Recovering Conditionals From a Joint Density

From a Joint Density to Conditional Models

In fully conditional specification, an update model describes one variable given the current values of the others. When a joint density is available, the conditional distributions it implies are determined: integrate out the variable being conditioned on to obtain a marginal density, then divide the joint density by that marginal. This gives a direct way to check whether proposed FCS models agree with a specified joint distribution.

The calculation depends on the support as well as the algebraic expression for the density. The density-ratio formula determines the conditional density where the conditioning variable has positive marginal density. At values where the marginal is zero, a conditional distribution can be specified as a version, but it is not determined by the joint distribution. The density being marginalized must also be a properly normalized joint density; an unnormalized function can sometimes give the right conditional shape after division, but its integral is not a marginal probability density.

Definition: Let $X$ and $Y$ have joint density $f_{X,Y}(x,y)$ on support $S$. The marginal density of $Y$ is $$ f_Y(y)=\int_{\{x:(x,y)\in S\}} f_{X,Y}(x,y)\,dx. $$ For values $y$ such that $f_Y(y)>0$, the conditional density of $X$ given $Y=y$ is $$ f_{X\mid Y}(x\mid y)=\frac{f_{X,Y}(x,y)}{f_Y(y)} $$ on the slice $\{x:(x,y)\in S\}$, and is zero outside that slice. The reverse conditional is obtained by integrating over $y$: $f_X(x)=\int f_{X,Y}(x,y)\,dy$, followed by $f_{Y\mid X}(y\mid x)=f_{X,Y}(x,y)/f_X(x)$ whenever $f_X(x)>0$.

These formulas follow from the conditional-density definition. For fixed $y$ with $f_Y(y)>0$, division by $f_Y(y)$ rescales the joint density along the $y$-slice so that it integrates to one:

\int_{\{x:(x,y)\in S\}} f_{X\mid Y}(x\mid y)\,dx = \frac{\int_{\{x:(x,y)\in S\}} f_{X,Y}(x,y)\,dx}{f_Y(y)} = \frac{f_Y(y)}{f_Y(y)} =1.

This also yields the factorization $f_{X,Y}(x,y)=f_Y(y)f_{X\mid Y}(x\mid y)$ wherever the densities are defined. Thus, if a candidate joint density is specified, its conditionals are not choices: they must equal these ratios, apart from sets of probability zero. Compatibility of Conditional Models concerns whether proposed conditionals can arise from a common joint distribution; here the joint is given, so the check is a direct calculation.

Worked Examples

Worked Example: A Triangular Support

Suppose the hypothetical joint density is $f_{X,Y}(x,y)=2$ when $0<x<y<1$, and zero otherwise. Derive both conditionals. First verify that this is a density: the support is triangular, and

\int_0^1\int_0^y 2\,dx\,dy = \int_0^1 2y\,dy =1.

For fixed $y$, the allowed values of $x$ are $0<x<y$. Therefore

f_Y(y)=\int_0^y 2\,dx=2y,\qquad 0<y<1, \qquad f_{X\mid Y}(x\mid y)=\frac{2}{2y}=\frac{1}{y},\quad 0<x<y.

The conditional integrates to $\int_0^y (1/y)\,dx=1$. In the other direction, for fixed $x$, the allowed values are $x<y<1$, so

f_X(x)=\int_x^1 2\,dy=2(1-x),\qquad f_{Y\mid X}(y\mid x)=\frac{2}{2(1-x)}=\frac{1}{1-x},\quad x<y<1.

This conditional also integrates to one: $\int_x^1 [1/(1-x)]\,dy=1$. The changing support is essential: $X\mid Y=y$ ranges only from 0 to $y$, whereas $Y\mid X=x$ ranges from $x$ to 1.

Worked Example: Independent Exponential Variables

Consider the hypothetical density $f_{X,Y}(x,y)=6e^{-2x-3y}$ for $x>0,\ y>0$. It is normalized because

\int_0^\infty\int_0^\infty 6e^{-2x-3y}\,dy\,dx = 6\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=1.

Integrating over $x$ gives $f_Y(y)=3e^{-3y}$ for $y>0$, and integrating over $y$ gives $f_X(x)=2e^{-2x}$ for $x>0$. Hence

f_{X\mid Y}(x\mid y) = \frac{6e^{-2x-3y}}{3e^{-3y}} = 2e^{-2x},\quad x>0, \qquad f_{Y\mid X}(y\mid x) = \frac{6e^{-2x-3y}}{2e^{-2x}} = 3e^{-3y},\quad y>0.

Each conditional is unchanged by the value of the other variable. These results agree with the product factorization $f_{X,Y}(x,y)=(2e^{-2x})(3e^{-3y})$. This agreement is a consequence of the density’s factorization, not an assumption to make before calculating the marginals.

Worked Example: Checking Proposed Conditionals Against a Joint Density

Suppose a hypothetical candidate joint density is $f_{X,Y}(x,y)=x+y$ on the unit square $0<x<1,\ 0<y<1$, and zero elsewhere. Its total integral is

\int_0^1\int_0^1(x+y)\,dx\,dy = \int_0^1\left(\frac{1}{2}+y\right)dy =1,

so it is normalized. For fixed $y$, integrating over $x$ gives $f_Y(y)=1/2+y$. For fixed $x$, integrating over $y$ gives $f_X(x)=x+1/2$. Thus the implied conditionals are

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

For instance, at $x=1/2$, the implied density of $Y\mid X=1/2$ is $y+1/2$, not the uniform density 1. Its conditional mean is $\int_0^1 y(y+1/2)\,dy=1/3+1/4=7/12$, about $0.5833$, whereas a uniform conditional would have mean $1/2$. A proposed FCS pair that uses the derived $X\mid Y$ model but assumes $Y\mid X$ is uniform does not agree with this candidate joint density. Matching the support and normalizing each conditional are not sufficient; the full conditional density must match.

A Practical Check and Its Limits

When a joint density is available, use the same sequence for each variable: identify the joint support, integrate over the variable being removed, divide by the resulting marginal, and check that the conditional integrates to one on its own support. Then compare the complete formula and support with the proposed FCS update model. Equality of conditional means, or agreement at a few selected values, does not establish equality of distributions.

Check the joint density.
Confirm that it is nonnegative on its stated support and integrates to one.

Marginalize over the correct slice.
For a conditional given $Y=y$, integrate over all $x$ allowed at that $y$; do not automatically use fixed limits.

Form and compare the ratio.
Divide by the marginal density and check the proposed conditional’s formula, support, and normalization.

An equivalent route, when a factorization is apparent, is to write the joint density as a marginal times a conditional and identify the factors. Direct integration is usually more reliable when the support has boundaries that depend on the conditioning value. In either route, a conditional density must integrate to one for each relevant conditioning value.

This check has a defined limit: it establishes agreement with the particular joint density supplied, not whether that joint density is appropriate for a real data-generating process. If no candidate joint density is given, deriving ratios alone cannot settle whether a collection of proposed conditionals has a common joint distribution. Further compatibility analysis is then needed. Also, conditionals at values where the conditioning variable has zero marginal density are not determined by the ratio; their values there do not affect the joint distribution.

Common Mistakes and Diagnostics

Using the wrong integration limits. In triangular or otherwise restricted supports, limits depend on the fixed conditioning value. Draw or describe the support and take a slice before integrating.
Calling an unnormalized integral a marginal density. For example, the constant function 3 on $0<x<y<1$ integrates over the triangle to $3/2$, not 1. Its $y$-integral is $3y$, which is not a marginal density. The normalized joint density is 2 on that triangle, giving $f_Y(y)=2y$ and $f_{X\mid Y}(x\mid y)=1/y$. The ratio based on the unnormalized function happens to give the same conditional shape, but the intermediate quantity is not a probability density.
Checking only conditional normalization. Many different densities on the same support integrate to one. Compare the full conditional formula, not just its integral.
Ignoring values where the marginal is zero. Division by a zero marginal is undefined. State the conditional only for conditioning values with positive marginal density.

A calculation is likely going wrong if a supposed marginal integrates to something other than one, a conditional does not integrate to one over its slice, or a conditional is positive outside the joint support. Each is a signal to revisit normalization, integration limits, or the support before comparing models.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

A joint density is constant $c$ on $0<y<x<1$ and zero elsewhere. Find $c$, $f_X(x)$, and $f_{Y\mid X}(y\mid x)$.
For $f_{X,Y}(x,y)=4xy$ on the unit square, derive $f_{X\mid Y}(x\mid y)$.
For the same unit-square density $4xy$, derive $f_{Y\mid X}(y\mid x)$. Would a proposed uniform conditional agree?
State two checks that should be made before concluding that a proposed conditional agrees with one derived from a joint density.

Solutions. (1) Normalization requires $1=c\int_0^1\int_0^x dy\,dx=c\int_0^1x\,dx=c/2$, so $c=2$. Then $f_X(x)=\int_0^x2\,dy=2x$, and $f_{Y\mid X}(y\mid x)=2/(2x)=1/x$ for $0<y<x<1$. (2) The marginal is $f_Y(y)=\int_0^1 4xy\,dx=2y$, so $f_{X\mid Y}(x\mid y)=4xy/(2y)=2x$ for $0<x<1$ and $0<y<1$. (3) The marginal is $f_X(x)=\int_0^1 4xy\,dy=2x$, giving $f_{Y\mid X}(y\mid x)=4xy/(2x)=2y$ for $0<y<1$ and $0<x<1$. It is not uniform: its density is $2y$, which varies with $y$. (4) Check that the joint density is normalized and that the marginal uses the correct support and integrates to one; then check the conditional’s support and full formula, as well as its integral of one.

Recovering Conditionals From a Joint Density

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