Probabilities of Linear Threshold Events

Why Linear Thresholds Can Be Simpler Than They Look

A probability involving two correlated normal variables may appear to require a two-dimensional calculation. But if the event is defined by a single linear inequality, such as $2X-Y\leq c$, the relevant random quantity is the linear combination $2X-Y$. That quantity is itself normal, so the event can be evaluated with a univariate normal probability.

The key is to preserve the coefficients and signs when calculating the mean and variance. In particular, correlation contributes a covariance term to the variance. Omitting that term, or treating a negative coefficient as positive, can substantially change the result.

Definition: Suppose $(X,Y)$ has a bivariate normal distribution, with $E[X]=\mu_X$, $E[Y]=\mu_Y$, $\operatorname{Var}(X)=\sigma_X^2$, $\operatorname{Var}(Y)=\sigma_Y^2$, and $\operatorname{Corr}(X,Y)=\rho$. For constants $a,b$, define $L=aX+bY$. Then $L$ is normal with $$ E[L]=a\mu_X+b\mu_Y $$ and $$ \operatorname{Var}(L)=a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y. $$

If $\operatorname{Var}(L)>0$, the threshold probabilities follow by standardizing $L$. Write $\Phi(z)$ for the cumulative distribution function of a standard normal variable. Then

P(aX+bY\leq c) = \Phi\left(\frac{c-(a\mu_X+b\mu_Y)} {\sqrt{a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y}}\right).

For a strict upper-tail event, use the complement:

P(aX+bY>c) = 1-\Phi\left(\frac{c-(a\mu_X+b\mu_Y)} {\sqrt{a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y}}\right).

A negative coefficient causes no special change to the procedure: it is included as a negative value of $a$ or $b$ in both the mean and variance calculations. If the variance of $L$ is positive, $L$ is continuous, so $P(L=c)=0$; consequently, replacing $<$ with $\leq$, or $>$ with $\geq$, does not change the probability. If $\operatorname{Var}(L)=0$, however, $L$ is constant and the event must be checked directly rather than standardized.

Why the Linear Combination Is Normal

The normality of $L$ is the reason the reduction works. The moment-generating function of a bivariate normal pair is

M_{X,Y}(s,t)= \exp\left( s\mu_X+t\mu_Y+ \frac12\left(s^2\sigma_X^2+t^2\sigma_Y^2 +2st\rho\sigma_X\sigma_Y\right) \right).

Substitute $s=ua$ and $t=ub$, because $uL=u(aX+bY)$. The moment-generating function of $L$ is therefore

\begin{aligned} M_L(u) &=E[e^{u(aX+bY)}]\\ &=M_{X,Y}(ua,ub)\\ &=\exp\left( u(a\mu_X+b\mu_Y) +\frac{u^2}{2} \left(a^2\sigma_X^2+b^2\sigma_Y^2 +2ab\rho\sigma_X\sigma_Y\right) \right). \end{aligned}

This is the moment-generating function of a normal variable with the stated mean and variance. The cross term $2ab\rho\sigma_X\sigma_Y$ comes from the joint distribution; it is not optional even when the event itself looks like a simple sum or difference.

A Consistent Calculation Method

Before using a standard normal table or calculator, identify the exact linear combination and calculate its parameters. The following sequence helps keep the event, signs, and tail direction aligned.

Name the linear combination.
Rewrite the event as $L=aX+bY\leq c$, $L>c$, or the corresponding strict or non-strict inequality. Keep negative coefficients negative.

Find its mean and variance.
Use $E[L]=a\mu_X+b\mu_Y$ and include the covariance term $2ab\rho\sigma_X\sigma_Y$ in $\operatorname{Var}(L)$.

Standardize the threshold.
Calculate $z=(c-E[L])/\sqrt{\operatorname{Var}(L)}$. This is the threshold measured in standard deviations above or below the mean of $L$.

Match the tail to the event.
For $L\leq c$, use $\Phi(z)$; for $L>c$, use $1-\Phi(z)$. Check that the probability is consistent with the threshold’s position relative to the mean.

Worked Example: A Negative Coefficient in a Lower-Tail Event

In a hypothetical monitoring system, the jointly normal variables $X$ and $Y$ represent two sensor readings. Assume $X\sim N(10,4)$, $Y\sim N(8,9)$, and $\operatorname{Corr}(X,Y)=0.5$. Find $P(2X-Y\leq 15)$. The second parameter in each normal distribution is the variance.

Method. Use the distribution of the linear combination $L=2X-Y$, retaining the coefficient $-1$ on $Y$.

First find the standard deviations and covariance:

\sigma_X=2,\qquad \sigma_Y=3,\qquad \operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y =(0.5)(2)(3)=3.

The mean and variance of $L$ are

\begin{aligned} E[L]&=2(10)-1(8)=12,\\ \operatorname{Var}(L) &=2^2(4)+(-1)^2(9)+2(2)(-1)(3)\\ &=16+9-12=13. \end{aligned}

Thus $L\sim N(12,13)$. Standardize the threshold $15$:

z=\frac{15-12}{\sqrt{13}} =\frac{3}{\sqrt{13}} \approx 0.8321.

Because the event is a lower-tail event,

P(2X-Y\leq15)=\Phi(0.8321)\approx 0.7972.

The answer is a little less than $0.8$, which is reasonable: the threshold is about $0.83$ standard deviations above the mean. The common mistake this calculation avoids is dropping the covariance term or changing its sign. Here $ab<0$, so the covariance contribution is $-12$, not $+12$.

Worked Example: An Upper-Tail Threshold at the Mean

A hypothetical portfolio has two jointly normal component changes, $X\sim N(40,25)$ and $Y\sim N(30,16)$, with correlation $-0.25$. Find the probability that the combined change $X+2Y$ exceeds $110$.

Method. Find the distribution of $L=X+2Y$, then use the upper tail. The model assumptions are that the pair is bivariate normal with the stated means, variances, and correlation.

The standard deviations are $5$ and $4$, so

\operatorname{Cov}(X,Y)=(-0.25)(5)(4)=-5.

The mean and variance of $L$ are

\begin{aligned} E[L]&=40+2(30)=100,\\ \operatorname{Var}(L) &=1^2(25)+2^2(16)+2(1)(2)(-5)\\ &=25+64-20=69. \end{aligned}

Therefore $L\sim N(100,69)$. The standardized threshold is

z=\frac{110-100}{\sqrt{69}} =\frac{10}{\sqrt{69}} \approx 1.2039.

The event is an upper tail, so

P(X+2Y>110)=1-\Phi(1.2039)\approx 0.1144.

The result is a relatively small probability because $110$ is about $1.20$ standard deviations above the mean. A common error is to report $\Phi(1.2039)$, which is the probability of being below the threshold, not above it.

Worked Example: A Difference with a Negative Threshold Shift

For a hypothetical quality-control study, let $X$ and $Y$ be jointly normal measurements with $X\sim N(12,9)$, $Y\sim N(7,4)$, and correlation $1/3$. Find $P(3X-2Y<20)$.

Method. Treat $3X-2Y$ as one normal variable. This difference has a negative coefficient on $Y$, but the given correlation is positive.

The standard deviations are $3$ and $2$, giving

\operatorname{Cov}(X,Y)=\frac13(3)(2)=2.

Calculate the mean and variance:

\begin{aligned} E[3X-2Y]&=3(12)-2(7)=36-14=22,\\ \operatorname{Var}(3X-2Y) &=3^2(9)+(-2)^2(4)+2(3)(-2)(2)\\ &=81+16-24=73. \end{aligned}

The threshold $20$ is below the mean $22$. Its standardized value and the required probability are

z=\frac{20-22}{\sqrt{73}} =\frac{-2}{\sqrt{73}} \approx -0.2341, \qquad P(3X-2Y<20)=\Phi(-0.2341)\approx 0.4075.

Because the variance is positive, the strict inequality has the same probability as $3X-2Y\leq20$. The answer is below $0.5$, as expected for a threshold below the mean. The covariance contribution is negative because $ab=3(-2)<0$, even though $\operatorname{Cov}(X,Y)$ is positive.

Common Traps and Boundary Cases

For a linear threshold, the most important checks are the algebraic sign of each coefficient, the sign of the covariance contribution, and the direction of the probability tail. These checks are especially useful when a problem describes the event in words rather than presenting it as an inequality.

Omitting covariance. In the first example, leaving out the cross term would give variance $16+9=25$, rather than $13$. The resulting standardization and probability would be wrong.
Using the wrong sign for the cross term. The term is $2ab\operatorname{Cov}(X,Y)$. Its sign depends on all three factors, including the coefficients in the linear combination.
Using a lower tail for an upper-tail event. If the event is $L>c$, the probability is $1-\Phi((c-E[L])/\sqrt{\operatorname{Var}(L)})$, not $\Phi$ of that value.
Thinking a negative coefficient reverses the whole event automatically. A coefficient’s sign is already handled by defining $L=aX+bY$. Do not reverse the inequality unless you multiply both sides of an inequality by a negative number.
Standardizing when the variance is zero. If $\operatorname{Var}(L)=0$, then $L=E[L]$ with probability one. For example, $P(L\leq c)=1$ if $E[L]\leq c$, and it is $0$ otherwise.

This method applies when the event is one linear inequality in a jointly normal pair. It does not by itself evaluate a region described by two simultaneous inequalities, such as a rectangle in the $X,Y$ plane; such an event generally requires joint-distribution methods. The advantage here is precisely that one linear threshold depends only on the one-dimensional distribution of $L$.

Key takeaway: For a single linear threshold in a bivariate normal pair, form $L=aX+bY$, calculate its mean and variance including covariance, and standardize the threshold. Then choose the normal tail that matches the original inequality.

Check Your Understanding

For each problem, identify the linear combination, calculate its mean and variance, and then select the correct tail.

Let $U$ and $V$ be jointly normal with $U\sim N(5,4)$, $V\sim N(2,9)$, and correlation $1/3$. Find $P(U-2V\leq0)$. (Answer: 0.4298)
Let $A$ and $B$ be jointly normal with $A\sim N(20,9)$, $B\sim N(15,16)$, and correlation $0.5$. Find $P(2A+B>60)$. (Answer: 0.2832)
Let $M$ and $N$ be jointly normal with $M\sim N(100,25)$, $N\sim N(80,36)$, and correlation $-0.4$. Find $P(M-N<15)$. (Answer: 0.2938)
Let $Q$ and $R$ be jointly normal with $Q\sim N(4,1)$, $R\sim N(10,4)$, and correlation $0.25$. Find $P(3Q+R\geq25)$. (Answer: 0.2266)
Let $C$ and $D$ be jointly normal with $C\sim N(50,100)$, $D\sim N(20,25)$, and correlation $0.6$. Find $P(0.5C-2D\leq-5)$. (Answer: 0.8925)

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