Conditional CDFs and Probabilities Over Regions

From a Joint Density to a Conditional CDF

A joint density describes how two continuous random variables vary together. To find the distribution of $Y$ when $X$ is fixed at a value $x$, focus on the vertical slice of the joint support at that $x$. The joint density along this slice must be rescaled so that its total integral is one. Integrating that conditional density up to a threshold gives the conditional cumulative distribution function (CDF).

The support matters at every step. It may give different lower or upper limits for different values of $x$, and a probability involving both variables can cut across that support. A reliable calculation first identifies the slice, then applies the density and limits appropriate to it.

Definition: Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)$, and let $f_X(x)$ be the marginal density of $X$. For a value $x$ with $f_X(x)>0$, the conditional density of $Y$ given $X=x$ is $$ f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)} $$ on the slice of the joint support at $x$, and is zero off that slice. The conditional CDF is $$ F_{Y\mid X}(t\mid x)=P(Y\leq t\mid X=x) =\int_{-\infty}^{t} f_{Y\mid X}(y\mid x)\,dy. $$

For a continuous $X$, $P(X=x)=0$. Thus the notation $P(Y\leq t\mid X=x)$ is understood through the conditional distribution obtained from the joint density, not by dividing $P(Y\leq t,\ X=x)$ by $P(X=x)$. At values where $f_X(x)=0$, this density-ratio formula does not apply; the conditional law at such points is not determined by the ratio.

Why the Density-Ratio Formula Works

The marginal density is the total joint density accumulated along the slice. Dividing by that total makes the density on the slice integrate to one. The following derivation also shows how probabilities in a thin vertical strip lead to the conditional density.

Compute the marginal density.
At a fixed $x$, integrate the joint density over all $y$ values allowed by the support. For a slice from $L(x)$ to $U(x)$, this gives $f_X(x)=\int_{L(x)}^{U(x)}f_{X,Y}(x,y)\,dy$.

Restrict the slice to the event.
The joint density accumulated where $Y\leq t$ is the integral over the allowed slice up to $t$, with the upper limit capped at $U(x)$.

Normalize and integrate.
Dividing the restricted slice integral by $f_X(x)$ gives the conditional CDF. In a narrow interval around $x$, both the restricted joint probability and the marginal probability are proportional to the interval's width; their ratio tends to this density ratio as the width shrinks.

When the slice is a single interval $[L(x),U(x)]$, the formula can be written as

F_{Y\mid X}(t\mid x) = \frac{\displaystyle\int_{L(x)}^{\min(t,U(x))} f_{X,Y}(x,y)\,dy} {\displaystyle\int_{L(x)}^{U(x)} f_{X,Y}(x,y)\,dy}

for $L(x)<t<U(x)$; the CDF is zero for $t\leq L(x)$ and one for $t\geq U(x)$. This compact form assumes one interval for the slice. If the slice has separate pieces, integrate over all allowed pieces below $t$, rather than filling gaps in the support.

Worked Example: A Conditional CDF on a Triangular Support

A hypothetical joint model has density $f_{X,Y}(x,y)=2$ on $0<x<1,\ 0<y<x$, and zero elsewhere. Find the conditional CDF of $Y$ given $X=x$, then calculate $P(3/10<Y<3/5\mid X=3/4)$. The model has no additional parameters.

Method: find the marginal, normalize the slice, and integrate over the requested interval. The support at a fixed $x$ is $0<y<x$, so

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

For $0<x<1$, the marginal is positive. Therefore the conditional density is

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

This density integrates to $\int_0^x(1/x)\,dy=1$, as a conditional density must. Integrating up to $t$ gives the piecewise CDF

F_{Y\mid X}(t\mid x)= \begin{cases} 0, & t\leq 0,\\[2pt] t/x, & 0<t<x,\\[2pt] 1, & t\geq x. \end{cases}

At $x=3/4$, the conditional density is $4/3$ on $0<y<3/4$. Both interval endpoints, $3/10$ and $3/5$, lie inside that slice. Thus

P(3/10<Y<3/5\mid X=3/4) =\int_{3/10}^{3/5}\frac{4}{3}\,dy =\frac{4}{3}\left(\frac{3}{5}-\frac{3}{10}\right) =\frac{4}{3}\cdot\frac{3}{10} =\frac{2}{5}=0.4000.

Check: the conditional density is $4/3$, and the interval has width $3/10$, so its probability is $4/3$ times $3/10$, or $0.4000$. Common mistake avoided: using the full support width $1$ instead of the slice width $3/4$ would incorrectly treat the conditional density as $1$.

Intervals and Nonrectangular Regions

Once the conditional CDF is known, an interval probability can be calculated by subtraction:

P(a<Y\leq b\mid X=x) =F_{Y\mid X}(b\mid x)-F_{Y\mid X}(a\mid x),\qquad a<b.

For a continuous conditional distribution, including or excluding an endpoint does not change the probability, since a single point has probability zero. If an endpoint lies outside the slice, use the CDF's value of zero or one there; do not integrate beyond the support as if the density continued.

A region involving both variables, such as $Y>X$ or $Y>2X$, becomes an interval or a union of intervals in $y$ when $x$ is fixed. For a region described by $g(x)<Y<h(x)$, its conditional probability is found by intersecting that interval with the slice allowed by the support. Equivalently, integrate the conditional density over the resulting slice. The boundary functions may change the integration limits with $x$, which is why the support and event must be considered together.

Worked Example: A Tilted Density and a Slanted Region

A hypothetical pair $(X,Y)$ has density $f_{X,Y}(x,y)=x+y$ on the square $0<x<1,\ 0<y<1$, and zero elsewhere. Find the conditional CDF of $Y$ given $X=x$. Then calculate (a) $P(1/4<Y\leq3/4\mid X=1/2)$ and (b) $P(Y>X\mid X=2/5)$.

Method: normalize the density at each fixed $x$, then integrate over the event's slice. First check that the specified joint density is normalized:

\int_0^1\int_0^1(x+y)\,dy\,dx =\int_0^1\left(x+\frac12\right)dx =\frac12+\frac12=1.

The marginal density of $X$ is

f_X(x)=\int_0^1(x+y)\,dy=x+\frac12,\qquad 0<x<1.

Hence $f_{Y\mid X}(y\mid x)=(x+y)/(x+1/2)$ for $0<y<1$. For $0<t<1$, integrate from zero to $t$:

F_{Y\mid X}(t\mid x) =\frac{\int_0^t(x+y)\,dy}{x+1/2} =\frac{xt+t^2/2}{x+1/2}.

The CDF is zero for $t\leq0$ and one for $t\geq1$. For part (a), at $x=1/2$ the denominator is $1$. The CDF at $3/4$ is $ (1/2)(3/4)+(3/4)^2/2=3/8+9/32=21/32$. At $1/4$, it is $ (1/2)(1/4)+(1/4)^2/2=1/8+1/32=5/32$. Therefore

P(1/4<Y\leq3/4\mid X=1/2) =\frac{21}{32}-\frac{5}{32} =\frac{16}{32} =\frac12=0.5000.

For part (b), fixing $x=2/5$ turns the nonrectangular region $Y>X$ into $2/5<y<1$. The marginal density at this $x$ is $2/5+1/2=9/10$. The joint-density integral over the event's slice is

\int_{2/5}^1\left(\frac25+y\right)dy =\frac25\left(1-\frac25\right)+\frac{1-(2/5)^2}{2} =\frac{6}{25}+\frac{21}{50} =\frac{33}{50}.

Normalize by the marginal density:

P(Y>X\mid X=2/5) =\frac{33/50}{9/10} =\frac{11}{15} \approx0.7333.

Check: the numerator $33/50$ is the mass accumulated on the event's slice, and division by $9/10$ makes it a conditional probability. Common mistake avoided: integrating over all $0<y<1$ would give the whole conditional distribution, not the probability of the slanted region $y>x$.

Curved Supports and Moving Slice Limits

For a curved support, the same process applies, but the support boundaries determine the slice. The next example has a lower boundary that depends quadratically on $x$. A probability region with a straight boundary must still be intersected with that curved slice before integrating.

Worked Example: A Conditional CDF Above a Curved Boundary

A hypothetical joint model has density $f_{X,Y}(x,y)=3/2$ on $0<x<1,\ x^2<y<1$, and zero elsewhere. Find the conditional CDF of $Y$ given $X=x$, and calculate $P(Y> X\mid X=1/2)$ and $P(1/2<Y<9/10\mid X=3/5)$.

Method: use the curved support to find the slice, then normalize its density. The support area is $\int_0^1(1-x^2)\,dx=1-1/3=2/3$, so the constant density $3/2$ integrates to one. The marginal density is

f_X(x)=\int_{x^2}^1\frac32\,dy =\frac32(1-x^2),\qquad 0<x<1.

The conditional density is therefore constant on the slice:

f_{Y\mid X}(y\mid x) =\frac{3/2}{(3/2)(1-x^2)} =\frac{1}{1-x^2},\qquad x^2<y<1.

Thus the conditional CDF is zero when $t\leq x^2$, is $(t-x^2)/(1-x^2)$ when $x^2<t<1$, and is one when $t\geq1$. At $x=1/2$, the slice is $1/4<y<1$. The event $Y>X$ retains $1/2<y<1$, so

P(Y>X\mid X=1/2) =\frac{1-1/2}{1-(1/2)^2} =\frac{1/2}{3/4} =\frac23 \approx0.6667.

At $x=3/5$, the support begins at $x^2=9/25$. The requested interval $1/2<y<9/10$ lies entirely above that lower boundary, so its width is $9/10-1/2=2/5$. The full conditional slice has width $1-9/25=16/25$. Consequently,

P(1/2<Y<9/10\mid X=3/5) =\frac{2/5}{16/25} =\frac{5}{8} =0.6250.

Check: both results are between zero and one, and each is the event length divided by the length of its conditional slice. Common mistake avoided: using zero as the lower support limit at $x=3/5$ would include values $y<9/25$, where the joint density is zero.

Common Exam Traps and Checks

Conditional probability calculations often fail because a correct integral is paired with the wrong slice or normalization. Use the following checks before accepting a result.

Check the marginal denominator. It is $f_X(x)$ when conditioning on $X=x$, obtained by integrating over the full allowed $y$-slice. It is not the joint density at a single point.
Check the limits against both support and event. For a region such as $Y>g(X)$, at a fixed $x$ the lower limit is determined by both $g(x)$ and the support boundary.
Check that the conditional density integrates to one. If it does not, a marginal, support limit, or algebraic simplification is likely wrong.
Check the CDF's endpoints. It must be nondecreasing, start at zero below the slice, and reach one at or above its upper boundary.
Do not use an ordinary ratio of point-event probabilities. Since $P(X=x)=0$ for continuous $X$, use the conditional density derived from the joint and marginal densities.

Key takeaway: A conditional CDF is an integral over the part of the joint-support slice below the threshold, divided by the full marginal mass of that slice. For a region involving both variables, fix $x$, intersect the region with the support, and integrate only over the resulting $y$-values.

Check Your Understanding

For each problem, identify the conditional slice before integrating. Probabilities are rounded to four decimals where needed.

A joint density is $2$ on $0<x<1,\ 0<y<x$. Find $P(3/20<Y<9/20\mid X=3/5)$. (Answer: $0.5000$)
A joint density is $x+y$ on the unit square. Find $P(Y\leq1/2\mid X=1/4)$. (Answer: $1/3$)
For the same unit-square density $x+y$, find $P(Y>3X\mid X=1/5)$. (Answer: $4/7$)
A joint density is $3/2$ on $0<x<1,\ x^2<y<1$. Find $P(1/2<Y<9/10\mid X=3/5)$. (Answer: $5/8$)
A joint density is $2$ on $0<x<1,\ 0<y<1-x$. Find $P(Y>2X\mid X=1/5)$. (Answer: $1/2$)

What You'll Learn

From a Joint Density to a Conditional CDF

Why the Density-Ratio Formula Works

Worked Example: A Conditional CDF on a Triangular Support

Worked Example: A Conditional CDF on a Triangular Support

Intervals and Nonrectangular Regions

Worked Example: A Tilted Density and a Slanted Region

Curved Supports and Moving Slice Limits

Worked Example: A Conditional CDF Above a Curved Boundary

Common Exam Traps and Checks

Check Your Understanding

Previous

Next