Marginal Distributions of a Bivariate Normal Pair

Why Marginal Distributions Matter

A bivariate model describes two random variables together, but many questions ask about just one of them: for example, the probability that a claim amount exceeds a threshold, or that a measurement lies in a specified range. Such a probability uses the marginal distribution of that variable. In a bivariate normal model, each variable has a normal marginal distribution with its own specified mean and variance.

This result is useful because it lets you answer a one-variable probability question without calculating a two-dimensional probability. It also clarifies an important distinction: the correlation is part of the joint model, but it does not alter either variable’s marginal mean or variance.

Definition: If $X$ and $Y$ have a joint distribution, the marginal distribution of $X$ describes probabilities involving $X$ alone, without restricting $Y$. The marginal distribution of $Y$ is defined in the corresponding way. For a continuous pair with joint density $f_{X,Y}(x,y)$, the marginal densities are obtained by integrating over the other variable: $$ f_X(x)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dy, \qquad f_Y(y)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx. $$

The Marginals of a Bivariate Normal Pair

Let $(X,Y)$ be a nondegenerate bivariate normal pair with means $\mu_X,\mu_Y$, standard deviations $\sigma_X>0,\sigma_Y>0$, and correlation $\rho$, where $-1<\rho<1$. Thus $\operatorname{Var}(X)=\sigma_X^2$, $\operatorname{Var}(Y)=\sigma_Y^2$, and $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$. Its 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 standard deviations and correlation describe the spread and dependence in the joint density. To find a marginal density, integrate this expression over the other variable. The key step is completing the square in the variable being integrated out.

Theorem: If $(X,Y)$ is a nondegenerate bivariate normal pair with parameters as above, then $$ X\sim N(\mu_X,\sigma_X^2) \qquad\text{and}\qquad Y\sim N(\mu_Y,\sigma_Y^2). $$ The marginal distributions retain their specified means and variances; they do not depend on $\rho$.

Why Integration Gives a Normal Marginal

To obtain $f_X(x)$, fix $x$ and integrate the joint density over every possible $y$. Write $a=(x-\mu_X)/\sigma_X$ and $b=(y-\mu_Y)/\sigma_Y$. Completing the square in $b$ gives

a^2-2\rho ab+b^2=(b-\rho a)^2+(1-\rho^2)a^2.

In the integral for $f_X(x)$, make the change of variable from $y$ to $b$; since $dy=\sigma_Y\,db$, the factor $\sigma_Y$ cancels the corresponding factor in the density. The completed-square expression separates into a factor depending on $x$ and a shifted standard normal integral:

\begin{aligned} f_X(x) &=\frac{1}{2\pi\sigma_X\sqrt{1-\rho^2}} \exp\left(-\frac{a^2}{2}\right) \int_{-\infty}^{\infty} \exp\left(-\frac{(b-\rho a)^2}{2(1-\rho^2)}\right)\,db\\ &=\frac{1}{2\pi\sigma_X\sqrt{1-\rho^2}} \exp\left(-\frac{a^2}{2}\right) \sqrt{2\pi(1-\rho^2)}\\ &=\frac{1}{\sigma_X\sqrt{2\pi}} \exp\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2}\right). \end{aligned}

The integral in the second line is $\sqrt{2\pi(1-\rho^2)}$: it is the area under a normal-shaped exponential with variance $1-\rho^2$, and its center $\rho a$ does not change the total area. The result is the $N(\mu_X,\sigma_X^2)$ density. Integrating over $x$ instead, and completing the square in $(x-\mu_X)/\sigma_X$, gives the $N(\mu_Y,\sigma_Y^2)$ density for $Y$. In both cases, the correlation-dependent terms cancel during integration.

This derivation requires $|\rho|<1$, so the displayed joint density exists. At $\rho=1$ or $\rho=-1$, the pair is degenerate and that density formula does not apply; the joint distribution lies on a line. In the usual degenerate bivariate normal model, each coordinate still has its stated univariate normal distribution.

Calculating Marginal Probabilities

Once the relevant marginal is identified, use the ordinary univariate normal calculation. Define $\Phi(z)=P(Z\le z)$ for $Z\sim N(0,1)$. If $X\sim N(\mu_X,\sigma_X^2)$, then standardizing gives

P(c<X<d)= \Phi\left(\frac{d-\mu_X}{\sigma_X}\right) -\Phi\left(\frac{c-\mu_X}{\sigma_X}\right).

The standard deviation, not the variance, is used in the denominator. For probabilities involving only $X$, neither the value of $\rho$ nor the distribution of $Y$ needs to enter the calculation, beyond establishing that the pair is bivariate normal with the stated marginal parameters.

Worked Example: Integrating Out the Second Measurement

A hypothetical instrument records two measurements $(X,Y)$ modeled as a nondegenerate bivariate normal pair with $\mu_X=4$, $\sigma_X=3$, $\mu_Y=10$, $\sigma_Y=4$, and $\rho=0.5$. Find the marginal distribution of $X$, then calculate $P(1<X<7)$.

Method. Integrating the joint density over all $y$ gives the marginal density of $X$. By the result just derived, it is normal with the specified $X$-mean and $X$-variance: $$ X\sim N(4,3^2)=N(4,9). $$ The variance is $9$, so the standard deviation used to standardize is $3$.

The lower and upper standardized endpoints are $$ \frac{1-4}{3}=-1, \qquad \frac{7-4}{3}=1. $$ Therefore $$ P(1<X<7)=\Phi(1)-\Phi(-1)\approx 0.6827. $$ The probability, rounded to four decimal places, is $0.6827$.

Check. The interval is centered at the marginal mean and extends one standard deviation on each side, so a probability of about $0.68$ is consistent. The correlation $0.5$ affects the joint density but not this probability.

Common mistake avoided. Do not replace the marginal variance $9$ with $9(1-\rho^2)=6.75$. That expression is not the marginal variance of $X$; the marginal variance remains $9$.

Worked Example: A Marginal Tail and an Interval Probability

For a hypothetical pair of jointly normal environmental readings, suppose $U\sim N(50,100)$ and $V\sim N(30,25)$, with correlation $-0.6$. Find $P(U>62)$ and $P(27<V<33)$.

Method. Use each marginal separately. The variance of $U$ is $100$, so its standard deviation is $10$. The variance of $V$ is $25$, so its standard deviation is $5$.

For the upper tail of $U$, the standardized threshold is $$ \frac{62-50}{10}=1.2. $$ Thus $$ P(U>62)=1-\Phi(1.2)\approx 1-0.8849303=0.1150697\approx 0.1151. $$ For $V$, the standardized endpoints are $$ \frac{27-30}{5}=-0.6, \qquad \frac{33-30}{5}=0.6. $$ Hence $$ P(27<V<33)=\Phi(0.6)-\Phi(-0.6) \approx 0.7257469-0.2742531 =0.4514938\approx 0.4515. $$ Both answers are rounded to four decimal places.

Check. The first threshold is $1.2$ standard deviations above the mean, so its upper-tail probability should be less than $0.5$. The second interval is centered at the mean and spans only $1.2$ standard deviations in total, so its probability should also be less than $0.68$. The calculated values meet both checks.

Common mistake avoided. The negative correlation does not change either of these marginal probabilities. It would matter for a question involving both $U$ and $V$, but each event here concerns one variable alone.

Worked Example: Recovering Marginal Parameters Before Standardizing

A hypothetical bivariate normal pair $(A,B)$ has means $\mu_A=20$ and $\mu_B=8$, variances $\operatorname{Var}(A)=36$ and $\operatorname{Var}(B)=4$, and correlation $\rho=0.75$. Find $P(A\ge26)$ and $P(6<B<11)$.

Method. First identify the standard deviations by taking the positive square roots of the variances: $$ \sigma_A=\sqrt{36}=6,\qquad \sigma_B=\sqrt{4}=2. $$ The marginal distributions are therefore $A\sim N(20,36)$ and $B\sim N(8,4)$.

For $A\ge26$, the standardized threshold is $$ \frac{26-20}{6}=1. $$ A continuous distribution assigns probability zero to a single endpoint, so $$ P(A\ge26)=1-\Phi(1)\approx 1-0.8413447=0.1586553\approx 0.1587. $$ For $B$, the standardized endpoints are $$ \frac{6-8}{2}=-1, \qquad \frac{11-8}{2}=1.5. $$ Thus $$ P(6<B<11)=\Phi(1.5)-\Phi(-1) \approx 0.9331928-0.1586553 =0.7745375\approx 0.7745. $$ The probabilities are rounded to four decimal places.

Check. The event $A\ge26$ is one standard deviation above its mean, giving an upper tail near $0.16$. The interval for $B$ includes its mean and extends farther above it than below it, so its probability is substantial, as calculated.

Common mistake avoided. The stated $36$ and $4$ are variances, not standard deviations. Standardizing with $36$ or $4$ directly would use the wrong scale.

Common Traps and Exam Checks

Marginal questions are often short, but parameter interpretation and rounding still matter. Before using a normal table or calculator, identify which variable the event concerns and write down that variable’s mean and standard deviation. Then standardize its endpoint or endpoints. Do not carry over the other variable’s parameters merely because the pair is described jointly.

A frequent incorrect calculation treats $\sigma_X^2(1-\rho^2)$ as the variance of $X$. For the example with $\sigma_X=3$ and $\rho=0.5$, this mistake would give a supposed variance of $9(1-0.25)=6.75$ and standard deviation $\sqrt{6.75}\approx2.598$. That would produce incorrect standardized endpoints for the event $1<X<7$. The correct marginal variance is $9$, the standard deviation is $3$, and the probability is $0.6827$. The marginal result follows from integrating the joint density, not by adjusting the marginal variance for correlation.

Another trap is subtracting normal-table entries that have already been rounded and then reporting the result as if it were the correctly rounded probability. For example, $\Phi(1)-\Phi(-1)$ is approximately $0.6826895$, which rounds to $0.6827$. Subtracting the four-decimal entries $0.8413-0.1587$ instead gives $0.6826$, an inaccurate final rounding. Keep adequate precision through the subtraction and round the probability only at the end.

Key takeaway: For a bivariate normal pair, $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$. A probability involving only one coordinate is calculated from that coordinate’s marginal distribution; the correlation does not change that marginal probability.

Check Your Understanding

For each problem, identify the relevant marginal parameters before calculating. Give probabilities to four decimal places.

A bivariate normal pair has $X\sim N(12,16)$, $Y\sim N(5,9)$, and correlation $0.4$. Find $P(X>16)$. (Answer: $0.1587$)
A pair $(R,S)$ is bivariate normal with $R\sim N(40,25)$, $S\sim N(18,4)$, and correlation $-0.7$. Find $P(15<S<21)$. (Answer: $0.8664$)
A nondegenerate bivariate normal pair has means $\mu_M=100,\mu_N=70$, variances $\operatorname{Var}(M)=64,\operatorname{Var}(N)=100$, and correlation $0.25$. Find $P(92<M<108)$. (Answer: $0.6827$)
For jointly normal $C,D$, suppose $C\sim N(6,9)$, $D\sim N(20,16)$, and $\rho=-0.5$. Find $P(C\le3)$ and $P(16<D<26)$. (Answer: $0.1587$ and $0.7745$)
A bivariate normal pair has $H\sim N(30,36)$, $J\sim N(50,25)$, and correlation $0.8$. Find $P(H\ge36)$ and $P(45<J<55)$, explaining briefly why the correlation is not used in either calculation. (Answer: $0.1587$ and $0.6827$)

What You'll Learn