Deriving Conditional Normal Distributions

Why Conditioning Changes a Normal Distribution

When two measurements are jointly normal, observing one gives information about the other whenever they are correlated. The conditional distribution describes that updated uncertainty: it identifies both the value around which $X$ is centered after observing $Y=y$, and the remaining spread of $X$. These are not obtained by simply substituting $y$ for $Y$; they follow from the joint density.

This tutorial derives the conditional distribution from that density. The derivation is useful because it shows why the correlation affects the conditional mean and why the conditional variance depends on the square of the correlation. The formulas apply to a nondegenerate bivariate normal pair; the perfectly correlated boundary cases do not have the ordinary two-dimensional density used here.

Model and assumptions: Let $(X,Y)$ have a bivariate normal distribution with means $\mu_X,\mu_Y$, positive standard deviations $\sigma_X,\sigma_Y$, and correlation $\rho$, where $-1<\rho<1$. Thus $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$. The second parameter of $N(\mu,\sigma^2)$ is the variance. Under these assumptions, the joint distribution has a density and $Y$ has a positive marginal density at every real $y$.

From the Joint Density to the Conditional Density

For a nondegenerate bivariate normal pair, 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 marginal distribution of $Y$ is normal with mean $\mu_Y$ and variance $\sigma_Y^2$, so its density is

f_Y(y)=\frac{1}{\sigma_Y\sqrt{2\pi}} \exp\left[-\frac{1}{2} \left(\frac{y-\mu_Y}{\sigma_Y}\right)^2\right].

For a continuous variable, $P(Y=y)=0$. Nevertheless, the conditional density at a specified value $y$ is defined through the density ratio $f_{X\mid Y}(x\mid y)=f_{X,Y}(x,y)/f_Y(y)$, wherever $f_Y(y)>0$. This is the density-based form of conditioning, not a division by the probability of the single-point event.

Completing the Square

Set $z_X=(x-\mu_X)/\sigma_X$ and $z_Y=(y-\mu_Y)/\sigma_Y$. The quadratic expression in the joint density can be rewritten as

\frac{z_X^2-2\rho z_Xz_Y+z_Y^2}{1-\rho^2} = z_Y^2+\frac{(z_X-\rho z_Y)^2}{1-\rho^2}.

To verify the identity, multiply the right-hand side by $1-\rho^2$. The result is $(1-\rho^2)z_Y^2+(z_X-\rho z_Y)^2 =z_Y^2-\rho^2z_Y^2+z_X^2-2\rho z_Xz_Y+\rho^2z_Y^2$, which simplifies to the numerator on the left. When the joint density is divided by $f_Y(y)$, the factor involving $z_Y^2$ cancels. The remaining expression is a normal density in $x$.

Theorem: For a nondegenerate bivariate normal pair with $-1<\rho<1$, the conditional distribution of $X$ given $Y=y$ is $$ X\mid Y=y\sim N\left( \mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y), \ \sigma_X^2(1-\rho^2) \right). $$ The conditional mean is $\mu_X+\rho(\sigma_X/\sigma_Y)(y-\mu_Y)$, and the conditional variance is $\sigma_X^2(1-\rho^2)$.

For additional detail, the completed-square term in the conditional density is \[ \frac{1}{2}\left( \frac{x-\mu_X-\rho(\sigma_X/\sigma_Y)(y-\mu_Y)} {\sigma_X\sqrt{1-\rho^2}} \right)^2. \] Its center is the conditional mean, and its scale in the denominator is the conditional standard deviation. The density’s leading constant is $1/(\sigma_X\sqrt{1-\rho^2}\sqrt{2\pi})$, so it is normalized as a normal density. This establishes both parameters, not just the shape.

Interpreting the Conditional Parameters

The conditional mean changes with the observed value $y$. If $y$ is above $\mu_Y$ and $\rho>0$, the conditional mean of $X$ is above $\mu_X$. If $\rho<0$, the direction reverses. The factor $\sigma_X/\sigma_Y$ accounts for the different measurement scales: a deviation of $y-\mu_Y$ is measured in $Y$'s units, while the adjustment is made in $X$'s units.

The conditional variance does not depend on the particular observed value $y$. It is at most the original variance, since $0<1-\rho^2\leq 1$ under the stated assumptions. When $\rho$ is close to zero, observing $Y$ makes little difference to the spread of $X$. As $|\rho|$ approaches 1, the conditional variance approaches zero. At exactly $|\rho|=1$, however, the pair is degenerate and the two-dimensional density derivation above no longer applies.

Worked Example: Conditioning on a Higher Sensor Reading

A hypothetical monitoring system produces two jointly normal readings, $X$ and $Y$, in the same measurement units. Assume $X\sim N(50,64)$, $Y\sim N(20,25)$, and correlation $\rho=0.6$. Find the conditional distribution of $X$ when $Y=25$, and calculate $P(X>61.2\mid Y=25)$. There is no interest-rate or mortality assumption in this measurement model.

Method. Apply the conditional normal result, using standard deviations $\sigma_X=\sqrt{64}=8$ and $\sigma_Y=\sqrt{25}=5$. The observed deviation of $Y$ from its mean is $25-20=5$.

The conditional mean is \[ 50+0.6\left(\frac{8}{5}\right)(5) =50+4.8 =54.8. \] The conditional variance is \[ 64(1-0.6^2)=64(1-0.36)=64(0.64)=40.96. \] Thus the conditional standard deviation is $\sqrt{40.96}=6.4$, and \[ X\mid Y=25\sim N(54.8,40.96). \]

Standardize the requested threshold using this conditional mean and standard deviation: \[ \frac{61.2-54.8}{6.4}=\frac{6.4}{6.4}=1. \] Therefore \[ P(X>61.2\mid Y=25)=P(Z>1)=1-\Phi(1)=0.1587, \] where $Z\sim N(0,1)$ and $\Phi$ is its cumulative distribution function. The probability is rounded to four decimals. As a check, $61.2$ is exactly one conditional standard deviation above $54.8$, so the upper-tail probability should be the standard normal upper tail at 1.

Common mistake avoided: The event is conditional on $Y=25$, so the standardization uses the conditional mean $54.8$ and conditional standard deviation $6.4$, not the marginal mean $50$ and standard deviation $8$.

Worked Example: A Negative Correlation Reverses the Direction

In a hypothetical quality-control model, $X$ is a calibration error and $Y$ is a separate instrument reading. Assume $X\sim N(100,144)$, $Y\sim N(40,100)$, and $\rho=-0.5$. Given $Y=34$, find the conditional distribution of $X$ and calculate the probability that $X$ exceeds its conditional mean by one conditional standard deviation. No interest rate or mortality assumption is relevant.

Method. First calculate the standard deviations: $\sigma_X=12$ and $\sigma_Y=10$. The observed deviation in $Y$ is $34-40=-6$. The conditional mean is \[ 100+(-0.5)\left(\frac{12}{10}\right)(-6) =100+3.6 =103.6. \] The conditional variance and standard deviation are \[ 144(1-(-0.5)^2)=144(0.75)=108, \qquad \sqrt{108}=6\sqrt{3}. \] Consequently, \[ X\mid Y=34\sim N(103.6,108). \]

The threshold one conditional standard deviation above the conditional mean is $103.6+\sqrt{108}$. Standardizing with the conditional parameters gives \[ P(X>103.6+\sqrt{108}\mid Y=34) =P(Z>1) =0.1587. \] The standard deviation need not be rounded to evaluate this probability because the threshold is expressed using the exact conditional standard deviation.

Common mistake avoided: One might see that $Y=34$ is below its mean and incorrectly expect $X$ to be below its mean as well. With negative correlation, the conditional adjustment is positive: the product $(-0.5)(34-40)$ is positive. Using $+6$ in place of the signed deviation $-6$ would incorrectly give $100+(-0.5)(12/10)(6)=96.4$.

Worked Example: Finding a Conditional Percentile

A hypothetical pair of jointly normal operating measurements has $X\sim N(30,16)$, $Y\sim N(10,9)$, and correlation $\rho=0.75$. Given $Y=14$, find the 90th percentile of $X\mid Y=14$. There is no interest-rate or mortality assumption; the model is specified entirely by these normal parameters.

Method. The standard deviations are $\sigma_X=4$ and $\sigma_Y=3$, and $14-10=4$. The conditional mean is \[ 30+0.75\left(\frac{4}{3}\right)(4)=30+4=34. \] The conditional variance and standard deviation are \[ 16(1-0.75^2)=16(1-0.5625)=16(0.4375)=7, \qquad \sqrt{7}. \] Thus $X\mid Y=14\sim N(34,7)$.

For a normal distribution, the 90th percentile is its mean plus $z_{0.90}$ times its standard deviation, where $P(Z\leq z_{0.90})=0.90$. Using $z_{0.90}=1.2816$, \[ q_{0.90}=34+1.2816\sqrt{7} =34+3.3908 =37.3908. \] The result is rounded to four decimals. As a check, standardizing $37.3908$ relative to the conditional distribution gives approximately $1.2816$, whose standard normal cumulative probability is $0.9000$.

Common mistake avoided: The second parameter is a variance, not a standard deviation. The conditional standard deviation here is $\sqrt{7}$, not $7$.

Common Traps and Scope

Using the marginal parameters after conditioning. Once $Y=y$ is given, use the conditional mean and variance for any probability or percentile involving $X$.
Dropping the sign of the deviation. Calculate $y-\mu_Y$ as a signed quantity before multiplying by $\rho$. Negative correlation can reverse the direction of the adjustment.
Confusing variance and standard deviation. The conditional variance is $\sigma_X^2(1-\rho^2)$; standardization requires its square root, $\sigma_X\sqrt{1-\rho^2}$.
Applying the density formula at perfect correlation. The derivation assumes $-1<\rho<1$. At $\rho=1$ or $\rho=-1$, the joint distribution is degenerate and does not have the stated two-dimensional density.
Treating a point condition as an ordinary positive-probability event. For continuous $Y$, $P(Y=y)=0$; the conditional density at $y$ is given by the joint-to-marginal density ratio.

The result gives the full conditional distribution, not only a conditional average. After identifying its parameters, an exam problem can be completed with ordinary normal standardization or a normal quantile. This is aligned with the SOA Exam P (Probability) syllabus.

Key takeaway: For a nondegenerate bivariate normal pair, conditioning on $Y=y$ gives a normal distribution for $X$. Its mean shifts by $\rho(\sigma_X/\sigma_Y)(y-\mu_Y)$, while its variance becomes $\sigma_X^2(1-\rho^2)$.

Check Your Understanding

For each problem, identify the conditional parameters before calculating the requested quantity. All models are hypothetical, with no interest-rate or mortality assumptions.

Let $X\sim N(20,25)$, $Y\sim N(4,16)$, and $\rho=0.4$. Find the conditional mean and variance of $X\mid Y=8$. (Answer: mean $22$, variance $21$)
Let $X\sim N(0,9)$, $Y\sim N(10,4)$, and $\rho=-0.5$. Find $P(X>1.5\mid Y=8)$. (Answer: $0.5000$)
Let $X\sim N(60,100)$, $Y\sim N(30,25)$, and $\rho=0.6$. Calculate $P(X>74\mid Y=35)$. (Answer: $0.1587$)
Let $X\sim N(5,4)$, $Y\sim N(2,9)$, and $\rho=1/3$. Find the 75th percentile of $X\mid Y=5$, using $z_{0.75}=0.6745$. (Answer: $6.9385$)

What You'll Learn