Why Piecewise Joint Models Need Care
A joint probability model may use different formulas in different regions, and its support may change from one region to another. To find a conditional law, it is not enough to divide formulas mechanically: first identify the slice of the joint support that remains after conditioning, then use the appropriate marginal or event probability to normalize that slice.
This distinction matters especially at boundaries. A line that bounds a continuous region usually has probability zero, so changing whether that line is included does not change the continuous probabilities. But a model can also place positive probability directly on a boundary. Such an atom must be included using probabilities, not treated as though it were part of a density.
The continuous formula follows the same normalization principle as the discrete one, but it conditions on an exact value of a continuous variable. Since \(P(X=x)=0\) for a typical value \(x\), this is not an ordinary event-probability ratio. It is a conditional density defined through the joint and marginal densities. Values assigned at isolated \(x\)-points where the marginal density is zero, or at exceptional boundary points, do not affect the joint distribution.
A Normalization Justification
Suppose a continuous joint density is given and fix a value \(x\) with \(f_X(x)>0\). The marginal density \(f_X(x)\) is obtained by integrating the joint density over all \(y\) values allowed at that \(x\). Dividing the joint slice by this marginal must therefore produce a density with total integral one.
Here \(\mathcal{S}_x\) is the conditional support of \(Y\) at the fixed value \(X=x\). The last equality holds because integrating over that full slice gives the marginal density \(f_X(x)\). If part of the support is omitted, the result will generally fail to integrate to one. For discrete variables, the same argument uses a sum over all allowed \(y\)-values instead of an integral.
Worked Example: A Support That Changes With the Conditioning Value
Worked Example: A Support That Changes With the Conditioning Value
A pair \((X,Y)\) has joint density \(f_{X,Y}(x,y)=1/3\) for \(0<x<1,\ 0<y<1\), and \(f_{X,Y}(x,y)=4/9\) for \(1<x<2,\ 0<y<x\), with density zero elsewhere. Find the conditional density of \(Y\) given \(X=x\), then calculate \(P(Y<1/2\mid X=3/2)\). The two regions have total probabilities \(1/3\) and \(2/3\), respectively, so the stated density is normalized.
Method. Find the marginal density by integrating across the full vertical slice at each \(x\), then divide the joint density by that marginal. This avoids carrying over the wrong support from one region to the other.
For \(0<x<1\), the permitted \(y\)-values are \(0<y<1\). Thus \(f_X(x)=\int_0^1 (1/3)\,dy=1/3\), and \(f_{Y\mid X}(y\mid x)=(1/3)/(1/3)=1\) on \(0<y<1\).
For \(1<x<2\), the permitted values are instead \(0<y<x\). Therefore \(f_X(x)=\int_0^x (4/9)\,dy=4x/9\), and \(f_{Y\mid X}(y\mid x)=(4/9)/(4x/9)=1/x\) on \(0<y<x\).
At \(x=3/2\), the conditional support is \(0<y<3/2\), and the conditional density there is \(1/(3/2)=2/3\). Hence \(P(Y<1/2\mid X=3/2)=\int_0^{1/2}(2/3)\,dy=(2/3)(1/2)=1/3=0.3333\). As a check, the full conditional density integrates to \(\int_0^{3/2}(2/3)\,dy=1\). The common mistake is to use the \(0<y<1\) support from the first region when \(x=3/2\); that would leave out valid conditional values.
Read the Support as a Slice
For a piecewise model, the conditional support is not necessarily the same for every value of the conditioning variable. A reliable method is to hold that variable fixed and trace the part of the joint support that remains. Only after finding that slice should you integrate or sum to obtain the marginal used in the denominator.
Determine which region or regions of the joint model contain that value. Check whether the value lies on an ordinary boundary or at a possible atom.
Keep only the allowed values of the other variable on the relevant slice. Write the bounds explicitly before evaluating a probability or density.
Use the conditional pmf ratio for a positive-probability discrete event or the density ratio where a marginal density is positive.
Sum the conditional probabilities or integrate the conditional density over its full support. The result must be one.
Worked Example: A Conditional Support With Two Formulas
Worked Example: A Conditional Support With Two Formulas
Suppose the joint density is \(2/3\) on \(0<y<1,\ 0<x<y\), and \(2/3\) on \(1<y<2,\ 0<x<1\), with density zero elsewhere. Find the conditional density of \(X\) given \(Y=y\), then calculate \(P(X>0.3\mid Y=0.6)\) and \(P(X>0.3\mid Y=1.4)\).
Method. Slice horizontally at the given \(y\), then integrate across the permitted \(x\)-values to find \(f_Y(y)\). The two regions each contribute probability \(1/3\) and \(2/3\), respectively, so the joint density integrates to one.
For \(0<y<1\), the horizontal slice is \(0<x<y\), so \(f_Y(y)=\int_0^y (2/3)\,dx=2y/3\). The conditional density is \((2/3)/(2y/3)=1/y\) on \(0<x<y\). For \(1<y<2\), the slice is \(0<x<1\), so \(f_Y(y)=\int_0^1(2/3)\,dx=2/3\). The conditional density is \(1\) on \(0<x<1\).
When \(y=0.6\), the conditional support is \(0<x<0.6\). Therefore \(P(X>0.3\mid Y=0.6)=\int_{0.3}^{0.6}(1/0.6)\,dx=0.3/0.6=0.5000\). When \(y=1.4\), the support is \(0<x<1\), so \(P(X>0.3\mid Y=1.4)=\int_{0.3}^{1}1\,dx=0.7000\). Each conditional density integrates to one on its own support. Using the first region’s bound \(x<y\) for \(y=1.4\) would incorrectly extend the support beyond \(x=1\).
Boundary Values: Zero-Probability Edges and Atoms
A boundary in a density-defined region does not automatically carry probability. For example, a joint density over \(0<x<y<1\) assigns probability zero to the line \(x=y\); including or excluding that line leaves all integrals unchanged. However, the model may explicitly assign mass to a line or point. That mass is part of the probability distribution and must be handled separately.
At a boundary value of a continuous conditioning variable, a conditional distribution may not be uniquely determined by the joint distribution at that single value. The density-ratio formula specifies the conditional law for almost every conditioning value, not necessarily every boundary point. If the model places an atom at that value, condition using the atom’s probabilities instead.
Worked Example: Conditioning at a Boundary Atom
Worked Example: Conditioning at a Boundary Atom
Consider a mixed model constructed as follows. With probability \(1/2\), let \(X\) and \(Y\) be independent uniform random variables on \((0,1)\). With probability \(1/2\), set \(X=1\) and choose \(Y=0\) or \(Y=1\), each with probability \(1/2\). Find \(P(Y=0\mid X=1)\) and \(P(Y<0.3\mid X=0.4)\).
Method. Separate the continuous component from the point masses. At \(X=1\), use the positive-probability event ratio; at an interior value \(X=0.4\), use the density ratio.
The two boundary points have joint probabilities \(P(X=1,Y=0)=(1/2)(1/2)=1/4\) and \(P(X=1,Y=1)=(1/2)(1/2)=1/4\). Thus \(P(X=1)=1/4+1/4=1/2\), giving \(P(Y=0\mid X=1)=(1/4)/(1/2)=0.5000\).
For \(0<x<1\), the continuous joint density is \(1/2\) on \(0<y<1\), and the marginal density is \(f_X(x)=\int_0^1(1/2)\,dy=1/2\). Hence \(f_{Y\mid X}(y\mid 0.4)=(1/2)/(1/2)=1\) for \(0<y<1\), so \(P(Y<0.3\mid X=0.4)=\int_0^{0.3}1\,dy=0.3000\). The check is that the interior conditional density integrates to one, while the two conditional boundary probabilities sum to \(0.5+0.5=1\).
A tempting but incorrect calculation at \(X=1\) is to substitute the continuous density values at \(x=1\) into a density ratio, obtaining a uniform-looking conditional law. That ignores the actual atoms: all probability conditional on \(X=1\) is concentrated at \(Y=0\) and \(Y=1\). A density value at a boundary cannot replace the probability mass there.
Common Checks Before Finalizing a Conditional Law
A piecewise conditional result should make clear which formula applies and where. State the support alongside each formula; a formula without its bounds can describe the wrong distribution. Then check its total probability. For a continuous conditional law, integrate over the entire conditional support. For a discrete law, sum over all allowed values. In a mixed model, account for both the continuous part and any atoms.
- Do not normalize over the wrong slice. A support bound may depend on the conditioning value.
- Do not omit a region with positive conditional probability. If a fixed value intersects multiple pieces, include every allowed part of the slice.
- Do not treat a boundary atom as density. Use a probability ratio for a positive-probability conditioning event.
- Do not force a unique answer at a zero-probability boundary. For continuous conditioning, the conditional law at an isolated value may be unspecified unless a version or extra model convention is given.
Check Your Understanding
For each problem, state or use the correct conditional support before calculating.
- A joint density is \(1/2\) on \(0<x<1,\ 0<y<1\), and \(1/4\) on \(1<x<3,\ 0<y<1\), and zero elsewhere. Find \(P(Y>0.7\mid X=2)\). (Answer: \(0.3000\))
- A joint density is \(2\) on \(0<y<x<1\), and zero elsewhere. Find \(P(Y<0.2\mid X=0.8)\). (Answer: \(0.2500\))
- A joint density is \(2/3\) on \(0<y<1,\ 0<x<y\), and \(2/3\) on \(1<y<2,\ 0<x<1\). Find \(P(X<0.4\mid Y=1.5)\). (Answer: \(0.4000\))
- Use the mixed model from the boundary-atom example. Find \(P(Y=1\mid X=1)\). (Answer: \(0.5000\))
- A discrete pair has joint probabilities \(P(X=0,Y=0)=0.10\), \(P(X=0,Y=1)=0.20\), \(P(X=0,Y=2)=0.10\), \(P(X=1,Y=0)=0.15\), \(P(X=1,Y=1)=0.05\), and \(P(X=1,Y=2)=0.40\). Find the conditional pmf of \(X\) given \(Y=2\). (Answer: \(P(X=0\mid Y=2)=0.2000,\ P(X=1\mid Y=2)=0.8000\))