Conditional Distributions From Joint Probabilities

Why Conditional Distributions Matter

A joint loss model describes how two random quantities behave together. In risk analysis, the practical question is often conditional: how is one loss distributed when another loss has been observed to take a particular value or fall in a particular range? The answer is a conditional distribution. It updates the probabilities for one variable using the information supplied by the other.

The calculation depends on whether the conditioning variable is discrete or continuous. For a discrete condition, divide joint probabilities by the probability of the condition. For a continuous variable observed at a point, divide joint densities by the corresponding marginal density. If the condition is an interval, first integrate the joint density over that interval. In every case, the resulting conditional law must be a valid probability distribution.

Definition: For discrete random variables \(X\) and \(Y\), the conditional probability mass function of \(X\) given \(Y=y\) is \(p_{X\mid Y}(x\mid y)=P(X=x\mid Y=y)=p_{X,Y}(x,y)/p_Y(y)\), provided \(p_Y(y)>0\). Here \(p_{X,Y}(x,y)=P(X=x,Y=y)\) is the joint mass function and \(p_Y(y)=P(Y=y)\) is the marginal mass function.

For continuous random variables, use densities instead of point probabilities. A continuous variable has probability zero of equaling any one exact value, so the ratio \(P(X\in A,Y=y)/P(Y=y)\) is not the calculation. Conditional densities are defined through the joint and marginal densities.

Definition: If \(X,Y\) have joint density \(f_{X,Y}(x,y)\) and marginal density \(f_Y(y)>0\), a conditional density of \(X\) given \(Y=y\) is \(f_{X\mid Y}(x\mid y)=f_{X,Y}(x,y)/f_Y(y)\). This formula applies where the joint density is supported; it is a conditional distribution for almost every \(y\) with positive marginal density. A value at a particular \(y\) is meaningful when a suitable version of the conditional density has been specified there.

Conditioning on a Discrete Loss

The discrete formula follows directly from the definition of conditional probability. The joint event \(X=x,Y=y\) is contained in the event \(Y=y\), so its probability, divided by \(P(Y=y)\), is the fraction of cases with \(Y=y\) that also have \(X=x\). Summing over all possible \(x\) gives \(\sum_x p_{X\mid Y}(x\mid y)=\sum_x p_{X,Y}(x,y)/p_Y(y)=p_Y(y)/p_Y(y)=1\). This sum-to-one check is an efficient way to detect a missing or incorrect denominator.

Worked Example: Conditional Claim Severity Given a Portfolio Loss

A hypothetical model records two losses, \(X\) and \(Y\), in thousands of dollars. Their joint probability mass function is given below; all combinations not shown have probability zero. Find the conditional mass function of \(X\) given \(Y=3\), and calculate \(P(X\geq 4\mid Y=3)\).

\(X\)	\(Y=0\)	\(Y=3\)	Marginal \(p_X(x)\)
0	0.10	0.05	0.15
4	0.20	0.25	0.45
10	0.15	0.25	0.40
Marginal \(p_Y(y)\)	0.45	0.55	1.00 total

Method: condition each joint probability on the observed value of \(Y\). First check the joint table: \(0.10+0.05+0.20+0.25+0.15+0.25=1.00\). The probability of the conditioning event is the \(Y=3\) column total: \(P(Y=3)=0.05+0.25+0.25=0.55\).

The three conditional probabilities are \(P(X=0\mid Y=3)=0.05/0.55=1/11\), \(P(X=4\mid Y=3)=0.25/0.55=5/11\), and \(P(X=10\mid Y=3)=0.25/0.55=5/11\). Thus the conditional mass function assigns probabilities \(1/11,5/11,5/11\) to \(X=0,4,10\), respectively. Their sum is \((1+5+5)/11=1\).

Since \(X\geq4\) includes \(X=4\) and \(X=10\), \(P(X\geq4\mid Y=3)=5/11+5/11=10/11=0.9091\) to four decimal places. Check: this is the combined conditional mass for the two qualifying outcomes and is no greater than one. The method avoids the common mistake of dividing by the total probability of the entire table rather than by \(P(Y=3)\).

Conditioning on an Exact Continuous Observation

For continuous variables, the marginal density is obtained by integrating the joint density over the variable not being conditioned on. The conditional density is then the joint density divided by that marginal density. It must integrate to one over the possible values of the variable being conditioned.

Justification: For a small interval of values near \(y\), the joint probability that \(X\) lies in a set \(A\) and \(Y\) lies in that interval is approximately \(\int_A f_{X,Y}(x,y)\,dx\,dy\). The probability that \(Y\) lies in the interval is approximately \(f_Y(y)\,dy\). Their ratio approaches \(\int_A f_{X,Y}(x,y)/f_Y(y)\,dx\) as the interval narrows. This is why the ratio of densities gives the conditional probability over \(A\), and hence the conditional density.

Worked Example: Loss Given an Observed Exposure Outcome

For a hypothetical pair of continuous losses, suppose the joint density is \(f_{X,Y}(x,y)=x+y\) on \(0<x<1,\ 0<y<1\), and zero elsewhere. Find \(f_{X\mid Y}(x\mid 0.5)\), then calculate \(P(X>0.75\mid Y=0.5)\).

Method: integrate to obtain the marginal, then divide. First verify that the proposed joint density is normalized: \(\int_0^1\int_0^1(x+y)\,dx\,dy=\int_0^1(1/2+y)\,dy=1/2+1/2=1\).

For \(0<y<1\), the marginal density of \(Y\) is \(f_Y(y)=\int_0^1(x+y)\,dx=1/2+y\). At \(y=0.5\), this gives \(f_Y(0.5)=1/2+1/2=1\). Therefore \(f_{X\mid Y}(x\mid0.5)=(x+0.5)/1=x+0.5\) for \(0<x<1\), and it is zero outside that interval.

The density check is \(\int_0^1(x+0.5)\,dx=1/2+1/2=1\). Now integrate over the requested tail: \(P(X>0.75\mid Y=0.5)=\int_{0.75}^1(x+0.5)\,dx\). The two terms are \((1^2-0.75^2)/2=0.21875\) and \(0.5(1-0.75)=0.125\). Their sum is \(0.34375=11/32\), or \(0.3438\) to four decimal places. Check: the result is between zero and one, and the conditional density integrates to one. The avoided mistake is inserting a normalization factor when the marginal density at \(0.5\) is already exactly one.

Conditioning on an Interval

Sometimes the available information is not an exact observation \(Y=y\), but rather an event such as \(a<Y<b\). This event has positive probability in many continuous models. Integrate the joint density over the allowed \(y\)-values to obtain the unnormalized density for \(X\), then divide by the probability of the conditioning event.

Result: If \(P(a<Y<b)>0\), then a conditional density of \(X\) given \(a<Y<b\) is \(f_{X\mid a<Y<b}(x)=\frac{\int_a^b f_{X,Y}(x,y)\,dy}{P(a<Y<b)}\), where \(P(a<Y<b)=\int_a^b\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\,dy\). The numerator is zero for values of \(x\) outside the joint support.

Restrict to the conditioning event.
Integrate the joint density over the specified interval of \(Y\), keeping \(x\) fixed.

Find the event probability.
Integrate the resulting function over all supported \(x\)-values, or integrate the joint density over the full conditioning region.

Normalize and answer the requested probability.
Divide by the event probability, check that the conditional density integrates to one, and integrate over any further range of \(X\).

Worked Example: Loss Given a Range of the Related Loss

Use the hypothetical joint density \(f_{X,Y}(x,y)=x+y\) on the unit square, zero elsewhere. Find the conditional density of \(X\) given \(0<Y<0.25\), and then find \(P(X>0.5\mid0<Y<0.25)\).

Method: marginalize over the conditioning interval and normalize. For fixed \(x\) in \(0<x<1\), the integrated joint density over \(0<y<0.25\) is \(\int_0^{0.25}(x+y)\,dy=0.25x+(0.25)^2/2=x/4+1/32\). This is not yet a conditional density; its integral over \(x\) is the probability of the conditioning event.

That event probability is \(\int_0^1(x/4+1/32)\,dx=(1/4)(1/2)+1/32=1/8+1/32=5/32\). Consequently, \(f_{X\mid0<Y<0.25}(x)=(x/4+1/32)/(5/32)=(8x+1)/5\), for \(0<x<1\), and zero otherwise.

Check normalization: \(\int_0^1(8x+1)/5\,dx=(4+1)/5=1\). For the requested probability, \(\int_{0.5}^1(8x+1)/5\,dx=[4x^2+x]_{0.5}^1/5\). At \(x=1\), the bracket is \(5\); at \(x=0.5\), it is \(4(0.25)+0.5=1.5\). The difference is \(3.5\), so the probability is \(3.5/5=0.7000\).

Check: the same result follows from the unnormalized integrals: the numerator is \(\int_{0.5}^1(x/4+1/32)\,dx=7/64\), while the denominator is \(5/32\), and \((7/64)/(5/32)=7/10\). The common error is to use the point-conditioned density at a selected \(y\), rather than integrating the joint density across the whole conditioning interval.

Common Errors and Practical Checks

Conditional calculations are short, but the denominator determines what population or probability mass is being retained. A correct numerator with the wrong denominator still produces the wrong conditional law. Before calculating, state the condition in words and identify its probability or marginal density.

Using a joint probability as the conditional probability. For discrete variables, divide by the marginal probability of the observed conditioning value.
Using \(P(Y=y)\) for a continuous variable. At an exact point this probability is zero; use the marginal density \(f_Y(y)\), when positive, in the density ratio.
Forgetting to integrate over an interval condition. Given \(a<Y<b\), integrate over all \(y\) in that interval before normalizing.
Skipping normalization checks. Conditional masses must sum to one and conditional densities must integrate to one over their support.
Ignoring support. A formula obtained from a density ratio is used only where the joint model permits the values; outside the support the conditional density is zero.

The distinction between exact-value conditioning and interval conditioning is especially important. The former uses the ratio \(f_{X,Y}(x,y)/f_Y(y)\) at a specified \(y\). The latter averages the joint density over the allowed range of \(Y\), then normalizes by the probability of that range. These operations generally give different distributions.

Key takeaway: A conditional law is the joint law restricted to the conditioning information and then normalized. Identify the correct conditioning event, use its probability or marginal density as the denominator, and verify that the result is a probability distribution.

Check Your Understanding

For each problem, calculate the conditional law or probability requested and check that the conditioning denominator matches the stated information.

A hypothetical discrete model has \(P(X=0,Y=1)=0.12\), \(P(X=3,Y=1)=0.18\), and \(P(X=8,Y=1)=0.30\). The remaining probability, \(0.40\), has \(Y\ne1\). Find the conditional mass function of \(X\) given \(Y=1\) and \(P(X\geq3\mid Y=1)\). (Answer: masses \(0.20, 0.30, 0.50\) at \(X=0,3,8\); \(P(X\geq3\mid Y=1)=0.8000\).)
A joint density is \(f_{X,Y}(x,y)=x+y\) on the unit square and zero elsewhere. Find \(f_{X\mid Y}(x\mid0.25)\) and \(P(X>0.5\mid Y=0.25)\). (Answer: \(f_{X\mid Y}(x\mid0.25)=(x+0.25)/0.75\) for \(0<x<1\), zero otherwise; probability \(2/3=0.6667\).)
For the same unit-square density, find the conditional density of \(X\) given \(0<Y<0.5\), and calculate \(P(X<0.25\mid0<Y<0.5)\). (Answer: density \((4x+1)/3\) for \(0<x<1\), zero otherwise; probability \(1/8=0.1250\).)
A hypothetical continuous model has joint density \(f_{X,Y}(x,y)=2\) for \(0<x<y<1\), and zero elsewhere. Find \(f_{X\mid Y}(x\mid0.6)\) and \(P(X<0.2\mid Y=0.6)\). (Answer: density \(1/0.6=5/3\) for \(0<x<0.6\), zero otherwise; probability \(1/3=0.3333\).)
For the unit-square density \(x+y\), find \(P(X>0.75\mid0<Y<0.25)\). (Answer: \(P=(\int_{0.75}^1(x/4+1/32)\,dx)/(5/32)=(29/512)/(5/32)=29/80=0.3625\).)

What You'll Learn