean. When two random variables are added, their deviations can reinforce one another or partly cancel. Independence gives a precise reason why, on average, the cross-product of those deviations contributes zero.

This result is useful whenever a total is made from separate, independent random contributions. It lets us find the variance of a sum without first listing every possible value of that sum. The condition matters: if the variables are dependent, their deviations may tend to move together, and simply adding their variances can give the wrong answer.

Definition: For a discrete random variable $X$ with finite mean $\mu_X=E(X)$ and finite second moment, its variance is the expected squared deviation from its mean: $$ \operatorname{Var}(X)=E\left[(X-\mu_X)^2\right]. $$ For discrete outcomes $x$ with probabilities $P(X=x)$, this is $$ \operatorname{Var}(X)=\sum_x (x-\mu_X)^2P(X=x). $$

Suppose $X$ and $Y$ are discrete random variables whose sum is meaningful, and both have finite variances. For example, if they are counts, both should use the same unit of count. Independence means that their joint probabilities factor: for every pair of possible values $x,y$, $P(X=x,Y=y)=P(X=x)P(Y=y)$. This factorization is the ingredient that makes the variance rule work.

Theorem: If $X$ and $Y$ are independent discrete random variables with finite variances, then $$ \operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y). $$

Deriving the Addition Rule

The mean of $X+Y$ is $\mu_X+\mu_Y$, by linearity of expectation. Therefore, the deviation of the sum from its mean is $(X+Y)-(\mu_X+\mu_Y)=(X-\mu_X)+(Y-\mu_Y)$. To calculate its variance, square this expression. The square contains a cross-product term, and independence is what eliminates that term after taking its expectation.

\begin{aligned} \operatorname{Var}(X+Y) &=E\left[\left((X-\mu_X)+(Y-\mu_Y)\right)^2\right]\\ &=E\left[(X-\mu_X)^2\right] +2E\left[(X-\mu_X)(Y-\mu_Y)\right] +E\left[(Y-\mu_Y)^2\right]. \end{aligned}

The first and third expectations are $\operatorname{Var}(X)$ and $\operatorname{Var}(Y)$. To evaluate the middle expectation, use the discrete joint probability mass function. Independence lets each joint probability be written as a product of marginal probabilities, so the double sum separates into two sums:

\begin{aligned} E\left[(X-\mu_X)(Y-\mu_Y)\right] &=\sum_x\sum_y (x-\mu_X)(y-\mu_Y)P(X=x,Y=y)\\ &=\sum_x\sum_y (x-\mu_X)(y-\mu_Y)P(X=x)P(Y=y)\\ &=\left(\sum_x (x-\mu_X)P(X=x)\right) \left(\sum_y (y-\mu_Y)P(Y=y)\right)\\ &=E(X-\mu_X)E(Y-\mu_Y)\\ &=0\cdot 0=0. \end{aligned}

Each centered variable has expected value zero because $E(X-\mu_X)=E(X)-\mu_X=0$, and likewise for $Y$. Substituting the zero cross-product expectation into the expansion gives the stated rule:

\operatorname{Var}(X+Y) =\operatorname{Var}(X)+2(0)+\operatorname{Var}(Y) =\operatorname{Var}(X)+\operatorname{Var}(Y).

The derivation pinpoints the role of independence. It is not the act of adding two variances that makes the rule true; it is the zero expected cross-product of the centered variables. Independence guarantees that zero by factoring the joint probabilities.

Worked Example: The Total Number of Two Independent Successes

Suppose $X$ is an indicator for success in one trial, with probability $0.30$, and $Y$ is an indicator for success in a separate, independent trial, with probability $0.40$. Each indicator is 1 for success and 0 otherwise. Let $S=X+Y$, the total number of successes. Find $\operatorname{Var}(S)$.

Check the conditions. Both variables are discrete and take only the values 0 and 1, so their variances are finite. The trials are specified to be independent, and both indicators count successes, so their sum is meaningful.

For $X$, the mean is $0.30$. From the definition of variance,

\begin{aligned} \operatorname{Var}(X) &=0.30(1-0.30)^2+0.70(0-0.30)^2\\ &=0.30(0.49)+0.70(0.09)\\ &=0.147+0.063=0.210. \end{aligned}

For $Y$, the mean is $0.40$, so

\begin{aligned} \operatorname{Var}(Y) &=0.40(1-0.40)^2+0.60(0-0.40)^2\\ &=0.40(0.36)+0.60(0.16)\\ &=0.144+0.096=0.240. \end{aligned}

The addition rule now gives

\operatorname{Var}(S)=\operatorname{Var}(X)+\operatorname{Var}(Y) =0.210+0.240=0.450.

Thus, the variance in the total number of successes is $0.450$ successes squared. As a check, independence gives $P(S=0)=0.70(0.60)=0.42$, $P(S=1)=0.30(0.60)+0.70(0.40)=0.46$, and $P(S=2)=0.30(0.40)=0.12$. These probabilities give $E(S)=0.46+2(0.12)=0.70$ and $E(S^2)=0.46+4(0.12)=0.94$. Hence $\operatorname{Var}(S)=E(S^2)-[E(S)]^2=0.94-0.70^2=0.450$, in agreement.

Worked Example: The Sum of Two Independent Four-Sided Dice

Let $X$ and $Y$ be the outcomes of two independent, fair four-sided dice, each numbered 1 through 4. Let $T=X+Y$, measured in points. Find the variance of $T$.

Check the conditions. Each variable is discrete with four possible values, so its variance is finite. The dice are stipulated to be independent, and their outcomes use the same point scale, so adding them is meaningful.

For either die, each outcome has probability $1/4$, and the mean is $(1+2+3+4)/4=2.5$. Thus,

\begin{aligned} \operatorname{Var}(X) &=\frac{(1-2.5)^2+(2-2.5)^2+(3-2.5)^2+(4-2.5)^2}{4}\\ &=\frac{2.25+0.25+0.25+2.25}{4}\\ &=\frac{5.00}{4}=1.25. \end{aligned}

The same calculation applies to $Y$, so $\operatorname{Var}(Y)=1.25$ points squared. By the addition rule,

\operatorname{Var}(T)=1.25+1.25=2.50

The sum has variance $2.50$ points squared. This calculation does not require finding the probability of every possible total from 2 to 8; the individual variances and independence are enough.

Why Independence Cannot Be Ignored

If the variables are dependent, the cross-product expectation in the derivation need not be zero. When high values of one variable tend to occur with high values of the other, their deviations tend to have the same sign. When one tends to be high while the other is low, the deviations tend to have opposite signs. Either pattern can change the variance of the sum.

Worked Example: A Dependent Pair Does Not Follow the Rule

Let $Z$ be the outcome of one fair six-sided die, and define $X=Z$ and $Y=Z$. Both variables have outcomes 1 through 6, but they are not independent: knowing $X$ determines $Y$. Find the variance of $X+Y$ and compare it with $\operatorname{Var}(X)+\operatorname{Var}(Y)$.

The mean of each variable is $3.5$. Its variance is

\begin{aligned} \operatorname{Var}(X)=\operatorname{Var}(Y) &=\frac{(1-3.5)^2+(2-3.5)^2+(3-3.5)^2 +(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}{6}\\ &=\frac{6.25+2.25+0.25+0.25+2.25+6.25}{6}\\ &=\frac{17.50}{6}=\frac{35}{12}\approx 2.917. \end{aligned}

But $X+Y=2Z$, so its mean is $7$. Each deviation of $X+Y$ from 7 is twice the corresponding deviation of $Z$; squaring multiplies each squared deviation by 4. Therefore,

\operatorname{Var}(X+Y)=4\operatorname{Var}(Z) =4\left(\frac{35}{12}\right)=\frac{35}{3}\approx 11.67.

Simply adding the individual variances would give $35/12+35/12=35/6\approx5.833$, not $35/3$. The discrepancy occurs because the deviations of $X$ and $Y$ are identical, not independent; the cross-product term is positive rather than zero.

Common Mistake and Key Takeaway

A common mistake is to reason, “The expected values of sums add, so the variances of sums must add too.” The conclusion does not follow from linearity of expectation alone. Expected values add without an independence condition, but variances involve a squared deviation, and expanding that square produces a cross-product. Independence is the condition used here to show that the cross-product has expectation zero.

Another error is to apply the addition rule merely because $X$ and $Y$ are described separately. Separate descriptions do not establish independence. The rule is justified when independence is part of the model or is otherwise established. If it is not, the derivation warns that the omitted cross-product may affect the answer.

Key takeaway: For independent discrete random variables with finite variances, expand the squared deviation of their sum. Independence makes the expected cross-product of the centered variables zero, leaving $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$.

Check Your Understanding

Try these questions using the definition and the independence condition.

In the variance expansion for $X+Y$, what is the cross-product term before taking its expectation?
Why does independence make $E[(X-\mu_X)(Y-\mu_Y)]$ equal to zero?
Two independent indicators have success probabilities $0.20$ and $0.50$. Find the variance of their total number of successes.
Two variables each have variance 4. What is the variance of their sum if they are independent? What assumption justifies your calculation?
If $X=Y$ for every outcome, can you use the independent-variance addition rule? Explain why or why not.

What You'll Learn