Simplifying Variances Under Independence

Why Independence Simplifies a Variance

When a total is built from several random quantities, its variance depends not only on the variance of each quantity but also on how the quantities vary together. Independence can simplify the calculation: it eliminates the cross-covariance terms, leaving a sum of individually weighted variances. The simplification is useful only when the independence and moment assumptions actually hold.

For example, a hypothetical operation might combine random costs, recoveries, or event counts. A negative coefficient can represent a credit or offset. In each case, the variance concerns the variability of the total, not simply the variability of each component considered separately.

Definition: Let $X_1,\ldots,X_n$ be random variables and let $a_1,\ldots,a_n$ be fixed constants. The weighted sum is $T=\sum_{i=1}^n a_iX_i$. If each $X_i$ has a finite second moment, its variance is $$ \operatorname{Var}(T) = \sum_{i=1}^n a_i^2\operatorname{Var}(X_i) + 2\sum_{1\leq i<j\leq n}a_i a_j\operatorname{Cov}(X_i,X_j). $$ If every distinct pair $X_i,X_j$ is independent, then each cross-covariance is zero, so $$ \operatorname{Var}(T)=\sum_{i=1}^n a_i^2\operatorname{Var}(X_i). $$

The second-moment condition means $E[X_i^2]<\infty$ for every variable included in the sum. It ensures that each variance and covariance in the formula exists and is finite. Pairwise independence is sufficient to eliminate all the cross terms; mutual independence is a stronger condition than this variance calculation requires. Independence does not require the variables to have the same distributions or variances.

The formula is for fixed coefficients. If a coefficient is itself random, it cannot simply be treated as a constant and pulled outside the variance. Likewise, if some component has an infinite second moment, the finite-variance formula cannot be applied as though all terms were ordinary finite numbers.

Why the Weighted-Sum Formula Holds

The general formula follows by expanding the centered weighted sum. This algebraic step does not assume independence; dependence is represented by the covariance terms.

\begin{aligned} \operatorname{Var}(T) &=E\left[\left(T-E[T]\right)^2\right]\\ &=E\left[\left(\sum_{i=1}^n a_i(X_i-E[X_i])\right)^2\right]\\ &=\sum_{i=1}^n a_i^2E[(X_i-E[X_i])^2] +2\sum_{1\leq i<j\leq n}a_i a_j E[(X_i-E[X_i])(X_j-E[X_j])]\\ &=\sum_{i=1}^n a_i^2\operatorname{Var}(X_i) +2\sum_{1\leq i<j\leq n}a_i a_j\operatorname{Cov}(X_i,X_j). \end{aligned}

The cross terms appear twice in the square: once as the product involving $X_i$ followed by $X_j$, and once in the reverse order. Since multiplication is commutative, the two contributions combine into the factor of $2$. A standard result is that independent variables with finite second moments have covariance zero. Thus, when each distinct pair is independent, every cross term in the expansion vanishes.

A fixed constant added to the total does not affect its variance. In particular, for $T=c+\sum_i a_iX_i$, the same variance formula applies: centering subtracts $c$ along with the expected value, so the constant disappears.

Applying Independence Carefully

For independent variables, the practical calculation has three main steps: verify the variables and their moment conditions, calculate each component variance, and multiply each variance by the square of its coefficient. Do not add standard deviations, and do not use the unsquared coefficients for the individual variance terms.

Check the model.
Confirm that the quantities are random variables with fixed coefficients, that every required second moment is finite, and that each pair used in the calculation is independent.

Find the component variances.
Use the appropriate distribution and parameters. For a Bernoulli variable with success probability $p$, the variance is $p(1-p)$; for a Poisson variable with parameter $\lambda$, it is $\lambda$.

Square the weights and add.
Compute $a_i^2\operatorname{Var}(X_i)$ for each component and sum the results. A negative coefficient is squared in these terms.

The sign of a coefficient still matters in the general formula when variables are dependent: a cross term contains $a_i a_j$, not $a_i^2$ or $a_j^2$. Under pairwise independence the cross terms vanish, so a negative weight has no effect on its component's contribution to the variance. It can still affect the expected value of the weighted sum.

Worked Example: Combining Independent Event Costs and a Recovery

A hypothetical one-period operating budget records three event indicators. Let $I_1\sim\operatorname{Bernoulli}(0.20)$, $I_2\sim\operatorname{Bernoulli}(0.10)$, and $I_3\sim\operatorname{Bernoulli}(0.40)$, and assume these indicators are mutually independent. The net amount, in dollars, is $C=1000I_1+2000I_2-500I_3$. The last term is a possible recovery. This is not an insurance or mortality model: no interest rate or mortality assumption is involved.

Method and assumptions. Apply the independent weighted-sum variance formula. A Bernoulli variable with parameter $p$ has variance $p(1-p)$; all three variables have finite second moments.

Component variances.

\begin{aligned} \operatorname{Var}(I_1)&=0.20(1-0.20)=0.16,\\ \operatorname{Var}(I_2)&=0.10(1-0.10)=0.09,\\ \operatorname{Var}(I_3)&=0.40(1-0.40)=0.24. \end{aligned}

Variance of the net amount. The coefficients are $1000$, $2000$, and $-500$, so their squares are $1{,}000{,}000$, $4{,}000{,}000$, and $250{,}000$.

\begin{aligned} \operatorname{Var}(C) &=1000^2(0.16)+2000^2(0.09)+(-500)^2(0.24)\\ &=160{,}000+360{,}000+60{,}000\\ &=580{,}000\text{ dollars}^2. \end{aligned}

As a check, the expected net amount is $1000(0.20)+2000(0.10)-500(0.40)=200$ dollars. This does not change the variance. The calculation avoids a common error: the recovery coefficient is negative, but its individual variance contribution is positive because the coefficient is squared.

Worked Example: Weighted Independent Poisson Counts

A hypothetical monitoring system records independent counts $N_1\sim\operatorname{Poisson}(2)$, $N_2\sim\operatorname{Poisson}(3)$, and $N_3\sim\operatorname{Poisson}(1)$ over one fixed observation period. Define $R=4N_1+2N_2-3N_3$, where each unit of $R$ represents \$100. There is no interest accumulation or mortality assumption in this count model.

Method and assumptions. For $N\sim\operatorname{Poisson}(\lambda)$, $E[N]=\operatorname{Var}(N)=\lambda$. Each Poisson parameter here is finite, so each count has a finite second moment. The counts are independent, so there are no cross-covariance terms.

The component variances are $2$, $3$, and $1$, respectively. Squaring the weights gives $16$, $4$, and $9$.

\begin{aligned} \operatorname{Var}(R) &=4^2(2)+2^2(3)+(-3)^2(1)\\ &=32+12+9\\ &=53\text{ units}^2. \end{aligned}

Since one unit represents \$100, the variance in dollars squared is $53(100)^2=530{,}000$ dollars squared. The expected total in the original units is $4(2)+2(3)-3(1)=11$ units, or \$1,100. As a check, the negative weight reduces the expected total but contributes $9$, not $-9$, units squared to the variance.

Independence Must Apply to the Relevant Pairs

A collection of variables may contain independent pairs without every pair being independent. In that case, remove only the cross terms for independent pairs. For any pair that is dependent, its covariance term remains unless its covariance is otherwise known to be zero. Do not declare a whole sum independent merely because some of its components are independent.

The following example shows why checking each pair matters. It also illustrates that the independent-sum shortcut does not apply to every pair just because some terms in the total are independent.

Worked Example: One Dependent Pair in a Weighted Sum

Let $X$ and $Y$ be independent $\operatorname{Bernoulli}(0.5)$ variables, and define $Z=Y$. Consider $Q=2X+3Y-Z$. The model is a hypothetical one-period indicator calculation; no interest rate or mortality assumption is involved. Both $Y$ and $Z$ have the $\operatorname{Bernoulli}(0.5)$ distribution, but they are the same random variable, not independent copies.

Method and assumptions. Use the full weighted-sum formula. Each variable has variance $0.5(0.5)=0.25$. Because $X$ is independent of $Y$, and $Z=Y$, the pairs $(X,Y)$ and $(X,Z)$ are independent. But $Y$ and $Z$ are dependent: $Z=Y$, so $\operatorname{Cov}(Y,Z)=\operatorname{Var}(Y)=0.25$.

With coefficients $2$, $3$, and $-1$, the three individual variance contributions sum to $2^2(0.25)+3^2(0.25)+(-1)^2(0.25)=1+2.25+0.25=3.5$. The $X,Y$ and $X,Z$ cross terms are zero. The remaining cross term is $2(3)(-1)(0.25)=-1.5$.

\operatorname{Var}(Q)=3.5+0+0-1.5=2.

A direct check is possible because $Q=2X+3Y-Y=2X+2Y$. Its four equally likely values are $0,2,2,4$, with mean $2$; the average squared deviation from $2$ is $(4+0+0+4)/4=2$. Treating $Y$ and $Z$ as independent would incorrectly discard their nonzero cross term.

Common Mistakes and Exam Traps

Using the weight instead of its square. For independent variables, the component contribution is $a_i^2\operatorname{Var}(X_i)$, not $a_i\operatorname{Var}(X_i)$. For instance, a coefficient of $4$ multiplies a variance by $16$, not $4$.
Adding standard deviations. Independent variances add after each has been weighted by the squared coefficient. Standard deviations do not generally add.
Assuming independence without support from the model. Knowing the separate distributions does not establish that the variables are independent. If dependence is present or unspecified, do not automatically set covariance terms to zero.
Dropping every covariance term when only some pairs are independent. Check the relationship for each pair in the sum. A dependent pair can retain a nonzero cross term.
Applying a finite-variance formula without finite second moments. Finite means alone do not guarantee that variances are finite. Check $E[X_i^2]<\infty$ for every component before using the formula as a finite numerical calculation.

Key takeaway: For a finite weighted sum with finite component second moments, expand the variance into squared-weight variance terms and pairwise covariance terms. Pairwise independence removes those cross terms, giving $\operatorname{Var}(\sum_i a_iX_i)=\sum_i a_i^2\operatorname{Var}(X_i)$. Verify which pairs are independent before simplifying.

Check Your Understanding

For each problem, identify the applicable variance rule and check the independence and moment assumptions before calculating.

Let $X\sim\operatorname{Bernoulli}(0.30)$ and $Y\sim\operatorname{Bernoulli}(0.60)$ be independent. Find $\operatorname{Var}(5X-2Y+7)$. (Answer: $6.21$)
Independent counts $N_1\sim\operatorname{Poisson}(1.5)$ and $N_2\sim\operatorname{Poisson}(4)$ are combined as $S=3N_1+N_2$. Each unit represents \$10. Find the variance in units squared and in dollars squared. (Answer: $17.5$ units squared, or $1{,}750$ dollars squared)
Independent variables satisfy $X_1\sim\operatorname{Binomial}(4,0.25)$ and $X_2\sim\operatorname{Binomial}(2,0.50)$. Find $\operatorname{Var}(2X_1-X_2)$. (Answer: $3.5$)
Let $A$ and $B$ be independent $\operatorname{Bernoulli}(0.40)$ variables, and set $C=A$. Find $\operatorname{Var}(A+2B-C)$, taking account of the dependence between $A$ and $C$. (Answer: $0.96$)
Suppose $U$ and $V$ are independent, $E[U]$ is finite but $E[U^2]=\infty$, and $V$ has finite variance. Is the finite independent weighted-sum variance formula valid as a finite calculation for $3U-4V$? (Answer: No; $U$ does not have a finite second moment, so the finite-variance formula does not apply, and $\operatorname{Var}(3U-4V)$ is infinite.)

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