Distributions of Linear Combinations

Why Linear Combinations Need the Covariance

In a bivariate normal model, many quantities of interest are weighted totals, differences, or other linear combinations of two random variables. Examples include a portfolio value built from two normally distributed positions or a net amount formed by subtracting one random cost from another. The result is still normally distributed, but its variance depends on how the original variables move together. Ignoring that dependence can substantially change a probability calculation.

Throughout, let $X$ and $Y$ have a bivariate normal distribution, with means $\mu_X$ and $\mu_Y$, variances $\sigma_X^2$ and $\sigma_Y^2$, and covariance $\operatorname{Cov}(X,Y)$. Let $a$ and $b$ be fixed real numbers, and define $W=aX+bY$. The coefficients are constants, not random variables.

Result: For a bivariate normal pair $(X,Y)$, the linear combination $W=aX+bY$ is normally distributed, with $$ E[W]=a\mu_X+b\mu_Y $$ and $$ \operatorname{Var}(W)=a^2\sigma_X^2+b^2\sigma_Y^2+2ab\operatorname{Cov}(X,Y). $$ If the variance is positive, $W\sim N(E[W],\operatorname{Var}(W))$. The second parameter in this notation is the variance.

Why the Formula Holds

The mean follows from linearity of expectation. For the variance, subtract the mean of $W$ and expand the square. This makes the cross-term, which accounts for dependence, explicit:

\begin{aligned} W-E[W]&=a(X-\mu_X)+b(Y-\mu_Y),\\ \operatorname{Var}(W) &=E\left[\left(a(X-\mu_X)+b(Y-\mu_Y)\right)^2\right]\\ &=a^2\sigma_X^2+b^2\sigma_Y^2 +2abE[(X-\mu_X)(Y-\mu_Y)]\\ &=a^2\sigma_X^2+b^2\sigma_Y^2+2ab\operatorname{Cov}(X,Y). \end{aligned}

The normality assertion can be seen from a standard representation of a bivariate normal pair. When $|\rho|<1$, where $\rho$ is the correlation, the pair can be represented using independent standard normal variables $U$ and $V$:

X=\mu_X+\sigma_XU,\qquad Y=\mu_Y+\sigma_Y\bigl(\rho U+\sqrt{1-\rho^2}\,V\bigr).

Substituting these expressions into $W$ writes it as a constant plus a linear combination of the independent standard normals $U$ and $V$. The moment-generating function of that combination is the product of their moment-generating functions:

E[e^{tW}] =\exp\left(tE[W]+\frac{t^2}{2} \left((a\sigma_X+b\rho\sigma_Y)^2+ b^2\sigma_Y^2(1-\rho^2)\right)\right).

This is the moment-generating function of a normal variable. Expanding the variance expression in parentheses gives $a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y$, and $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$. If $\rho=1$ or $\rho=-1$, the pair is degenerate and the same conclusion follows because $Y$ is an affine function of $X$; the linear combination is then normal or constant. Thus the result also covers the boundary correlations.

Using Covariance and Correlation Correctly

The variance formula requires covariance, not correlation by itself. If a problem gives correlation $\rho$, first convert it using $ \operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y $. Then substitute the covariance and the signed coefficients into the formula. The term $2ab\operatorname{Cov}(X,Y)$ may be positive or negative, depending on both the coefficients and the covariance.

Identify the combination.
Write the target variable as $W=aX+bY$, preserving any minus signs.

Find the covariance if needed.
If correlation is supplied, calculate $\rho\sigma_X\sigma_Y$. Do not insert $\rho$ directly into the variance formula.

Calculate the mean and variance.
Use the original means, variances, and covariance. Square each coefficient in its variance term, but not in the cross-term.

Standardize if a probability is required.
For positive variance, use $Z=(W-E[W])/\sqrt{\operatorname{Var}(W)}$, which has the standard normal distribution.

Worked Example: A Difference With Negative Covariance

In a hypothetical model, two normal measurements satisfy $X\sim N(10,4)$ and $Y\sim N(4,9)$, with $\operatorname{Cov}(X,Y)=-1.5$. Define $W=3X-2Y$. Find its distribution and calculate $P(W>31)$.

Method and assumptions. The pair $(X,Y)$ is bivariate normal, the coefficients are fixed, and the supplied second parameters are variances. Therefore $W$ is normal. Its mean is $3(10)-2(4)=30-8=22$.

The coefficients are $a=3$ and $b=-2$. The variance is

\begin{aligned} \operatorname{Var}(W) &=3^2(4)+(-2)^2(9)+2(3)(-2)(-1.5)\\ &=36+36+18\\ &=90. \end{aligned}

Thus $W\sim N(22,90)$. Standardizing the requested threshold gives $z=(31-22)/\sqrt{90}=9/9.4868\approx0.9487$. Consequently, $P(W>31)=1-\Phi(0.9487)\approx0.1714$, to four decimal places.

Check. The covariance is negative and the coefficients have opposite signs, so their product with the covariance is positive. The variance increases by 18 in the cross-term, consistent with the calculation. A common mistake is to assume that negative covariance must always reduce the variance; its effect also depends on the signs of the coefficients.

Worked Example: Converting Correlation Before Calculating

A hypothetical model for two amounts, measured in thousands of dollars, has $X\sim N(50,100)$, $Y\sim N(30,64)$, and correlation $\rho=-0.25$. The amounts are jointly normal. Let $W=X-2Y$. Find the distribution of $W$ and $P(W>20)$.

Method and assumptions. First convert correlation to covariance. The standard deviations are $\sigma_X=10$ and $\sigma_Y=8$, so $\operatorname{Cov}(X,Y)=(-0.25)(10)(8)=-20$, in squared thousands of dollars.

The mean is $E[W]=50-2(30)=-10$. Its variance is

\begin{aligned} \operatorname{Var}(W) &=1^2(100)+(-2)^2(64)+2(1)(-2)(-20)\\ &=100+256+80\\ &=436. \end{aligned}

Therefore $W\sim N(-10,436)$. For the threshold 20, $z=(20-(-10))/\sqrt{436}=30/20.8806\approx1.4367$. It follows that $P(W>20)=1-\Phi(1.4367)\approx0.0754$, to four decimal places.

Check. The coefficient product $1(-2)$ is negative, as is the covariance, so the cross-term is positive. The common mistake is to use $-0.25$ as though it were the covariance, producing a cross-term with the wrong scale.

Worked Example: An Interval Probability for a Weighted Combination

A hypothetical pair of jointly normal measurements has $X\sim N(8,9)$, $Y\sim N(5,4)$, and $\operatorname{Cov}(X,Y)=3$. Set $W=2X-4Y$. Find the distribution of $W$ and the probability that $-10<W<3$.

Method and assumptions. The variables are jointly normal and the weights are fixed, so the linear combination is normal. Its mean is $2(8)-4(5)=16-20=-4$. Its variance is

\begin{aligned} \operatorname{Var}(W) &=2^2(9)+(-4)^2(4)+2(2)(-4)(3)\\ &=36+64-48\\ &=52. \end{aligned}

Hence $W\sim N(-4,52)$. Standardize both endpoints using the same mean and standard deviation:

\begin{aligned} P(-10<W<3) &=P\left(\frac{-10-(-4)}{\sqrt{52}}< \frac{W-(-4)}{\sqrt{52}}< \frac{3-(-4)}{\sqrt{52}}\right)\\ &=P(-0.8321<Z<0.9707)\\ &=\Phi(0.9707)-\Phi(-0.8321)\\ &\approx0.8341-0.2027=0.6314. \end{aligned}

The probability is $0.6314$, to four decimal places. Check. The covariance is positive, but the coefficients have opposite signs, so the cross-term is negative. The mistake of making it positive would overstate the variance and alter both standardized endpoints.

Common Traps and When the Result Applies

Dropping the cross-term. The shortcut $a^2\sigma_X^2+b^2\sigma_Y^2$ is valid only when the covariance is zero. A bivariate normal pair with zero covariance is independent, but general pairs with zero covariance need not be.
Losing a coefficient sign. The cross-term uses $ab$, including its sign. In a difference $aX-bY$, the second coefficient is negative.
Confusing variance and standard deviation. The formula takes $\sigma_X^2$ and $\sigma_Y^2$ as inputs; only standardization divides by the square root of the resulting variance.
Assuming normality without a joint model. The mean-variance formulas hold for any pair with finite second moments, but the conclusion that $aX+bY$ is normal is guaranteed here by the bivariate normal assumption. Normal marginals alone do not guarantee a bivariate normal pair.

If the calculated variance is zero, $W$ is constant with probability one, rather than a nondegenerate normal variable; probabilities should then be evaluated from that constant directly. Otherwise, the result provides a complete distribution and reduces one-dimensional threshold or interval questions to standard normal probabilities. It is aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each problem, use the stated joint normal model and retain the signs of the coefficients.

A jointly normal pair has $X\sim N(2,4)$, $Y\sim N(1,9)$, and $\operatorname{Cov}(X,Y)=1$. For $T=X+2Y$, find its distribution. (Answer: $T\sim N(4,44)$.)
A jointly normal pair has $X\sim N(6,16)$, $Y\sim N(3,25)$, and correlation $0.2$. For $W=2X-Y$, find $P(W<0)$. (Answer: approximately $0.1461$.)
A jointly normal pair has $X\sim N(4,9)$, $Y\sim N(2,4)$, and covariance $-1.5$. For $Q=X-2Y$, find $P(Q>5)$. (Answer: approximately $0.1847$.)
A jointly normal pair has $X\sim N(20,25)$, $Y\sim N(10,16)$, and correlation $0.5$. For $T=X+Y$, find $P(T<24)$. (Answer: approximately $0.2212$.)
A jointly normal pair has $X\sim N(0,4)$, $Y\sim N(0,9)$, and covariance $3$. Find all real $c$ for which $\operatorname{Var}(2X+cY)=25$. (Answer: $c=(-2+\sqrt{13})/3$ or $c=(-2-\sqrt{13})/3$.)

What You'll Learn