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Paired-Loss Dependence Problem-Solving Workshop

Learn a reliable step-by-step method for solving paired-loss dependence problems and checking each calculation against the joint model.

Intermediate 10 min read

What You'll Learn

  • Read a joint probability table or density and verify that it defines a valid joint law.
  • Derive marginal and conditional distributions from a joint law.
  • Test independence using factorization, including support restrictions.
  • Calculate joint event probabilities from the correct region of a joint distribution.
  • Interpret covariance in the units of the paired losses.
  • Distinguish conditional independence given a shared driver from unconditional independence.

Introduction

A paired-loss problem is solved from the joint law: the probability model describing the two losses together. Marginals alone do not tell the whole story. Two losses can have the same individual distributions under different joint laws, with very different probabilities of experiencing large losses at the same time.

A reliable solution therefore follows a fixed sequence: identify the joint model and its support, verify it, derive the marginals or conditional distributions needed, and then calculate the requested event or summary. This tutorial applies that sequence to discrete, continuous, and shared-driver examples. Its methods are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: The joint law of losses \(X\) and \(Y\) specifies probabilities for pairs of values or regions of values. For discrete losses, it is given by the joint probability mass function \(p_{X,Y}(x,y)=P(X=x,Y=y)\). For continuous losses, it is given by a joint density \(f_{X,Y}(x,y)\), for which probabilities are integrals over regions in the joint support.

The marginal law of \(X\) is obtained by summing over all possible values of \(Y\), in the discrete case, or integrating over \(Y\), in the continuous case. Conditional distributions describe one loss after the value or event for the other loss is known. For discrete variables, for example, \(P(Y=y\mid X=x)=P(X=x,Y=y)/P(X=x)\), provided \(P(X=x)>0\).

A Practical Solution Sequence

Before calculating, make the support and the conditioning event explicit. A probability table must have nonnegative entries that sum to one. A density must be nonnegative on its support and integrate to one. Only after these checks should marginalization, conditioning, or event calculations begin.

1
Identify the joint model and support.
Record whether the law is a table, a density, or a specified mixture, and state where its probabilities or density are nonzero.
2
Check that the model is valid.
Verify that probabilities sum to one or that the density integrates to one over the full support.
3
Derive what the question needs.
Find marginals by summing or integrating, and find conditional laws by dividing the joint law by the relevant marginal probability or density.
4
Calculate and interpret.
For an event, translate its inequalities into a region of the joint support. For a covariance, calculate the joint product expectation and both marginal means.

Worked Example: A Joint Probability Table

Worked Example: A Joint Probability Table

A hypothetical insurer models two annual loss amounts, \(X\) and \(Y\), in thousands of dollars. Their joint probability mass function is shown below. Find the marginal distributions, test independence, calculate \(P(X\geq 1,Y=2)\), find \(E[X]\), \(E[Y]\), and \(\operatorname{Cov}(X,Y)\), and determine the conditional distribution of \(Y\) given \(X=2\).

\(X\backslash Y\)\(Y=0\)\(Y=1\)\(Y=2\)Marginal \(P(X=x)\)
\(X=0\)0.100.050.050.20
\(X=1\)0.100.200.100.40
\(X=2\)0.050.100.250.40
Marginal \(P(Y=y)\)0.250.350.401.00

Validity and marginals. The row totals are \(0.20+0.40+0.40=1.00\), so the table is a valid joint probability distribution. The marginal probabilities are the row and column totals shown in the table.

Independence test. Independence requires every joint probability to equal the product of its corresponding marginals. For the cell \(X=0,Y=0\), the joint probability is \(0.10\), but the product of the marginals is \(0.20(0.25)=0.05\). Since these are unequal, \(X\) and \(Y\) are not independent.

Joint event. The event \(X\geq 1,Y=2\) includes the cells \((1,2)\) and \((2,2)\), so

$$ P(X\geq 1,Y=2)=0.10+0.25=0.35. $$

Means and covariance. The means are

$$ E[X]=0(0.20)+1(0.40)+2(0.40)=1.20, \qquad E[Y]=0(0.25)+1(0.35)+2(0.40)=1.15. $$

For the joint product expectation, only cells with both coordinates nonzero contribute:

$$ E[XY]=1(1)(0.20)+1(2)(0.10)+2(1)(0.10)+2(2)(0.25) =0.20+0.20+0.20+1.00=1.60. $$

Therefore,

$$ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y] =1.60-(1.20)(1.15)=1.60-1.38=0.22. $$

Since each loss is measured in thousands of dollars, this covariance is in squared thousands of dollars. It is positive, indicating positive average co-movement as summarized by covariance; it is not a probability.

Conditional distribution. Given \(X=2\), divide each probability in the \(X=2\) row by the row total \(P(X=2)=0.40\):

$$ P(Y=0\mid X=2)=\frac{0.05}{0.40}=0.125,\quad P(Y=1\mid X=2)=\frac{0.10}{0.40}=0.250,\quad P(Y=2\mid X=2)=\frac{0.25}{0.40}=0.625. $$

These conditional probabilities sum to \(0.125+0.250+0.625=1.000\), as required. The method avoids two common errors: using marginal probabilities to calculate a joint event, and dividing a row’s entries by the wrong marginal total.

Continuous Joint Laws: Marginals, Conditioning, and Events

For a continuous pair, the support is part of the model. A density that is constant on a triangle, for example, is not constant on the entire square that contains it. The integration bounds must follow the actual support and the event being calculated.

Conditional-density result: If \(f_{X,Y}(x,y)\) is a joint density and \(f_Y(y)>0\), then \(f_{X\mid Y}(x\mid y)=f_{X,Y}(x,y)/f_Y(y)\). The marginal density is \(f_Y(y)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx\). The corresponding formulas for \(f_X\) and \(f_{Y\mid X}\) follow by exchanging the roles of \(X\) and \(Y\).

This conditional-density formula follows from the same factorization as conditional probability. For small intervals around \(x\) and \(y\), the joint probability is approximately \(f_{X,Y}(x,y)\,dx\,dy\), while the probability of \(Y\) being in its small interval is approximately \(f_Y(y)\,dy\). Dividing the joint probability by the probability of the conditioning interval and taking the limiting density gives \(f_{X,Y}(x,y)/f_Y(y)\).

Worked Example: A Triangular Support and a Joint Exceedance

For a hypothetical paired-loss severity model, let \(X\) and \(Y\) be continuous variables with joint density \(f_{X,Y}(x,y)=2\) on \(0<x<y<1\), and zero elsewhere. Verify the density, derive both marginals and the conditional density of \(X\) given \(Y=y\), test independence, calculate \(P(X>1/2,Y>3/4)\), and find the covariance.

Verify the model. The density is nonnegative. Its integral over the triangular support is

$$ \int_0^1\int_0^y 2\,dx\,dy =\int_0^1 2y\,dy =\left[y^2\right]_0^1=1. $$

Thus it is a valid joint density.

Marginal densities. For fixed \(x\), the support requires \(x<y<1\); for fixed \(y\), it requires \(0<x<y\). Therefore,

$$ f_X(x)=\int_x^1 2\,dy=2(1-x),\quad 0<x<1, \qquad f_Y(y)=\int_0^y 2\,dx=2y,\quad 0<y<1. $$

Each marginal integrates to one: \(\int_0^1 2(1-x)\,dx=1\) and \(\int_0^1 2y\,dy=1\). For \(0<y<1\), the conditional density is

$$ f_{X\mid Y}(x\mid y)=\frac{2}{2y}=\frac{1}{y},\quad 0<x<y. $$

This is uniform over the interval \((0,y)\), and its integral over that interval is \((1/y)y=1\). The pair is not independent: the joint density is zero outside \(x<y\), even though the product \(f_X(x)f_Y(y)\) is positive for some points with \(x>y\). Thus the joint density does not factor as the product of the marginal densities.

Joint event. The event \(X>1/2,Y>3/4\), within the support, has \(3/4<y<1\) and \(1/2<x<y\). Integrate over that region:

$$ \begin{aligned} P(X>1/2,Y>3/4) &=\int_{3/4}^{1}\int_{1/2}^{y}2\,dx\,dy\\ &=\int_{3/4}^{1}(2y-1)\,dy\\ &=\left[y^2-y\right]_{3/4}^{1} =0-\left(\frac{9}{16}-\frac{12}{16}\right) =\frac{3}{16}=0.1875. \end{aligned} $$

The inner integral is \(2y-1\): the density is \(2\), and the interval of integration in \(x\) has length \(y-1/2\). Multiplying these gives \(2(y-1/2)=2y-1\).

Covariance. First calculate the means and product expectation from the joint law or marginals:

$$ \begin{aligned} E[X]&=\int_0^1 x\,2(1-x)\,dx =2\left(\frac12-\frac13\right)=\frac13,\\ E[Y]&=\int_0^1 y\,2y\,dy =2\left(\frac13\right)=\frac23,\\ E[XY]&=\int_0^1\int_0^y 2xy\,dx\,dy =\int_0^1 y^3\,dy=\frac14. \end{aligned} $$

Hence \(\operatorname{Cov}(X,Y)=1/4-(1/3)(2/3)=1/4-2/9=1/36\). The positive covariance is consistent with the support constraint that \(Y\) always exceeds \(X\), though that constraint alone is not a general rule for the sign of covariance.

Common-Driver Models and Unconditional Dependence

A shared risk driver can make losses dependent even when they are conditionally independent once the driver is known. To solve such a problem, calculate within each driver state first, then average over the states to obtain unconditional probabilities. Do not confuse a conditional model with the unconditional joint law.

Worked Example: Conditional Independence Does Not Remove Unconditional Dependence

A hypothetical portfolio has a quiet state with probability \(0.75\) and a stressed state with probability \(0.25\). Let \(X\) and \(Y\) be indicators that two distinct coverages generate a positive loss during the year. Conditional on the state, the indicators are independent. In the quiet state, \(P(X=1)=0.10\) and \(P(Y=1)=0.20\); in the stressed state, \(P(X=1)=0.60\) and \(P(Y=1)=0.70\). Find the unconditional marginals, the joint table, \(P(Y=1\mid X=1)\), and the covariance.

Joint probability of two positive losses. Conditional independence gives the product of the conditional probabilities within each state. Averaging over the states,

$$ P(X=1,Y=1) =0.75(0.10)(0.20)+0.25(0.60)(0.70) =0.015+0.105=0.120. $$

The unconditional marginal probabilities are

$$ P(X=1)=0.75(0.10)+0.25(0.60)=0.075+0.150=0.225, $$
$$ P(Y=1)=0.75(0.20)+0.25(0.70)=0.150+0.175=0.325. $$

Subtract the joint probability from each marginal to obtain the other cells:

$$ \begin{aligned} P(X=1,Y=0)&=0.225-0.120=0.105,\\ P(X=0,Y=1)&=0.325-0.120=0.205,\\ P(X=0,Y=0)&=1-0.120-0.105-0.205=0.570. \end{aligned} $$

The four cells sum to \(0.120+0.105+0.205+0.570=1.000\). The joint table is therefore

\(Y=0\)\(Y=1\)Total
\(X=0\)0.5700.2050.775
\(X=1\)0.1050.1200.225
Total0.6750.3251.000

Conditional probability and dependence. The requested conditional probability is

$$ P(Y=1\mid X=1)=\frac{0.120}{0.225}=\frac{8}{15}\approx 0.5333. $$

If \(X\) and \(Y\) were unconditionally independent, their joint probability of both being one would be \((0.225)(0.325)=0.073125\), not \(0.120\). They are therefore dependent without conditioning on the state.

Covariance. Since \(X\) and \(Y\) are indicators, \(E[X]=0.225\), \(E[Y]=0.325\), and \(E[XY]=P(X=1,Y=1)=0.120\). Thus

$$ \operatorname{Cov}(X,Y) =0.120-(0.225)(0.325) =0.120-0.073125 =0.046875. $$

The positive covariance reflects that both indicators are more likely to be one in the stressed state. Conditional independence within each state does not imply unconditional independence across the mixture of states.

Common Calculation Trap

A frequent error in a continuous joint-event calculation is to integrate the density over the correct bounds but then evaluate the inner integral incorrectly. For the triangular-density example, the inner integral is

$$ \int_{1/2}^{y}2\,dx=2\left(y-\frac12\right)=2y-1, $$

not \(y^2-1/4\). The latter expression incorrectly treats the constant density as though it depended on \(x\) in a way that produces a difference of squares. The corrected event probability is

$$ \int_{3/4}^{1}(2y-1)\,dy =\left[y^2-y\right]_{3/4}^{1} =\frac{3}{16}=0.1875. $$

A useful check is to interpret the inner integral geometrically: a constant density multiplied by the width of the allowable \(x\)-interval must give the integrand for the outer integral. When the upper bound depends on \(y\), that width depends on \(y\) too.

Check Your Understanding

For each problem, begin with the joint model and keep the support or conditioning event explicit.

  1. A discrete pair has probabilities \(P(0,0)=0.30\), \(P(0,2)=0.20\), \(P(1,0)=0.10\), and \(P(1,2)=0.40\), where each ordered pair is \((X,Y)\). Find \(P(Y=2\mid X=1)\), test independence, and calculate \(P(X=0\text{ or }Y=2)\). (Answer: \(0.8000\); not independent; \(0.9000\).)
  2. For the density \(f_{X,Y}(x,y)=2\) on \(0<x<y<1\), find \(P(X<1/2\mid Y=3/4)\). (Answer: \(2/3\approx 0.6667\).)
  3. A shared state is quiet with probability \(0.80\) and stressed with probability \(0.20\). Conditional on the state, binary indicators \(X,Y\) are independent; their probabilities of one are \(0.20,0.10\) in quiet conditions and \(0.70,0.50\) in stressed conditions, respectively. Find \(P(X=1,Y=1)\), \(P(X=1)\), and \(P(Y=1\mid X=1)\). (Answer: \(0.0860,\ 0.3000,\ 0.2867\).)
  4. A binary pair has \(P(0,0)=0.40\), \(P(0,1)=0.20\), \(P(1,0)=0.10\), and \(P(1,1)=0.30\). Calculate \(\operatorname{Cov}(X,Y)\) and \(P(X=1\text{ or }Y=1)\). (Answer: \(0.1000,\ 0.6000\).)
  5. For the density \(f_{X,Y}(x,y)=2\) on \(0<x<y<1\), calculate \(P(X>1/3,Y>2/3)\). (Answer: \(1/3\approx 0.3333\).)

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