Linear Combinations of Independent Normal Variables

Why Weighted Normal Sums Are Useful

A normal model can describe several sources of random variation that contribute to a total. For example, a hypothetical insurer might model two independent components of a claim-related cost as normal variables, then form a net amount by applying different weights and adding a fixed adjustment. To calculate a probability for that net amount, the key task is to find its distribution—not to approximate it by adding probabilities or to treat a fixed adjustment as random.

This tutorial develops the distribution of a weighted sum of independent normal variables. The coefficients may be positive, negative, or zero, and a fixed intercept may be included. Once the mean and variance have been found, probabilities follow by standardizing the resulting normal variable.

Definition: Suppose \(X_1,\ldots,X_n\) are mutually independent, with \(X_j\sim N(\mu_j,\sigma_j^2)\) and \(\sigma_j^2>0\). For fixed real coefficients \(a_1,\ldots,a_n\) and fixed constant \(c\), define \(W=c+\sum_{j=1}^n a_jX_j\). If at least one coefficient is nonzero, then \(W\sim N\left(c+\sum_{j=1}^n a_j\mu_j,\ \sum_{j=1}^n a_j^2\sigma_j^2\right)\). If every coefficient is zero, then \(W=c\) with probability 1; it is a constant, not a nondegenerate normal variable.

How Coefficients Affect the Distribution

The mean and variance follow different rules. Each coefficient \(a_j\) multiplies the corresponding mean, so its sign matters: a negative coefficient reverses the contribution to the mean. In the variance, the coefficient is squared. Thus a negative coefficient still contributes a nonnegative amount to the variance.

E[W]=c+\sum_{j=1}^n a_j\mu_j, \qquad \operatorname{Var}(W)=\sum_{j=1}^n a_j^2\sigma_j^2.

The fixed constant \(c\) shifts every possible value by the same amount. It changes the mean by \(c\), but adds no variance. A coefficient of zero removes that variable from the sum altogether.

For probability calculations, let \(\Phi(z)\) denote the cumulative distribution function of a standard normal variable, \(Z\sim N(0,1)\). If \(W\sim N(\mu_W,\sigma_W^2)\) with \(\sigma_W>0\), then \(\frac{W-\mu_W}{\sigma_W}\sim N(0,1)\). In particular,

P(W\leq w)=\Phi\left(\frac{w-\mu_W}{\sigma_W}\right), \qquad P(W>w)=1-\Phi\left(\frac{w-\mu_W}{\sigma_W}\right).

For a continuous normal variable, \(P(W>w)=P(W\geq w)\), because the probability of equaling any single value is zero. For an interval, subtract the two cumulative probabilities: \(P(r<W<s)=\Phi\left(\frac{s-\mu_W}{\sigma_W}\right)-\Phi\left(\frac{r-\mu_W}{\sigma_W}\right)\).

Why the Weighted Sum Is Normal

The normal result can be justified with moment generating functions (MGFs). The MGF of \(X_j\sim N(\mu_j,\sigma_j^2)\) is \(M_{X_j}(t)=\exp(\mu_jt+\sigma_j^2t^2/2)\), for real \(t\). For a fixed coefficient, \(M_{a_jX_j}(t)=M_{X_j}(a_jt)\). Adding the intercept multiplies the MGF by \(e^{ct}\).

Apply the intercept and coefficients.
For each term, the MGF contribution is \(M_{a_jX_j}(t)=\exp(a_j\mu_jt+a_j^2\sigma_j^2t^2/2)\). The intercept contributes \(e^{ct}\).

Use mutual independence.
The MGF of the sum is the product of the individual MGFs, so \(M_W(t)=e^{ct}\prod_{j=1}^n M_{X_j}(a_jt)\).

Collect terms in the exponent.
Combining the factors gives \(M_W(t)=\exp\left[\left(c+\sum_{j=1}^n a_j\mu_j\right)t+\frac{t^2}{2}\sum_{j=1}^n a_j^2\sigma_j^2\right]\). This is the MGF of the stated normal distribution.

Because this MGF exists for all real \(t\) and matches the MGF of a normal distribution, it identifies that distribution. The squared coefficients in the exponent also show directly why negative weights do not make variance contributions negative.

Worked Applications

Worked Example: A Positive Weight, a Negative Weight, and a Tail Probability

In a hypothetical cost model, \(X\) and \(Y\) are independent, with \(X\sim N(50,9)\) and \(Y\sim N(30,4)\), where the second parameter is the variance. A net measure is \(W=2X-Y+5\). Find its distribution and calculate \(P(W>85)\).

Method. Apply the weighted-sum result, then standardize the threshold. The common mistake to avoid is subtracting \(Y\)'s variance because its coefficient is negative.

Mean. The intercept is 5, and the coefficients of \(X\) and \(Y\) are 2 and \(-1\): \(E[W]=2(50)-1(30)+5=100-30+5=75\).

Variance. Square each coefficient before multiplying by its variable's variance: \(\operatorname{Var}(W)=2^2(9)+(-1)^2(4)=36+4=40\). Thus \(W\sim N(75,40)\), with standard deviation \(\sqrt{40}\).

Probability. Standardize 85: \(\frac{85-75}{\sqrt{40}}=\frac{10}{\sqrt{40}}\approx1.5811\). Therefore \(P(W>85)=1-\Phi(1.5811)\approx0.0569\), to four decimal places.

Check. The computed mean is 75, and 85 is above that mean, so a probability below 0.5 is consistent. The variance is positive, as required.

Worked Example: Several Terms with a Negative Coefficient

Let \(X,Y,Z\) be mutually independent, with \(X\sim N(10,4)\), \(Y\sim N(6,9)\), and \(Z\sim N(2,1)\). Define \(Q=X-2Y+3Z-4\). Find the distribution of \(Q\) and calculate \(P(Q\leq-7)\).

Method. Compute the mean using signed coefficients and the variance using squared coefficients, then standardize. The main trap is to use \(-2\), rather than \((-2)^2\), in the variance calculation.

Mean. The intercept is \(-4\): \(E[Q]=10-2(6)+3(2)-4=10-12+6-4=0\).

Variance. \(\operatorname{Var}(Q)=1^2(4)+(-2)^2(9)+3^2(1)=4+36+9=49\). Therefore \(Q\sim N(0,49)\), with standard deviation 7.

Probability. The standardized cutoff is \(\frac{-7-0}{7}=-1\). Hence \(P(Q\leq-7)=\Phi(-1)\approx0.1587\), to four decimal places.

Check. The cutoff is exactly one standard deviation below the mean, so its lower-tail probability is less than 0.5, as the answer indicates.

Worked Example: An Intercept and a Two-Standard-Deviation Tail

Suppose independent measurements \(X\) and \(Y\) satisfy \(X\sim N(100,25)\) and \(Y\sim N(40,4)\). A hypothetical combined index is \(R=0.8X+1.5Y+12\). Find its distribution and calculate \(P(R>162)\).

Method. Form the weighted mean and variance, including the fixed intercept only in the mean. Then use the standard normal upper tail.

Mean. \(E[R]=0.8(100)+1.5(40)+12=80+60+12=152\).

Variance. \(\operatorname{Var}(R)=0.8^2(25)+1.5^2(4)=0.64(25)+2.25(4)=16+9=25\). Thus \(R\sim N(152,25)\), with standard deviation 5.

Probability. \(\frac{162-152}{5}=2\), so \(P(R>162)=1-\Phi(2)\approx0.0228\), to four decimal places.

Check. The cutoff 162 is two standard deviations above the mean, so a small upper-tail probability is reasonable. The intercept shifts the mean from 140 to 152 but does not change the variance.

Assumptions, Dependence, and Common Exam Traps

Mutual independence is a sufficient condition for the formula above: it guarantees both the stated variance formula and the normal distribution of the sum. Independence is not necessary for either result in every setting. If variables are dependent, the sum need not be normal in general. If the inputs are jointly normal, however, a linear combination is normal, and its variance includes covariance terms.

\operatorname{Var}\left(c+\sum_{j=1}^n a_jX_j\right) = \sum_{j=1}^n a_j^2\operatorname{Var}(X_j) + 2\sum_{1\leq j<k\leq n}a_ja_k\operatorname{Cov}(X_j,X_k).

Thus, for dependent inputs, omitting covariance terms is generally incorrect. Zero covariance is enough for the displayed variance to reduce to the sum of the individual variance terms, but zero covariance alone does not ensure that a dependent sum is normal.

A frequent exam error is to calculate the variance as though coefficients affected it linearly. For example, with independent \(X\sim N(50,9)\) and \(Y\sim N(30,4)\), the variance of \(2X-Y+5\) is not \(2(9)-4=14\). That wrong calculation fails because it does not square the coefficients and even produces a negative contribution from the negative weight. The correct calculation is \(\operatorname{Var}(2X-Y+5)=2^2(9)+(-1)^2(4)=40\). The intercept contributes no variance.

Before evaluating a probability, check that the distribution is nondegenerate, identify whether the requested event is an upper or lower tail, and use the standard deviation—not the variance—in the denominator when standardizing. These checks catch several common errors at once.

Key takeaway: For a weighted sum of mutually independent normal variables, add the coefficient-weighted means and add the squared-coefficient-weighted variances. Add an intercept to the mean only. A negative coefficient changes the mean contribution's sign but is squared in the variance.

Check Your Understanding

For each problem, identify the relevant weighted-sum distribution before evaluating any requested probability.

Let independent \(X\sim N(4,9)\) and \(Y\sim N(1,4)\), and define \(T=X+2Y-3\). Find \(T\)'s distribution and \(P(T>8)\). (Answer: \(T\sim N(3,25)\); \(P(T>8)=0.1587\).)
Let independent \(A\sim N(20,16)\) and \(B\sim N(5,9)\), and define \(R=3A-2B+1\). Find \(R\)'s distribution and \(P(R<39)\). (Answer: \(R\sim N(51,180)\); \(P(R<39)=0.1855\).)
Let independent \(C\sim N(12,4)\) and \(D\sim N(8,1)\), and define \(S=-C+4D+6\). Find \(S\)'s distribution and \(P(S\geq30)\). (Answer: \(S\sim N(26,20)\); \(P(S\geq30)=0.1855\).)
Let \(X,Y,Z\) be mutually independent standard normal variables, and set \(V=2X-Y+2Z+4\). Find \(V\)'s distribution and \(P(1<V<7)\). (Answer: \(V\sim N(4,9)\); \(P(1<V<7)=0.6827\).)
Suppose \(U,V\) are jointly normal with means 3 and \(-1\), variances 4 and 9, and covariance 2. For \(H=2U-V+5\), find its distribution. (Answer: \(H\sim N(12,17)\).)

What You'll Learn