Conditioning on Events Versus Conditioning on Values

Why the Conditioning Statement Matters

A statement such as “the probability of a large claim, given a high exposure” is incomplete until “high exposure” is made precise. It could mean that exposure belongs to a range, or that it equals a particular value. Those two conditions generally produce different conditional distributions. The distinction is especially important when the variable being conditioned on is continuous: an exact value has probability zero, even though a conditional distribution at that value can be described using a density.

The key is to match the conditioning method to the condition. For a positive-probability event, conditional probability is a ratio of probabilities. For a value of a continuous random variable, the useful object is a conditional density, obtained from the joint density and the marginal density.

Definition: If $A$ and $B$ are events with $P(B)>0$, the conditional probability of $A$ given $B$ is $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ The condition $P(B)>0$ is essential for this ratio definition.

When $X$ is discrete and $P(X=x)>0$, the event $\{X=x\}$ has positive probability, so $P(Y\in C\mid X=x)$ is an ordinary event-conditioned probability. When $X$ has a continuous distribution, $P(X=x)=0$ for every individual value $x$. The ratio definition would then have a zero denominator. Conditional densities provide the appropriate description at continuous values.

Worked Example: Conditioning on a Discrete Value or a Discrete Event

A hypothetical inspection study records a rating $X\in\{0,1,2\}$ and an indicator $Y$, where $Y=1$ means that a part requires rework. The joint probabilities are shown below. Find (a) $P(Y=1\mid X=1)$ and (b) $P(Y=1\mid X\geq 1)$.

$X$	$P(X=x,Y=0)$	$P(X=x,Y=1)$	$P(X=x)$
0	0.10	0.10	0.20
1	0.20	0.30	0.50
2	0.10	0.20	0.30

(a) Condition on the value $X=1$. The joint probability of $Y=1$ and $X=1$ is $0.30$, and $P(X=1)=0.50$. Thus

P(Y=1\mid X=1)=\frac{P(Y=1,X=1)}{P(X=1)} =\frac{0.30}{0.50}=0.60.

(b) Condition on the event $X\geq 1$. The event includes both $X=1$ and $X=2$. Its probability is $0.50+0.30=0.80$. The joint probability of rework and this event is $0.30+0.20=0.50$. Therefore

P(Y=1\mid X\geq 1)=\frac{0.50}{0.80}=0.625.

The two answers differ because $X\geq1$ combines two rating groups, whereas $X=1$ selects only one. Method and trap avoided: add the joint probabilities and the conditioning probabilities over all values included in the event; do not substitute the result for one value.

Conditioning on a Continuous Value

Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)$, and $X$ has marginal density $f_X(x)$. For a continuous variable, $P(X=x)=0$, so the event ratio cannot define $P(Y\in C\mid X=x)$. Instead, the conditional density describes the distribution of $Y$ at the specified value $x$.

Definition: At values $x$ for which $f_X(x)>0$, a conditional density of $Y$ given $X=x$ is $$ f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)}. $$ Consequently, for a set $C$ of possible $Y$-values, $$ P(Y\in C\mid X=x)=\int_C f_{Y\mid X}(y\mid x)\,dy, $$ with the integral taken over the possible values of $Y$ under the condition.

This formula applies where the marginal density is positive. A joint density equal to zero outside its support contributes no conditional probability there. Conditional densities are defined up to sets of $x$-values having marginal probability zero; at a particular such value, a formula for a conditional law may not be uniquely determined by the joint distribution. In routine continuous models, the density ratio gives the natural conditional version wherever $f_X(x)>0$.

Why the Density Ratio Works

The ratio is not obtained by dividing by $P(X=x)$. Instead, it is the limiting conditional law over increasingly narrow intervals around $x$. This argument explains both the role of the marginal density and the condition that it be positive.

Condition on a short interval.
Take an interval $(x,x+h)$ and a set $C$ of possible $Y$-values. By the joint-density formula, the probability that $Y\in C$ and $x<X<x+h$ is the integral of the joint density over that region.

Form an ordinary event-conditioned probability.
For $P(x<X<x+h)>0$, divide the joint probability by $P(x<X<x+h)$. The denominator is the integral of $f_X$ over the same interval.

Shrink the interval.
If the relevant densities are continuous near $x$ and $f_X(x)>0$, the ratio of these integrals tends to the ratio of their integrands at $x$. This gives the conditional probability for $C$ by integrating $f_{X,Y}(x,y)/f_X(x)$ over $C$.

In equations, for a suitable set $C$, the interval-conditioned probability is

P(Y\in C\mid x<X<x+h) = \frac{\displaystyle\int_x^{x+h}\int_C f_{X,Y}(u,y)\,dy\,du} {\displaystyle\int_x^{x+h} f_X(u)\,du}.

As $h$ decreases to zero, continuity and $f_X(x)>0$ give

\lim_{h\downarrow0}P(Y\in C\mid x<X<x+h) = \frac{\displaystyle\int_C f_{X,Y}(x,y)\,dy}{f_X(x)} = \int_C f_{Y\mid X}(y\mid x)\,dy.

Thus the conditional density is a local limiting description. It is not an ordinary conditional probability formed by dividing by $P(X=x)$.

Worked Example: A Point Condition and a Positive-Probability Event

Let $(X,Y)$ have joint density $f_{X,Y}(x,y)=2$ on $0<x<1,\ 0<y<1-x$, and zero elsewhere. Find (a) $P(Y>0.6\mid X<0.4)$ and (b) $P(Y>0.6\mid X=0.1)$. Give probabilities to four decimal places.

First find the marginal density of $X$. At a fixed $x$, the allowed values of $y$ run from $0$ to $1-x$. Therefore

f_X(x)=\int_0^{1-x}2\,dy=2(1-x),\qquad 0<x<1.

(a) Condition on $X<0.4$. The conditioning event has positive probability:

P(X<0.4)=\int_0^{0.4}2(1-x)\,dx =[2x-x^2]_0^{0.4} =0.80-0.16=0.64.

For $Y>0.6$, the triangular support requires $y<1-x$. When $0<x<0.4$, the allowed $y$-range in the numerator is $0.6<y<1-x$. Hence

P(Y>0.6,\ X<0.4) =\int_0^{0.4}\int_{0.6}^{1-x}2\,dy\,dx =2\int_0^{0.4}(0.4-x)\,dx =2(0.16-0.08)=0.16.

Divide the joint probability by the probability of the conditioning event:

P(Y>0.6\mid X<0.4)=\frac{0.16}{0.64}=0.2500.

(b) Condition on the continuous value $X=0.1$. The conditional density is

f_{Y\mid X}(y\mid 0.1) =\frac{2}{2(1-0.1)} =\frac{1}{0.9},\qquad 0<y<0.9.

Thus $Y$, under this conditional law, is uniform over an interval of length $0.9$. The part of the interval above $0.6$ has length $0.9-0.6=0.3$, so

P(Y>0.6\mid X=0.1)=\int_{0.6}^{0.9}\frac{1}{0.9}\,dy =\frac{0.3}{0.9} =\frac{1}{3} \approx 0.3333.

The event-conditioned answer is $0.2500$; the value-conditioned answer is $0.3333$. Method and trap avoided: use a probability ratio for $X<0.4$, but use a conditional density for $X=0.1$. The exact-value event has probability zero, so it cannot be the denominator of the event ratio.

Conditioning on a Range Is Not Conditioning at One Value

A condition such as $a<X<b$ has positive probability when the density assigns positive mass to that interval. The distribution of $Y$ under this condition combines the conditional distributions across all $x$-values in the interval. It is therefore generally not the same as the conditional distribution at one selected value of $X$.

Worked Example: Conditioning on an Interval of Continuous Values

Use the joint density $f_{X,Y}(x,y)=2$ on $0<x<1,\ 0<y<1-x$. Find $P(Y>0.5\mid 0.2<X<0.6)$. Give the final probability to four decimal places.

Method. Calculate the joint probability of the event in the numerator and divide by the probability of the interval condition. The marginal density is $f_X(x)=2(1-x)$.

P(0.2<X<0.6) =\int_{0.2}^{0.6}2(1-x)\,dx =[2x-x^2]_{0.2}^{0.6} =(1.20-0.36)-(0.40-0.04) =0.84-0.36 =0.48.

For $Y>0.5$, the support requires $0.5<y<1-x$, which is possible only when $x<0.5$. Combining this with $0.2<x<0.6$, the numerator has $0.2<x<0.5$:

P(Y>0.5,\ 0.2<X<0.6) =\int_{0.2}^{0.5}\int_{0.5}^{1-x}2\,dy\,dx =2\int_{0.2}^{0.5}(0.5-x)\,dx.

Evaluate the integral:

2\left[0.5x-\frac{x^2}{2}\right]_{0.2}^{0.5} =2\left((0.25-0.125)-(0.10-0.02)\right) =2(0.125-0.08) =0.09.

Therefore

P(Y>0.5\mid 0.2<X<0.6) =\frac{0.09}{0.48} =0.1875.

As a comparison, at the single value $X=0.4$, the conditional density of $Y$ is uniform on $(0,0.6)$, giving $P(Y>0.5\mid X=0.4)=0.1/0.6\approx0.1667$. The value $0.1667$ does not replace the interval-conditioned probability $0.1875$. Method and trap avoided: impose the support limits in the numerator and integrate over every $x$-value in the conditioning interval.

Common Mistake: Treating a Density as a Probability

A density value is not the probability of the variable equalling that value, nor is it the probability of an interval. For a continuous $X$, probabilities over intervals are integrals of $f_X$. In particular, $f_X(c)$ cannot be substituted for $P(X<c)$ or $P(X=c)$.

In the triangular model above, $f_X(0.4)=2(1-0.4)=1.2$, while $P(X<0.4)=0.64$. The incorrect calculation $P(Y>0.6\mid X<0.4)=0.16/1.2$ uses a density value in place of the event probability and gives approximately $0.1333$. The correct denominator is the integral of the marginal density:

P(X<0.4)=\int_0^{0.4}f_X(x)\,dx=0.64, \qquad P(Y>0.6\mid X<0.4)=\frac{0.16}{0.64}=0.2500.

A related error is to use the event formula with $B=\{X=c\}$ for a continuous $X$. Since $P(X=c)=0$, that ratio is undefined. Use the conditional-density formula instead, provided $f_X(c)>0$.

Key takeaway: If the condition is a positive-probability event, divide the relevant joint probability by the event probability. If the condition specifies a value of a continuous variable, use the conditional density $f_{X,Y}(x,y)/f_X(x)$ where the marginal density is positive. Conditioning on a range generally averages over the values in that range; it is not conditioning at a single value.

Check Your Understanding

For each problem, identify whether the condition is an event or a continuous value, then calculate as appropriate.

A discrete joint distribution has $P(X=0,Y=1)=0.12$, $P(X=0)=0.30$, $P(X=1,Y=1)=0.28$, and $P(X=1)=0.70$. Find $P(Y=1\mid X=1)$. (Answer: $0.4000$)
A joint density is $2$ on $0<x<1,\ 0<y<1-x$. Find $P(Y>0.4\mid X=0.25)$. (Answer: $0.4667$)
For the same joint density, find $P(Y>0.5\mid X<0.25)$. (Answer: $0.4286$)
A joint density is $2$ on $0<y<x<1$. Find $P(Y<0.3\mid X=0.8)$. (Answer: $0.3750$)
For the density $2$ on $0<x<1,\ 0<y<1-x$, find $P(Y>0.5\mid 0.2<X<0.6)$. (Answer: $0.1875$)

What You'll Learn