Discrete and Mixed Conditional-Law Calculations

Why Conditional-Law Calculations Need Care

A conditional law is often introduced through a ratio: the probability of two events occurring divided by the probability of the conditioning event. That ratio works when the conditioning event has positive probability. It does not directly define a conditional law given $Y=y$ when $P(Y=y)=0$, as happens at every individual value of a continuously distributed random variable.

Regular conditional distributions provide a systematic way to handle both situations. Their values at positive-probability atoms are determined by ordinary conditional probabilities. In continuous settings, a density calculation gives a useful version of the conditional law, even though each individual conditioning value has probability zero. Such values are only determined up to changes on a set of marginal probability zero.

Conditional Laws as Kernels

Let $X$ and $Y$ be real-valued random variables. A regular conditional distribution of $X$ given $Y$ is a probability kernel $K$ from the state space of $Y$ to the state space of $X$. Thus, for each $y$, $K(y,\cdot)$ is a probability measure on the Borel sets of $\mathbb R$; for each Borel set $A$, $y\mapsto K(y,A)$ is measurable. If $\mu_Y$ is the law of $Y$, the defining identity is

P(X\in A,\;Y\in B)=\int_B K(y,A)\,\mu_Y(dy)

for all Borel sets $A,B\subseteq\mathbb R$. In particular, the identity is an integral statement about the kernel, not a requirement that each value $K(y,\cdot)$ be determined uniquely.

Definition: A version of a regular conditional distribution may be changed on any $\mu_Y$-null set of conditioning values without changing the defining identity. Consequently, its values at an individual $y$ with $P(Y=y)=0$ are not specified by that identity alone.

The calculations below construct convenient versions. Atoms are handled by dividing the joint mass on the relevant slice by the mass of the atom. For a density, the corresponding calculation divides the joint density by the marginal density. In each case, the denominator and the set on which the formula applies matter.

Discrete Joint Laws

Suppose $X$ and $Y$ take values in countable sets, with joint probabilities $p_{ij}=P(X=x_i,Y=y_j)$. Write $p_Y(j)=P(Y=y_j)=\sum_i p_{ij}$. Whenever $p_Y(j)>0$, the conditional mass function is

P(X=x_i\mid Y=y_j)=\frac{p_{ij}}{p_Y(j)}.

This is the ordinary conditional-probability ratio, applied to each possible value of $X$. If $p_Y(j)=0$, every $p_{ij}$ in that column is zero, and the ratio is undefined. A probability distribution still needs to be assigned to $K(y_j,\cdot)$ to complete a kernel, but any such choice has no effect on the defining identity.

Worked Example: A Discrete Joint Probability Table

Let $X$ take values $0,1,2$, and let $Y$ take values $a,b,c$. Suppose the joint masses, with rows indexed by $X$, are

\begin{array}{c|ccc} & a & b & c\\ \hline X=0 & 1/10 & 1/10 & 0\\ X=1 & 2/10 & 2/10 & 0\\ X=2 & 1/10 & 3/10 & 0 \end{array}

The entries sum to $1$. The marginal probabilities are $P(Y=a)=4/10$, $P(Y=b)=6/10$, and $P(Y=c)=0$. At $a$, dividing the first column by $4/10$ gives

K(a,\{0,1,2\})=(1/4,\;1/2,\;1/4).

Indeed, the entries sum to $1$, and, for example, $P(X=1,Y=a)=(1/2)(4/10)=2/10$. At $b$, the same calculation gives $K(b,\{0,1,2\})=(1/6,\;1/3,\;1/2)$; these entries sum to $1$, and multiplying them by $6/10$ recovers $1/10,2/10,3/10$. At $c$, the ratio is undefined. Choosing $K(c,\cdot)=\delta_0$, for example, completes the kernel, but choosing another probability distribution there would give the same joint law.

Proposition: The discrete ratio formula, together with any probability distribution assigned at zero-mass values of $Y$, defines a regular conditional distribution of $X$ given $Y$.

Proof. For any sets $A$ of values of $X$ and $B$ of values of $Y$, sum over the points of $B$ with positive marginal mass. At each such point, the definition gives $K(y_j,A)p_Y(j)=\sum_{x_i\in A}p_{ij}$. If $p_Y(j)=0$, then $p_{ij}=0$ for every $i$, so that point contributes zero to the joint probability, regardless of the assigned $K(y_j,\cdot)$. Summing these equalities over $j\in B$ gives the kernel identity. The state spaces are countable, so measurability in $y$ holds. $\square$

Joint Densities and Zero-Probability Conditioning Values

Now suppose the joint law of $(X,Y)$ has a density $f_{X,Y}(x,y)$ with respect to two-dimensional Lebesgue measure. Its marginal density for $Y$ is

f_Y(y)=\int_{\mathbb R}f_{X,Y}(x,y)\,dx.

At values where $0<f_Y(y)<\infty$, 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)}.

This formula is useful even though $P(Y=y)=0$ at each individual $y$. The relevant conditional distribution is a value of a measurable kernel satisfying an integral identity. The formula gives one version of that kernel on the set where the marginal density is positive and finite. On values where $f_Y(y)=0$, the ratio cannot be used; a probability measure must be chosen separately.

Theorem: Let $f_{X,Y}$ be a nonnegative joint density integrating to $1$, and define $f_Y$ by integrating it over $x$. On the set $G=\{y:0<f_Y(y)<\infty\}$, the ratio above specifies a conditional density for almost every $y$ with respect to the law of $Y$. Assign any fixed probability measure to $K(y,\cdot)$ outside $G$. The resulting kernel is a regular conditional distribution.

Proof. For almost every $y\in G$, the integral of the proposed conditional density over $x$ equals $f_Y(y)/f_Y(y)=1$. Integrating a nonnegative measurable function over $x$ shows that $y\mapsto K(y,A)$ is measurable for every Borel set $A$; the fixed assignment off $G$ preserves measurability. For Borel sets $A,B$, Tonelli's theorem gives $$ \int_B K(y,A)f_Y(y)\,dy =\int_{B\cap G}\int_A f_{X,Y}(x,y)\,dx\,dy =P(X\in A,\;Y\in B). $$ The equality holds because values outside $G$ have zero $Y$-marginal measure: where $f_Y=0$, the density contributes no joint mass, and the set where $f_Y=\infty$ is Lebesgue-null and also has zero marginal measure. Thus the chosen values off $G$ do not affect the integral identity. $\square$

Worked Example: A Density on a Triangle

Suppose the joint density is $f_{X,Y}(x,y)=2$ when $0<x<y<1$, and is $0$ otherwise. Its total integral is $\int_0^1\int_0^y2\,dx\,dy=\int_0^1 2y\,dy=1$. The marginal density of $Y$ is $f_Y(y)=2y$ for $0<y<1$, and zero outside that interval. For $0<y<1$,

f_{X\mid Y}(x\mid y)= \begin{cases} 1/y,&0<x<y,\\ 0,&\text{otherwise}. \end{cases}

The conditional density integrates to $\int_0^y(1/y)\,dx=1$, and its product with $f_Y(y)$ recovers the joint density: $(1/y)(2y)=2$ on the triangle. Although $P(Y=y)=0$ for every fixed $y$, the formula defines a useful version for every $y$ in $(0,1)$. At $y=0$, the denominator vanishes and the formula does not apply. Assigning, for instance, $\delta_0$ there is valid; the value at that single point does not change the joint law.

Mixed Laws: Treat Atoms and Density Parts Separately

A mixed marginal law may have both atoms and a density. The two components require separate calculations. Suppose, for simplicity, that the law of $Y$ consists of atoms at values $y_j$, together with a component absolutely continuous with respect to Lebesgue measure. Write $\nu_j(A)=P(X\in A,Y=y_j)$ for the joint slice at an atom, and suppose the continuous-in-$y$ part of the joint law can be written as $\nu_y(dx)\,dy$. Here each $\nu_y$ is a finite measure on the $X$-space, and $y\mapsto\nu_y(A)$ is measurable for each Borel $A$.

The atomic marginal mass is $m_j=\nu_j(\mathbb R)$; the density of the continuous marginal part is $q(y)=\nu_y(\mathbb R)$. At an atom with $m_j>0$, the conditional law is $\nu_j/m_j$. At a continuous-component value with $0<q(y)<\infty$, it is $\nu_y/q(y)$. These formulas are not interchangeable: an atom uses the mass of its joint slice, whereas the continuous part uses the density of its joint slice.

Result: The slice formulas define a conditional-law version on positive-mass atoms and, almost everywhere under the continuous marginal component, wherever $q(y)>0$. At any remaining values, choices can be made arbitrarily, subject to measurability.

To check the claim, integrate over a Borel set $B$ of conditioning values. At each atom $y_j\in B$, multiplying $\nu_j(A)/m_j$ by $m_j$ recovers $\nu_j(A)$. For the continuous part, multiplying $\nu_y(A)/q(y)$ by its marginal density $q(y)$ recovers $\nu_y(A)$, and integration over $B$ gives the continuous joint mass. Adding the atomic and continuous contributions gives $P(X\in A,Y\in B)$. Values where the marginal mass or density is zero contribute nothing.

Worked Example: An Atom and a Continuous Component

Define a joint law by placing mass $\tfrac14\delta_{-1}(dx)+\tfrac14U(dx)$ on the slice $Y=0$, and by using the continuous component

\nu_y(dx)\,dy = \frac12\left(\frac13\delta_2(dx)+\frac23U(dx)\right)dy, \qquad 0<y<1,

where $U$ is the uniform probability measure on $[0,1]$, and there is no continuous component outside $(0,1)$. The mass on the atom $Y=0$ is $\tfrac14+\tfrac14=\tfrac12$. The continuous component has total mass $\int_0^1\tfrac12\,dy=\tfrac12$, so the joint law has total mass $1$. At the atom,

K(0,\cdot)=\frac12\delta_{-1}+\frac12U.

For $0<y<1$, the marginal density is $q(y)=\tfrac12$, and division of $\nu_y$ by $\tfrac12$ gives $K(y,\cdot)=\tfrac13\delta_2+\tfrac23U$. For example, on the set $A=\{2\}$, the continuous joint slice assigns mass density $\nu_y(A)=\tfrac16$; the conditional probability is $\tfrac16/(\tfrac12)=\tfrac13$, as stated. At $Y=0$, in contrast, the conditional probability of $X=2$ is zero, since the atomic slice contains neither a mass at $2$ nor any uniform mass there. The atom and the continuous component therefore give different conditional laws at $y=0$, and the atom determines the kernel value at that positive-probability conditioning point. At any fixed $y\in(0,1)$, the displayed continuous formula is a natural version, but its value could be changed at that single point without changing the joint law.

Interpreting Values on Null Sets

There are two distinct reasons a conditional-law formula may fail at a conditioning value. At an atom of zero mass, the discrete ratio has denominator zero. In a density calculation, the marginal density may vanish, so the density ratio is unavailable. In either case, the defining kernel identity does not determine that value. More generally, even where a convenient formula is available, the kernel can be altered on a marginal-null set and remain a version of the same conditional law.

This does not make continuous conditional distributions useless or arbitrary in practice. A density ratio or a specified model often supplies a natural, regular version that can be evaluated at particular values. The important distinction is that this useful value comes from the chosen version or model, not from conditioning on a positive-probability event. In a mixed law, keep the distinction especially clear: use slice probabilities at atoms and density ratios on the continuous part.

Common pitfall: Do not apply the discrete ratio $P(X\in A,Y=y)/P(Y=y)$ at a continuous conditioning value, where both numerator and denominator are zero. Use a regular conditional distribution, typically obtained from a joint-density ratio or a continuous slice formula.

Check Your Understanding

Use the calculations and distinctions above to answer the following questions.

Why does assigning any probability measure at a zero-mass value of a discrete $Y$ leave the joint law unchanged?
In the triangular-density example, what is the marginal density of $Y$, and why does the conditional density integrate to one?
In a mixed law, why must the atomic slice at $Y=y_j$ be normalized by its mass rather than by the continuous marginal density?
What does it mean for two regular conditional distributions to differ only on a marginal-null set?
In the last example, why is the conditional law at $Y=0$ not the same as the continuous-component formula used for $0<y<1$?