Deriving Marginal Distributions From Joint Laws

Why Marginal Distributions Matter

A joint probability law describes two losses together. A marginal distribution describes the probability behavior of one loss on its own, without retaining the value of the other. Marginals are needed whenever a question asks for the distribution of a single component of a risk pair, or when a proposed marginal law needs to be checked against the joint model.

The operation is simple in principle: for a discrete pair, add the joint probabilities associated with every possible value of the other variable; for a continuous pair, integrate the joint density across every value of the other variable allowed by the support. The main source of error is not the summation or integration itself. It is including too few values, or using bounds that do not match the joint support.

Definition: For discrete random variables $X$ and $Y$, their joint probability mass function is $p_{X,Y}(x,y)=P(X=x,Y=y)$. The marginal mass functions are $$ p_X(x)=\sum_y p_{X,Y}(x,y), \qquad p_Y(y)=\sum_x p_{X,Y}(x,y), $$ where each sum includes all possible values of the variable being summed out.

Definition: For continuous random variables $X$ and $Y$ with joint density $f_{X,Y}(x,y)$, the marginal densities are $$ f_X(x)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dy, \qquad f_Y(y)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx. $$ In practice, integrate only across values where the joint density is nonzero. Bounds may depend on the value being held fixed.

Why Marginalization Works

For discrete variables, the events $\{X=x,Y=y\}$, as $y$ ranges over all its possible values, are disjoint and together make up the event $\{X=x\}$. Adding their probabilities therefore gives $P(X=x)$. The same argument applies to $Y$.

For continuous variables, the probability that $X$ lies in an interval $A$ is obtained by integrating the joint density over all points whose $x$-coordinate lies in $A$. Integrating first over $y$ gives

P(X\in A) =\int_A\left(\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\right)dx.

The expression in parentheses is the density of $X$, which justifies the marginal-density formula. This argument assumes a joint density exists; it does not describe models with probability mass concentrated on curves or points.

Identify the variable to retain.
For $f_X(x)$, hold $x$ fixed and sum or integrate over $y$.

Read the joint support.
Determine which values of the variable being eliminated are actually allowed for the fixed value.

Marginalize and state the support.
Sum the joint masses or integrate the joint density, and specify where the resulting marginal is nonzero.

Check the result.
Marginal probabilities must sum to one, and a marginal density must integrate to one.

Marginal Probability Mass Functions

In a discrete joint law, each row or column of a probability table contributes to one marginal probability. The table must represent a valid joint distribution: its entries are nonnegative and their total is one. A marginal is formed by adding entries, not by renormalizing an individual row or column.

Worked Example: Marginals From a Joint Loss Table

In a hypothetical one-period model, $X$ is a small-claim count loss and $Y$ is a service-interruption loss category. Both are discrete. Their joint probabilities are shown below. Derive both marginal mass functions and find $P(X\geq 1)$.

$X\backslash Y$	$Y=0$	$Y=1$
$X=0$	0.10	0.15
$X=1$	0.20	0.10
$X=2$	0.15	0.30

Method. Sum across each row to obtain $p_X$, and down each column to obtain $p_Y$. The joint entries are nonnegative, and their total is $0.10+0.15+0.20+0.10+0.15+0.30=1.00$, so the table is normalized.

For $X$, the row sums are

p_X(0)=0.10+0.15=0.25,\quad p_X(1)=0.20+0.10=0.30,\quad p_X(2)=0.15+0.30=0.45.

For $Y$, the column sums are

p_Y(0)=0.10+0.20+0.15=0.45,\qquad p_Y(1)=0.15+0.10+0.30=0.55.

The check is $0.25+0.30+0.45=1.00$ for $X$ and $0.45+0.55=1.00$ for $Y$. Finally, $P(X\geq1)=p_X(1)+p_X(2)=0.30+0.45=0.75$. This method avoids the common mistake of treating one table entry as the probability of a marginal event; a marginal event includes every joint outcome with the specified value of $X$.

Marginal Densities and Nonrectangular Support

For continuous variables, the support is the set of $(x,y)$ values where the joint density can be positive. If this set is not a rectangle, the integration bounds for one variable may depend on the value of the other. The density being integrated must also be a valid joint density: it is nonnegative on its support and has total integral one.

Worked Example: Marginals on a Triangular Support

Suppose two hypothetical normalized loss measures $X$ and $Y$ have joint density $f_{X,Y}(x,y)=2$ when $0<x<y<1$, and zero elsewhere. Derive both marginal densities and verify their normalization.

Method. For fixed $x$, the support requires $x<y<1$. For fixed $y$, it requires $0<x<y$. The joint density integrates to one because

\int_0^1\int_x^1 2\,dy\,dx =\int_0^1 2(1-x)\,dx =2\left(1-\frac12\right)=1.

To obtain $f_X(x)$, integrate over $y$ from $x$ to 1:

f_X(x)=\int_x^1 2\,dy=2(1-x),\qquad 0<x<1.

For $f_Y(y)$, integrate over $x$ from 0 to $y$:

f_Y(y)=\int_0^y 2\,dx=2y,\qquad 0<y<1.

Both marginal densities are zero outside $(0,1)$. Their checks are $\int_0^1 2(1-x)\,dx=1$ and $\int_0^1 2y\,dy=1$. The different bounds arise from the triangular support; using the same bounds for both integrations would integrate over the wrong region.

Worked Example: A Nonconstant Density With Reversed Bounds

For a hypothetical pair of normalized loss indicators, suppose $f_{X,Y}(x,y)=6y$ on $0<y<x<1$, and is zero elsewhere. Find both marginal densities and calculate $P(X<0.5)$.

Method. First check the joint density. At a fixed $x$, $y$ ranges from 0 to $x$, so

\int_0^1\int_0^x 6y\,dy\,dx =\int_0^1 3x^2\,dx=1.

For the $X$ marginal, integrate over the allowed values $0<y<x$:

f_X(x)=\int_0^x 6y\,dy=3x^2,\qquad 0<x<1.

For the $Y$ marginal, hold $y$ fixed. The support condition $y<x<1$ means $x$ ranges from $y$ to 1:

f_Y(y)=\int_y^1 6y\,dx=6y(1-y),\qquad 0<y<1.

The normalization checks are $\int_0^1 3x^2\,dx=1$ and $\int_0^1 6y(1-y)\,dy=3-2=1$. Then

P(X<0.5)=\int_0^{0.5}3x^2\,dx=(0.5)^3=0.125.

As a check, the marginal distribution assigns probability $0.125$ to that interval, with the boundary point having probability zero in this continuous model. A common error is to calculate $f_Y(y)$ using $\int_0^y 6y\,dx=6y^2$. That bound describes $x<y$, which is excluded by the given support $y<x$. The correct bounds are $y$ to 1.

Common Mistakes and Checks

Ignoring the support. The formal limits from negative infinity to infinity do not mean the density is nonzero everywhere. Use the actual support to set the effective bounds.
Holding the wrong variable fixed. To find $f_X(x)$, keep $x$ fixed and integrate with respect to $y$; to find $f_Y(y)$, keep $y$ fixed and integrate with respect to $x$.
Forgetting a piecewise marginal. If the support changes shape as the retained variable changes, the marginal’s integration bounds may change too. State the support and any resulting pieces explicitly.
Skipping normalization checks. A proposed marginal that sums to less than one or integrates to a value other than one usually signals an omitted outcome, an incorrect bound, or an error in the joint law.

Marginalization removes one coordinate while retaining all of its possible outcomes. This makes it a useful consistency check: a joint model should produce valid distributions for each component when the other component is summed or integrated out. These definitions and calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, derive the requested marginal quantities from the stated joint law and check the relevant support.

A discrete loss pair has joint probabilities $p(0,0)=0.12$, $p(0,1)=0.18$, $p(1,0)=0.28$, and $p(1,1)=0.42$, where the arguments are $X,Y$. Find both marginal mass functions and $P(X=1,Y=0)$. (Answer: $p_X(0)=0.30,\ p_X(1)=0.70;\ p_Y(0)=0.40,\ p_Y(1)=0.60;\ P(X=1,Y=0)=0.28$.)
A joint density is $f_{X,Y}(x,y)=2$ on $0<x<y<1$, and zero elsewhere. Find $f_X(x)$, $f_Y(y)$, and $f_X(0.25)$. (Answer: $f_X(x)=2(1-x)$, $f_Y(y)=2y$ on $0<x,y<1$; $f_X(0.25)=1.5$.)
A joint density is $f_{X,Y}(x,y)=3x$ on $0<y<x<1$, and zero elsewhere. Derive both marginals and calculate $P(Y>0.5)$. (Answer: $f_X(x)=3x^2$, $f_Y(y)=\frac{3}{2}(1-y^2)$ on $(0,1)$; $P(Y>0.5)=0.3125$.)
A joint density is $2/3$ on $0<x<1,\ x<y<2$, and zero elsewhere. Derive $f_X(x)$ and give $f_Y(y)$ on each of its support intervals. (Answer: $f_X(x)=\frac{2}{3}(2-x)$ for $0<x<1$; $f_Y(y)=\frac{2y}{3}$ for $0<y<1$, and $f_Y(y)=\frac{2}{3}$ for $1\leq y<2$.)
A joint density is $1/2$ on $0<x<2,\ 0<y<2-x$, and zero elsewhere. Derive $f_X(x)$ and calculate $P(X>1)$. (Answer: $f_X(x)=\frac{2-x}{2}$ for $0<x<2$; $P(X>1)=0.25$.)

\(X\backslash Y\)	\(Y=0\)	\(Y=1\)
\(X=0\)	0.10	0.15
\(X=1\)	0.20	0.10
\(X=2\)	0.15	0.30

What You'll Learn