Linear Combinations Of Correlated Normal Variables

Why Covariance Matters in a Normal Combination

A sum or difference of random variables often represents a quantity of direct interest: a portfolio total, a difference between two measurements, or a weighted combination of risks. If the variables are correlated, their dependence affects the variability of that result. Ignoring this dependence can substantially misstate a probability, even when each variable separately has a normal distribution.

The key setting for an exact normal calculation is that the variables are jointly normal. Under that assumption, every linear combination is normal, and its variance is determined by both the individual variances and the covariances. This tutorial develops the calculation for two variables and for a vector of variables, then applies it to probability questions. It is aligned with the SOA Exam P (Probability) syllabus.

For two variables, the covariance is \(\operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])]\). If the standard deviations are \(\sigma_X\) and \(\sigma_Y\), their correlation is \(\rho_{X,Y}=\operatorname{Cov}(X,Y)/(\sigma_X\sigma_Y)\), so \(\operatorname{Cov}(X,Y)=\rho_{X,Y}\sigma_X\sigma_Y\). Correlation is dimensionless; covariance carries the product of the variables’ units.

Distribution of a Linear Combination

Suppose \(X\) and \(Y\) are jointly normal, with \(X\sim N(\mu_X,\sigma_X^2)\), \(Y\sim N(\mu_Y,\sigma_Y^2)\), and covariance \(\operatorname{Cov}(X,Y)\). For fixed constants \(a\) and \(b\), the combination \(W=aX+bY\) is normal. Its mean and variance are:

The sign of the covariance contribution depends on the coefficients as well as on the covariance itself. For a difference \(X-Y\), the coefficient product is negative, so a positive covariance reduces the variance of the difference. A negative covariance instead increases it.

The final step uses the definition of covariance. Joint normality then supplies the additional fact that \(aX+bY\) has a normal distribution with this mean and variance. The variance identity itself does not require normality; the claim that the combination is normal does.

Several Variables and the Covariance Matrix

For more than two variables, the same principle applies to every pair. Let the vector of jointly normal variables be \(\mathbf{X}=(X_1,\ldots,X_n)^\mathsf{T}\), with mean vector \(\boldsymbol{\mu}\) and covariance matrix \(\boldsymbol{\Sigma}\). The diagonal entries of \(\boldsymbol{\Sigma}\) are the variances; each off-diagonal entry is the covariance of the corresponding pair. For fixed coefficient vector \(\mathbf{a}\), the combination \(L=\mathbf{a}^\mathsf{T}\mathbf{X}\) has mean \(\mathbf{a}^\mathsf{T}\boldsymbol{\mu}\) and variance \(\mathbf{a}^\mathsf{T}\boldsymbol{\Sigma}\mathbf{a}\).

The matrix expression is a compact way to include each variance and covariance contribution exactly once. In the expanded sum, the factor of \(2\) appears because each covariance would arise twice when the quadratic expression is expanded.

When the Normal Result Applies—and When It Does Not

There are two separate questions to answer: what are the mean and variance of the combination, and is its distribution normal? The mean and variance formulas apply to variables with finite second moments whether or not they are normal. An exact normal distribution for the combination is guaranteed when the variables are jointly normal.

Normal marginals alone do not guarantee joint normality. Dependence can be arranged so that each variable separately has a normal distribution while a linear combination does not. Therefore, do not conclude that a sum or difference is normal merely because each component is described as normal; look for a joint normal model or an explicit statement that the variables are jointly normal.

For jointly normal variables, zero covariance between a pair implies that pair is independent. This equivalence is special to the joint normal model; in general, zero covariance does not establish independence. Conversely, independence of two variables with normal distributions implies they are jointly normal, so the same calculation applies.

Common Exam Traps

• Assuming independence without evidence. Independence sets the covariance to zero; it is not implied by having separate normal distributions.
• Dropping the coefficient signs. In a difference, a negative coefficient can reverse the sign of a covariance contribution.
• Using correlation as covariance. Convert using \(\operatorname{Cov}(X,Y)=\rho_{X,Y}\sigma_X\sigma_Y\) before inserting the value into a variance formula.
• Using variance where standard deviation is required. Standardization divides by the square root of the variance, not by the variance itself.
• Claiming an exact normal result too soon. The variance calculation is general, but the exact normal distribution requires joint normality or another sufficient model assumption.