Conditional Distributions in the Bivariate Normal Model

Why Conditional Normal Distributions Matter

A bivariate normal model describes two measurements that vary together. In actuarial work, one might use it as a hypothetical approximation for a claim-cost measure and a risk score, or for two financial measurements. Once one value is observed, the relevant question may be how the distribution of the other measurement changes. The bivariate normal model gives a particularly useful answer: the conditional distribution is also normal, with its mean adjusted according to the observed value and its variance reduced according to the strength of the association.

The adjustment depends on more than the correlation alone. It also depends on the relative standard deviations of the two variables. Keeping track of which variable is being conditioned on is therefore essential.

Definition: A pair $(X,Y)$ has a bivariate normal distribution with means $\mu_X,\mu_Y$, standard deviations $\sigma_X,\sigma_Y$, and correlation $\rho$ if every linear combination of $X$ and $Y$ is normally distributed. In the nonsingular case, assume $\sigma_X>0$, $\sigma_Y>0$, and $-1<\rho<1$. The marginal distributions are $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$, and their covariance is $\rho\sigma_X\sigma_Y$.

Under these assumptions, the joint density is

f_{X,Y}(x,y)= \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left\{-\frac{1}{2(1-\rho^2)} \left[ \left(\frac{x-\mu_X}{\sigma_X}\right)^2 -2\rho\left(\frac{x-\mu_X}{\sigma_X}\right) \left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2 \right]\right\}.

The key result is that fixing $Y=y$ leaves a normal distribution for $X$, but generally not the original marginal distribution of $X$. The observed $y$ shifts the conditional mean; the correlation determines how much it shifts.

The Conditional Normal Formula

Theorem: If $(X,Y)$ is nonsingular bivariate normal as defined above, then for any real $y$, $$ X\mid Y=y \sim N\left( \mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y), \ \sigma_X^2(1-\rho^2) \right). $$ Thus $$ E[X\mid Y=y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y), \qquad \operatorname{Var}(X\mid Y=y)=\sigma_X^2(1-\rho^2). $$

The variance in this formula is the conditional variance, not its square root. The conditional standard deviation is $\sigma_X\sqrt{1-\rho^2}$. In particular, the conditional variance does not depend on the observed value $y$; the conditional mean does.

Why the Formula Holds

A short standardization argument explains both the mean adjustment and the remaining variance. Define standardized variables $U=(X-\mu_X)/\sigma_X$ and $V=(Y-\mu_Y)/\sigma_Y$. They are jointly standard normal with correlation $\rho$. Now define $W=(U-\rho V)/\sqrt{1-\rho^2}$. The pair $(W,V)$ is jointly normal. Also, $E[W]=0$, $\operatorname{Var}(W)=1$, and $\operatorname{Cov}(W,V)=0$. For jointly normal variables, zero covariance implies independence, so $W$ and $V$ are independent standard normal variables.

Rearrange the definition of $W$ to obtain $U=\rho V+\sqrt{1-\rho^2}W$. When $Y=y$, the standardized value of $V$ is $v=(y-\mu_Y)/\sigma_Y$. Since $W$ is independent of $V$, conditioning on $V=v$ leaves $W$ standard normal. Therefore the conditional mean of $U$ is $\rho v$ and its conditional variance is $1-\rho^2$. Returning to the original scale gives

E[X\mid Y=y] =\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y} =\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y), \qquad \operatorname{Var}(X\mid Y=y)=\sigma_X^2(1-\rho^2).

This argument also shows why correlation alone is insufficient for the mean adjustment: converting the observed deviation $y-\mu_Y$ into the scale of $X$ requires the ratio $\sigma_X/\sigma_Y$. The result uses the bivariate normal assumption. For variables that are not jointly normal, the same means, variances, and correlation do not in general determine the conditional distribution or guarantee that it is normal.

Worked Example: A Claim-Cost Measure Given a Risk Score

Model and method. Suppose a hypothetical claim-cost measure $X$, in dollars, and a risk score $Y$ have a bivariate normal distribution with $\mu_X=\$500$, $\sigma_X=\$40$, $\mu_Y=12$, $\sigma_Y=6$, and $\rho=-0.50$. Given an observed risk score $Y=9$, find the conditional distribution of $X$, then calculate $P(X>\$550\mid Y=9)$. Use the conditional normal formula, followed by standardization.

Conditional mean. The score is $9-12=-3$ units below its mean. The adjustment is \[ \rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y) =(-0.50)\frac{40}{6}(9-12) =(-0.50)(6.6667)(-3)=10. \] Thus $E[X\mid Y=9]=500+10=\$510$.

Conditional variance and distribution. \[ \operatorname{Var}(X\mid Y=9)=40^2(1-(-0.50)^2) =1600(0.75)=1200. \] The conditional standard deviation is $\sqrt{1200}=\$34.6410$. Therefore $X\mid Y=9\sim N(510,1200)$, where the second parameter is the variance.

Conditional probability. Standardize the threshold: \[ z=\frac{550-510}{\sqrt{1200}} =\frac{40}{34.6410}=1.1547. \] Hence $P(X>550\mid Y=9)=1-\Phi(1.1547)\approx 1-0.8759=0.1241$. The conditional probability is 0.1241 to four decimals.

Check. Because the correlation is negative, a score below its mean raises the conditional mean of $X$ above $\$500$, as the calculated value $\$510$ does. The upper-tail probability is also below one half because $\$550$ is above the conditional mean.

Applying the Result in Either Direction

The formula is directional. To find $Y\mid X=x$, interchange the roles of $X$ and $Y$:

Y\mid X=x\sim N\left( \mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X), \ \sigma_Y^2(1-\rho^2) \right).

Although the correlation is unchanged when the variables are interchanged, the standard-deviation ratio changes. The two conditional means are therefore not obtained by using the same slope. The conditional variance for the variable being predicted uses that variable's marginal variance.

Worked Example: A Conditional Range for a Hypothetical Reserve Measure

Model and method. A hypothetical reserve measure $X$, in thousands of dollars, and an index $Y$ are bivariate normal with $\mu_X=300$, $\sigma_X=50$, $\mu_Y=20$, $\sigma_Y=4$, and $\rho=0.60$. Given $Y=24$, find the conditional probability that $X$ is between 290 and 370, and find the conditional 90th percentile. Use the conditional normal distribution and standard normal probabilities.

Conditional mean. The observed index is $24-20=4$ above its mean. The adjustment is \[ 0.60\frac{50}{4}(4)=0.60(12.5)(4)=30. \] Thus $E[X\mid Y=24]=300+30=330$ thousand dollars.

Conditional variance. \[ \operatorname{Var}(X\mid Y=24)=50^2(1-0.60^2) =2500(0.64)=1600. \] The conditional standard deviation is $\sqrt{1600}=40$ thousand dollars, so $X\mid Y=24\sim N(330,1600)$.

Conditional range probability. The endpoint standardized values are \[ z_{290}=\frac{290-330}{40}=-1,\qquad z_{370}=\frac{370-330}{40}=1. \] Consequently, \[ P(290<X<370\mid Y=24)=\Phi(1)-\Phi(-1) =0.8413-0.1587=0.6826, \] or 0.6827 using the more precise standard normal probabilities.

Conditional 90th percentile. The standard normal 90th percentile is approximately $1.2816$. Therefore the conditional percentile is \[ 330+1.2816(40)=330+51.264=381.264 \] thousand dollars, or $\$381.26$ thousand dollars to the nearest cent in the stated thousands scale.

Check. The conditional interval is symmetric around 330, and its endpoints are one conditional standard deviation from that mean. The calculated probability is the standard normal probability between $-1$ and $1$, as expected.

Worked Example: Conditioning in the Reverse Direction

Model and method. Let $X$ be a hypothetical operating index and $Y$ a hypothetical measured cost, in thousands of dollars. Suppose they are bivariate normal with $\mu_X=8$, $\sigma_X=3$, $\mu_Y=100$, $\sigma_Y=20$, and $\rho=-0.40$. Given $X=11$, find the conditional distribution of $Y$ and calculate $P(Y<80\mid X=11)$.

Conditional mean. Here $X$ is the conditioning variable, so the adjustment to the mean of $Y$ uses $\sigma_Y/\sigma_X$: \[ \rho\frac{\sigma_Y}{\sigma_X}(11-8) =(-0.40)\frac{20}{3}(3)=(-0.40)(20)=-8. \] Thus $E[Y\mid X=11]=100-8=92$ thousand dollars.

Conditional variance. \[ \operatorname{Var}(Y\mid X=11)=20^2(1-(-0.40)^2) =400(0.84)=336. \] The conditional standard deviation is $\sqrt{336}=18.3303$, so $Y\mid X=11\sim N(92,336)$.

Conditional probability. \[ z=\frac{80-92}{\sqrt{336}} =\frac{-12}{18.3303}\approx -0.6547. \] It follows that $P(Y<80\mid X=11)=\Phi(-0.6547)\approx0.2564$, to four decimals.

Check. The observed $X$ is above its mean and the correlation is negative, so the conditional mean of $Y$ must be below 100; the result, 92, has the correct direction.

Common Mistakes and Exam Traps

The conditional mean is a frequent source of sign and arithmetic errors. For example, if $\mu_X=500$, $\sigma_X=20/3$, $\sigma_Y=6$, $\rho=-0.50$, and $y-\mu_Y=-3$, the adjustment is \[ (-0.50)\left(\frac{20/3}{6}\right)(-3) =(-0.50)\left(\frac{10}{9}\right)(-3) =\frac{5}{3}, \] not 20. A different scale gives a different adjustment: in the first worked example, with $\sigma_X=40$ and $\sigma_Y=6$, the adjustment is $(-0.50)(40/6)(-3)=10$, so the conditional mean is 510, not 520. The reliable approach is to calculate the scale ratio first, then multiply by the correlation and the observed deviation.

Using the wrong direction. For $X\mid Y=y$, the mean adjustment uses $\sigma_X/\sigma_Y$, and the variance uses $\sigma_X^2$. For $Y\mid X=x$, both roles switch.
Dropping the sign. A negative correlation and an observation below the conditioning variable's mean produce a positive adjustment to the predicted mean.
Confusing variance and standard deviation. The conditional variance is $\sigma_X^2(1-\rho^2)$; take its square root only when a standard deviation is needed.
Assuming the mean changes the variance. Under this model, the conditional variance depends on the marginal variance and correlation, not on the particular observed value.
Applying the result without joint normality. Having normally distributed marginals alone does not establish the bivariate normal model. The conditional-normal result requires joint normality.

Key takeaway: To find $X\mid Y=y$, adjust $\mu_X$ by $\rho(\sigma_X/\sigma_Y)(y-\mu_Y)$, and multiply $\sigma_X^2$ by $1-\rho^2$ for the conditional variance. Check the conditioning direction, units, and sign before calculating probabilities.

Check Your Understanding

For each problem, assume the stated pair is nonsingular bivariate normal. Use standard normal probabilities where needed.

Suppose $X\sim N(200,30^2)$, $Y\sim N(10,4^2)$, and $\rho=0.50$. Find the conditional mean and variance of $X\mid Y=14$. (Answer: mean $215$; variance $675$.)
Suppose $X\sim N(80,12^2)$, $Y\sim N(50,10^2)$, and $\rho=-0.25$. Given $Y=42$, find the conditional mean and variance of $X$. (Answer: mean $82.4$; variance $135$.)
Suppose $X\sim N(40,8^2)$, $Y\sim N(5,2^2)$, and $\rho=0.75$. Find the conditional 95th percentile of $X\mid Y=7$, using $z_{0.95}=1.6449$. (Answer: approximately $54.70$.)
Suppose $X\sim N(100,20^2)$, $Y\sim N(30,5^2)$, and $\rho=-0.60$. Find $P(X<80\mid Y=35)$. (Answer: $0.3085$.)
Suppose $X\sim N(8,3^2)$, $Y\sim N(100,20^2)$, and $\rho=-0.40$. Given $X=11$, find $P(Y<80\mid X=11)$. (Answer: $0.2564$.)

What You'll Learn