Dependence Revealed By Conditional Distributions

Why Compare Conditional and Marginal Distributions?

A marginal distribution describes a loss without regard to another risk. A conditional distribution describes that loss after information about another variable is known. Comparing the two provides a direct way to see whether the risks are independent. It can reveal dependence that is invisible in a mean, a variance, or even the fact that two variables have identical marginal distributions.

For example, knowing that one portfolio’s loss has the same marginal distribution in two business environments does not show that the loss behaves the same within each environment. Nor does an unchanged conditional mean guarantee an unchanged conditional distribution. These distinctions matter when an insurer assesses paired risks under different observed conditions. The methods here are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: For random variables $X$ and $Y$, the marginal distribution of $X$ is its unconditional distribution. For an event $B$ with $P(B)>0$, the conditional distribution of $X$ given $B$ assigns probability $P(X\in A\mid B)$ to a set $A$, where $$ P(X\in A\mid B)=\frac{P(\{X\in A\}\cap B)}{P(B)}. $$ For discrete $Y$, the conditional law of $X$ given $Y=y$ is defined when $P(Y=y)>0$. For continuous variables, conditional distributions or densities are defined for almost every conditioning value $y$; the event $Y=y$ itself can have probability zero.

The Independence Test

The key comparison is between the law of $X$ given information about $Y$ and the marginal law of $X$. If conditioning on $Y$ changes probabilities for any relevant range of $X$, the variables are dependent. In discrete settings, it is enough to compare conditional probabilities at values of $Y$ that have positive probability. In continuous settings, compare conditional densities or conditional CDFs with their marginal counterparts.

Theorem: $X$ and $Y$ are independent if and only if, for every set $A$ and every conditioning event $B$ determined by $Y$ with $P(B)>0$, $$ P(X\in A\mid B)=P(X\in A). $$ Equivalently, for discrete $Y$, the conditional distribution of $X$ given each positive-probability value $Y=y$ equals the marginal distribution of $X$. For continuous $Y$, the conditional distribution equals the marginal distribution for almost every $y$.

The equivalence follows directly from the definition of conditional probability. If $X$ and $Y$ are independent, then $P(\{X\in A\}\cap B)=P(X\in A)P(B)$, so dividing by $P(B)>0$ gives $P(X\in A\mid B)=P(X\in A)$. Conversely, if this conditional probability equality holds for every $A$ and $B$, multiplying by $P(B)$ gives the factorization $P(\{X\in A\}\cap B)=P(X\in A)P(B)$, which is the definition of independence. A conditioning event with probability zero is not used in this argument because the conditional probability ratio is undefined.

In practice, compare probabilities across a collection of sets $A$, or compare the complete conditional and marginal CDFs. A single matching statistic is not enough. Equal conditional and marginal means, for example, establish only agreement of the means—not agreement of the distributions.

Worked Example: Equal Conditional Means, Different Loss Laws

A hypothetical insurer classifies a policy-year as either regime $A$ or regime $B$, each with probability $0.5$. The claim loss $X$, measured in thousands of dollars, has the following conditional laws:

Given $A$, $X=0$ with probability $0.5$ and $X=2$ with probability $0.5$.
Given $B$, $X=1$ with probability $1$.

Find the marginal distribution and mean of $X$, compare the conditional means, and decide whether $X$ and the regime indicator are independent.

Method. Apply the law of total probability to each possible loss value. Then compare conditional distributions with the resulting marginal distribution.

For $X=0$, only regime $A$ contributes, so $P(X=0)=0.5(0.5)=0.25$. For $X=1$, only regime $B$ contributes, giving $P(X=1)=0.5(1)=0.50$. For $X=2$, only regime $A$ contributes, giving $P(X=2)=0.5(0.5)=0.25$. The probabilities sum to $0.25+0.50+0.25=1$.

The marginal mean is $$ E[X]=0(0.25)+1(0.50)+2(0.25)=1. $$ The conditional means are $E[X\mid A]=0(0.5)+2(0.5)=1$ and $E[X\mid B]=1(1)=1$. Thus both conditional means match the marginal mean.

But the conditional laws differ: given $A$, $P(X=0\mid A)=0.5$; given $B$, $P(X=0\mid B)=0$; and marginally $P(X=0)=0.25$. The independence criterion fails, so the loss and regime are dependent. As a further check, the conditional variances differ: $\operatorname{Var}(X\mid A)=1$, while $\operatorname{Var}(X\mid B)=0$. Common mistake avoided: equal conditional means do not imply equal conditional distributions or independence.

Identical Marginals Can Still Be Dependent

A comparison between two losses can be especially misleading if it stops at the marginal distributions. The marginal laws of $X$ and $Y$ say how each loss behaves alone; independence requires an additional condition about how they behave together. The next example shows that even identical marginal laws provide no guarantee of independence.

Worked Example: Matching Marginals, Different Conditional Losses

For a hypothetical pair of coverages, let the losses $X$ and $Y$, in thousands of dollars, have joint probabilities $P(X=0,Y=0)=0.5$ and $P(X=100,Y=100)=0.5$. All other pairs have probability zero. Find both marginal distributions, compare $X\mid Y=100$ with the marginal distribution of $X$, and test independence.

Method. Sum the joint probabilities over the other variable to obtain each marginal. Then calculate the conditional probability by dividing the relevant joint probability by $P(Y=100)$.

For $X$, the marginal probabilities are $P(X=0)=0.5$ and $P(X=100)=0.5$. The same calculation gives $P(Y=0)=0.5$ and $P(Y=100)=0.5$. Each loss has mean $0(0.5)+100(0.5)=50$ thousand dollars, and the two marginal distributions are identical.

Since $P(Y=100)=0.5$, $$ P(X=100\mid Y=100)=\frac{P(X=100,Y=100)}{P(Y=100)} =\frac{0.5}{0.5}=1. $$ By contrast, $P(X=100)=0.5$. Thus the conditional distribution of $X$ given $Y=100$ is not its marginal distribution. Independence would require $P(X=100,Y=100)=P(X=100)P(Y=100)=0.5(0.5)=0.25$, but the specified joint probability is $0.5$. The losses are dependent. Common mistake avoided: matching marginal distributions and means does not establish independence.

Comparing Conditional Densities and CDFs

For continuous variables, a conditional density makes the same comparison possible. Suppose the joint density is $f_{X,Y}(x,y)$, and the marginal density $f_Y(y)$ is positive. The conditional density is the joint density divided by the marginal density of the conditioning variable. Its integral over $x$ must equal one for almost every relevant $y$.

Definition: Where $f_Y(y)>0$, the conditional density of $X$ given $Y=y$ is $$ f_{X\mid Y}(x\mid y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}. $$ The associated conditional CDF is $F_{X\mid Y}(a\mid y)=\int_{-\infty}^{a}f_{X\mid Y}(x\mid y)\,dx$. Independence requires this conditional CDF to equal the marginal CDF $F_X(a)$ for every $a$, for almost every $y$.

Worked Example: Uniform Marginals with Changing Conditional Laws

Let $X$ and $Y$ have joint density $$ f_{X,Y}(x,y)=1+\frac{1}{2}(2x-1)(2y-1), \qquad 0<x<1,\quad 0<y<1, $$ and density zero elsewhere. Verify the marginal densities, calculate two conditional CDF values, and determine whether $X$ and $Y$ are independent.

Method. Integrate the joint density over the other variable to obtain each marginal. Divide by $f_Y(y)$ to get the conditional density, then compare conditional and marginal probabilities.

For fixed $x$, integrating over $0<y<1$ gives $$ f_X(x)=\int_0^1 \left[1+\frac{1}{2}(2x-1)(2y-1)\right]dy=1, $$ because $\int_0^1(2y-1)\,dy=0$. By the same calculation, $f_Y(y)=1$. Both marginals are uniform on $(0,1)$. The joint density is valid: its smallest possible value on the unit square is $1-\frac12=\frac12$, so it is nonnegative, and its integral is one because both marginal densities integrate to one.

Since $f_Y(y)=1$, the conditional density is $$ f_{X\mid Y}(x\mid y)=1+\frac{1}{2}(2x-1)(2y-1). $$ For $0<a<1$, integrating from zero to $a$ yields $$ F_{X\mid Y}(a\mid y)=a+\frac{1}{2}(2y-1)(a^2-a). $$ At $a=0.25$ and $y=0.75$, this is $0.25+\frac12(0.5)(0.0625-0.25)=0.203125$, or $13/64$. The marginal CDF at $0.25$ is $0.25=16/64$. At $y=0.25$, the conditional CDF is $0.25+\frac12(-0.5)(-0.1875)=0.296875$, or $19/64$. Both conditional values differ from the marginal value, so the variables are dependent. Common mistake avoided: uniform marginal distributions do not imply independence.

A Practical Diagnostic and Common Traps

A reliable analysis states what is being conditioned on and compares probabilities, not just summaries. For discrete losses, choose a possible value or event for $Y$, calculate the conditional probabilities for values or ranges of $X$, and compare them with the marginal probabilities. For continuous losses, compare conditional densities or CDF values with the corresponding marginal quantities. If at least one comparison differs, independence is ruled out. If the full distributions agree across all relevant conditioning values (almost everywhere in the continuous case), the independence criterion is met.

A calculation such as $E[X\mid Y=y]=E[X]$ for each $y$ is not a full test. It checks only a summary. Likewise, equal marginal means, variances, or even complete marginal distributions do not establish independence. A conditional law can shift probability among possible losses while preserving its mean, or it can vary even though both variables have the same marginal distribution.

Key takeaway: Independence is a statement about complete conditional and marginal distributions. To diagnose dependence, compare conditional probabilities, CDFs, or densities with the marginal law; matching means or other isolated summaries is not enough.

Check Your Understanding

For each problem, compare the conditional law with the marginal law rather than relying only on means.

A regime $R$ is $A$ or $B$, each with probability $0.5$. A loss $X$, in thousands of dollars, is 0 or 4 with equal probabilities given $A$, and is 2 with probability 1 given $B$. Find the marginal distribution of $X$ and decide whether $X$ and $R$ are independent. (Answer: $P(X=0)=0.25,\ P(X=2)=0.50,\ P(X=4)=0.25$; dependent.)
Two losses have joint probabilities $P(X=0,Y=0)=0.3$, $P(X=0,Y=10)=0.2$, $P(X=10,Y=0)=0.2$, and $P(X=10,Y=10)=0.3$. Find $P(X=10\mid Y=10)$, compare it with $P(X=10)$, and determine whether the losses are independent. (Answer: $P(X=10\mid Y=10)=0.6$, while $P(X=10)=0.5$; dependent.)
Suppose $f_{X,Y}(x,y)=2x$ for $0<x<1,\ 0<y<1$, and zero elsewhere. Find $f_X(x)$, $f_Y(y)$, and $f_{X\mid Y}(x\mid y)$. Are the variables independent? (Answer: $f_X(x)=2x,\ f_Y(y)=1,\ f_{X\mid Y}(x\mid y)=2x$; independent.)
For the density in Question 3, calculate $P(X\leq 0.5\mid Y=0.8)$ and the marginal probability $P(X\leq 0.5)$. (Answer: both are $0.25$.)
A hypothetical loss has $E[X\mid A]=E[X\mid B]=30$ thousand dollars, but $P(X>50\mid A)=0.1$ and $P(X>50\mid B)=0.4$, where both regimes have positive probability. Can the loss be independent of the regime? (Answer: No; the conditional tail probabilities differ, so the conditional distributions are not equal.)

What You'll Learn