Probabilities of Sums and Differences

Why Sums and Differences Need Their Own Calculation

A sum or difference of two measurements can be useful when the quantities are correlated. For example, a hypothetical analyst might compare two normal estimates by taking their difference, or combine two normal components to model a total. The probability for that total or difference depends not only on each component’s variance, but also on how the components vary together.

The key procedure is to treat the sum or difference as a new random variable. Find its mean and variance using the means, variances, and covariance of the pair. Because a linear combination of a bivariate normal pair is normal, the resulting variable can then be standardized and evaluated with the standard normal cumulative distribution function, denoted by $\Phi$.

Definition: Let $(X,Y)$ have a bivariate normal distribution with $E[X]=\mu_X$, $E[Y]=\mu_Y$, variances $\sigma_X^2$ and $\sigma_Y^2$, and correlation $\rho$. Then $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$. The sum $S=X+Y$ and difference $D=X-Y$ are normal random variables, with $$ E[S]=\mu_X+\mu_Y,\qquad \operatorname{Var}(S)=\sigma_X^2+\sigma_Y^2+2\operatorname{Cov}(X,Y), $$ $$ E[D]=\mu_X-\mu_Y,\qquad \operatorname{Var}(D)=\sigma_X^2+\sigma_Y^2-2\operatorname{Cov}(X,Y). $$

Deriving the Mean and Variance Formulas

Expectation is linear, so adding or subtracting the means gives the mean of the new variable. The variance formulas follow by expanding the variance of a sum or difference. In particular, the covariance term has a positive sign for a sum and a negative sign for a difference.

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

For the difference, expanding the square of $(X-\mu_X)-(Y-\mu_Y)$ changes the sign of the cross term:

\operatorname{Var}(X-Y) =\sigma_X^2+\sigma_Y^2-2\operatorname{Cov}(X,Y).

These variance identities hold for any pair with finite variances. The additional bivariate normal assumption is what ensures that the sum or difference is itself normal, so a standard normal calculation gives its probabilities. If the pair is not jointly normal, knowing its means, variances, and covariance alone does not generally determine the distribution of the sum or difference.

Standardizing a Sum or Difference

Once the mean and variance are found, use the normal standardization rule. If $W$ is normal with mean $\mu_W$ and variance $\sigma_W^2>0$, then $Z=(W-\mu_W)/\sigma_W$ has the standard normal distribution. Remember that the standard deviation in the denominator is the square root of the variance.

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

For an interval, standardize both endpoints and subtract the lower-tail probabilities. The same procedure applies to either $S=X+Y$ or $D=X-Y$. For example, $P(a<D<b)$ is found by evaluating the standardized lower and upper endpoints of $D$; no bivariate integration is needed for this event.

Identify the new variable.
Write $S=X+Y$ or $D=X-Y$, matching the event in the question.

Find the covariance.
Use the stated covariance, or calculate $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$.

Calculate its mean and variance.
Use the plus covariance term for $S$ and the minus covariance term for $D$.

Standardize the event.
Use the standard deviation, not the variance, in the denominator. Apply $\Phi$ or its upper-tail complement as required.

Worked Probability Calculations

Worked Example: An Upper-Tail Probability for a Sum

Suppose a hypothetical pair of normal measurements $X$ and $Y$ is jointly bivariate normal, with means $8$ and $5$, variances $16$ and $9$, and correlation $0.5$. Find $P(X+Y>19)$.

Method and model. Set $S=X+Y$. The pair is jointly normal, so $S$ is normal. Its component standard deviations are $\sigma_X=4$ and $\sigma_Y=3$, and its covariance is $\operatorname{Cov}(X,Y)=0.5(4)(3)=6$.

The mean and variance of $S$ are

E[S]=8+5=13,\qquad \operatorname{Var}(S)=16+9+2(6)=37.

Thus $S\sim N(13,37)$, where the second parameter is the variance, and its standard deviation is $\sqrt{37}$. Standardize the threshold:

P(S>19) =1-\Phi\left(\frac{19-13}{\sqrt{37}}\right) =1-\Phi(0.9864) \approx 0.1620.

Answer: $P(X+Y>19)\approx 0.1620$, to four decimal places. The threshold is about $0.99$ standard deviations above the mean, so a probability below $0.5$ is reasonable. The common mistake avoided here is omitting the covariance term and using $16+9$ as the variance.

Worked Example: A Difference With Independent Components

A hypothetical model has jointly normal variables $X$ and $Y$, with means $9$ and $3$, variances $16$ and $36$, and correlation $0$. Let $D=X-Y$. Find $P(D\leq 13)$.

Method and model. Since the pair is jointly normal, $D$ is normal. Correlation $0$ gives covariance $0(4)(6)=0$. Therefore,

E[D]=9-3=6,\qquad \operatorname{Var}(D)=16+36-2(0)=52.

The standard deviation is $\sqrt{52}\approx 7.2111$, not $52$. The threshold is $13-6=7$ units above the mean, so its standardized value is

z=\frac{13-6}{\sqrt{52}}=\frac{7}{\sqrt{52}}\approx 0.9707.

Consequently,

P(D\leq 13) =\Phi\left(\frac{7}{\sqrt{52}}\right) =\Phi(0.9707) \approx 0.8342.

Answer: $0.8342$, to four decimal places. Since 13 is about $0.97$ standard deviations above the mean of 6, a probability comfortably above $0.5$ is reasonable. This calculation also illustrates why it is important to evaluate the stated standardized value carefully rather than substituting a rough description such as “one standard deviation.”

Worked Example: An Interval Probability for a Difference

Suppose $X$ and $Y$ are jointly bivariate normal with means $10$ and $6$, variances $9$ and $9$, and correlation $-1/3$. Find $P(-1<X-Y<9)$.

Method and model. Let $D=X-Y$. Both standard deviations are $3$, so the covariance is $\operatorname{Cov}(X,Y)=(-1/3)(3)(3)=-3$. The mean and variance of the difference are

E[D]=10-6=4,\qquad \operatorname{Var}(D)=9+9-2(-3)=24.

The negative covariance increases the variance of the difference: when $X$ and $Y$ tend to move in opposite directions, their difference tends to vary more. The standard deviation of $D$ is $\sqrt{24}$. Standardizing the lower and upper endpoints gives

\frac{-1-4}{\sqrt{24}}=-1.0206,\qquad \frac{9-4}{\sqrt{24}}=1.0206.

Therefore,

\begin{aligned} P(-1<D<9) &=\Phi(1.0206)-\Phi(-1.0206)\\ &=2\Phi(1.0206)-1\\ &\approx 2(0.8463)-1\\ &=0.6926. \end{aligned}

Answer: $0.6926$, to four decimal places. The endpoints are equally far from the mean, so symmetry about the mean provides a useful check. The common mistake avoided is using the sum formula for the difference: the covariance is negative, so subtracting twice that covariance adds $6$ to the component variances.

Common Mistakes and Exam Checks

The sign of the covariance term is the main source of errors. A positive correlation increases the variance of a sum and decreases the variance of a difference. A negative correlation does the reverse. This follows directly from the formulas, and it is a useful reasonableness check before calculating a probability.

Common mistake: For $D=X-Y$, a student with covariance $-3$ might incorrectly calculate $\operatorname{Var}(D)=9+9-2(3)=12$, treating the covariance as positive. The correction is to retain its sign: $$ \operatorname{Var}(D)=9+9-2(-3)=24. $$ The standard deviation is then $\sqrt{24}$, not $\sqrt{12}$.

Other checks help catch errors before they reach the final answer:

Do not assume independence from the marginal distributions. The covariance term is zero only when the variables are uncorrelated. In a jointly normal model, zero correlation implies independence, but a nonzero correlation must be incorporated.
Use variance for the second normal parameter and its square root for standardization. If $W\sim N(\mu_W,\sigma_W^2)$, divide by $\sigma_W$, not $\sigma_W^2$.
Check the event direction. A probability above a threshold uses the upper tail $1-\Phi(z)$; a probability below it uses $\Phi(z)$.
Check the scale of the result. If a threshold lies above the mean, its lower-tail probability should exceed $0.5$, while its upper-tail probability should be below $0.5$.

These calculations are aligned with the SOA Exam P (Probability) syllabus: the essential skills are recognizing a linear combination, including dependence correctly, and translating the resulting normal variable’s event into a standard normal probability.

Check Your Understanding

For each problem, first find the mean and variance of the sum or difference, then standardize. Give probabilities to four decimal places.

Let $X,Y$ be jointly bivariate normal with means $2,3$, variances $4,9$, and correlation $1/3$. Find $P(X+Y\leq 9)$. (Answer: $0.8340$)
Let $X,Y$ be jointly bivariate normal with means $6,2$, variances $9,4$, and correlation $0.5$. Find $P(X-Y>7)$. (Answer: $0.1284$)
Let $X,Y$ be jointly bivariate normal with means $0,1$, variances $1,4$, and correlation $-0.5$. Find $P(-2<X+Y<4)$. (Answer: $0.9167$)
Let $X,Y$ be jointly bivariate normal with means $5,1$, variances $9,4$, and correlation $-0.5$. Find $P(X-Y\leq 0)$. (Answer: $0.1795$)

What You'll Learn