Marginals of Mixed Discrete-Continuous Laws

Why Marginalizing a Mixed Law Requires Two Operations

For a purely discrete pair, a marginal probability is obtained by summing over the other coordinate; for a pair with a joint density, it is obtained by integrating. A joint law containing both kinds of probability requires both operations. The distinction matters: a point mass has a probability at a specific pair of values, whereas a density assigns probability through area integrals.

This tutorial treats joint laws whose entire probability consists of countably many point masses and an ordinary two-dimensional density. Marginalizing such a law means projecting each point mass onto the coordinate of interest and integrating the density over the other coordinate. The two resulting parts must remain distinct, even if a projected atom lies where the continuous marginal density is positive.

Joint Representation and Marginal Formula

Definition: Suppose a pair $(X,Y)$ has a law represented by a finite or countably infinite collection of distinct points $(x_r,y_r)$, with weights $p_r>0$, and a nonnegative measurable joint density $g(x,y)$ with respect to area. The representation means that, for every Borel set $A\subseteq\mathbb{R}^2$, $$ P((X,Y)\in A)=\sum_r p_r\,\mathbf{1}_{\{(x_r,y_r)\in A\}}+\int\!\!\int_A g(x,y)\,dx\,dy, $$ where an empty sum is allowed. The weights and density must satisfy $$ \sum_r p_r+\int_{\mathbb{R}^2}g(x,y)\,dx\,dy=1. $$

The points $(x_r,y_r)$ must be distinct, but their $x$-coordinates need not be distinct, and neither need their $y$-coordinates. If several joint points share an $x$-coordinate, all their weights contribute to the marginal atom there. The strict condition $p_r>0$ applies to every term in the finite or countable list; zero-weight terms are simply omitted.

Theorem: Define $$ h_X(x)=\int_{\mathbb{R}}g(x,y)\,dy,\qquad h_Y(y)=\int_{\mathbb{R}}g(x,y)\,dx. $$ These integrals define nonnegative measurable functions, possibly infinite on null sets, and each is finite almost everywhere. For every Borel set $B\subseteq\mathbb{R}$, $$ P(X\in B)=\sum_{r:x_r\in B}p_r+\int_B h_X(x)\,dx,\qquad P(Y\in B)=\sum_{r:y_r\in B}p_r+\int_B h_Y(y)\,dy. $$ In particular, $h_X$ and $h_Y$ are densities of the respective continuous components, not necessarily densities of the full marginals.

Proof. Apply the joint representation to $B\times\mathbb{R}$. The point $(x_r,y_r)$ belongs to this set exactly when $x_r\in B$, giving the stated sum for $X$. For the density part, Tonelli's theorem applies because $g(x,y)\mathbf{1}_{\{x\in B\}}$ is nonnegative and measurable. Therefore,

\int\!\!\int_{B\times\mathbb{R}}g(x,y)\,dx\,dy =\int_B\left(\int_{\mathbb{R}}g(x,y)\,dy\right)dx =\int_B h_X(x)\,dx.

Taking $B=\mathbb{R}$ shows $\int h_X(x)\,dx=\int\!\!\int g(x,y)\,dx\,dy\leq 1$. Thus $h_X$ is integrable and finite almost everywhere. Repeating the argument on $\mathbb{R}\times B$, integrating first with respect to $x$, gives the formula for $Y$ and the same integrability conclusion for $h_Y$. $\square$

Atoms, Densities, and Marginal CDFs

Theorem: For any $a\in\mathbb{R}$, the marginal point probabilities are $$ P(X=a)=\sum_{r:x_r=a}p_r,\qquad P(Y=a)=\sum_{r:y_r=a}p_r. $$ The area-density component contributes zero probability to every single coordinate value.

Proof. Set $B=\{a\}$ in the marginal formula. A singleton has Lebesgue measure zero, so $\int_{\{a\}}h_X(x)\,dx=0$ and $\int_{\{a\}}h_Y(y)\,dy=0$. The remaining terms are precisely the weights of points with the specified coordinate. $\square$

This theorem explains why atom weights that share a coordinate must be added, but density values at that coordinate must not be added to them. A density value is not a probability. The probability from the continuous component is obtained by integrating over a set with positive length, not by evaluating the density at one point.

Writing $F_X(t)=P(X\leq t)$ and $F_Y(t)=P(Y\leq t)$, the same formulas give

F_X(t)=\sum_{r:x_r\leq t}p_r+\int_{-\infty}^{t}h_X(x)\,dx,\qquad F_Y(t)=\sum_{r:y_r\leq t}p_r+\int_{-\infty}^{t}h_Y(y)\,dy.

The sums create jumps at projected atom locations. The integrals provide the continuous contributions. A coordinate can have both a positive atom and a positive density nearby; the density part still gives zero probability to the exact coordinate.

Worked Example: Marginals With Atoms at a Density Boundary

Suppose the joint law has atoms of weight $0.1$ at $(-1,0)$ and $0.2$ at $(0,2)$, together with density $$ g(x,y)=\frac{7x}{5}\,\mathbf{1}_{[0,1]\times[0,1]}(x,y). $$ First check its total mass. The atoms have mass $0.1+0.2=0.3$, and

\int_0^1\int_0^1\frac{7x}{5}\,dy\,dx =\int_0^1\frac{7x}{5}\,dx =\frac{7}{10}.

The total is $0.3+0.7=1$, as required. For $X$, integrating the density over $y$ gives $h_X(x)=7x/5$ on $[0,1]$. The projected atoms are at $-1$ and $0$, with weights $0.1$ and $0.2$. Hence

F_X(t)= \begin{cases} 0,&t<-1,\\ 0.1,&-1\leq t<0,\\ 0.3+\dfrac{7t^2}{10},&0\leq t<1,\\ 1,&t\geq 1. \end{cases}

At $t=0$, the CDF includes both atoms, so $F_X(0)=0.3$. Although the density component is supported on an interval beginning at $0$, it contributes no atom at $0$.

For $Y$, integrating over $x$ gives $h_Y(y)=\int_0^1(7x/5)\,dx=7/10$ on $[0,1]$. Its atoms are at $0$ and $2$, again with weights $0.1$ and $0.2$. Thus $F_Y(t)=0$ for $t<0$, $F_Y(t)=0.1+(7/10)t$ for $0\leq t<1$, $F_Y(t)=0.8$ for $1\leq t<2$, and $F_Y(t)=1$ for $t\geq2$. The continuous component has mass $0.7$, and the two jumps account for the remaining $0.3$.

Worked Example: Combining Weights at a Shared Coordinate

Consider atoms at $(-2,0)$, $(-2,3)$, and $(1,-1)$, with respective weights $0.12$, $0.08$, and $0.15$. Add the density $$ g(x,y)=\frac{13x}{40}\,\mathbf{1}_{[0,2]\times[0,1]}(x,y). $$ The atomic mass is $0.12+0.08+0.15=0.35$. The density mass is

\int_0^2\int_0^1\frac{13x}{40}\,dy\,dx =\int_0^2\frac{13x}{40}\,dx =\frac{13}{40}\cdot 2 =\frac{13}{20}=0.65.

Thus the total is $1$. For $X$, the first two atoms project to the same value, $-2$, so their weights combine to $0.20$. The third atom gives weight $0.15$ at $1$. The marginal density component is $h_X(x)=13x/40$ on $[0,2]$, since the $y$-interval has length $1$. For example, the interval $0<X<1$ excludes both atoms and has probability $$ P(0<X<1)=\int_0^1\frac{13x}{40}\,dx=\frac{13}{80}. $$

For $Y$, the atoms occur at $0$, $3$, and $-1$, with respective weights $0.12$, $0.08$, and $0.15$. Its density component is $h_Y(y)=\int_0^2(13x/40)\,dx=13/20$ on $[0,1]$. The atoms at $Y=0$ and $Y=3$ remain distinct; projection combines weights only when the projected coordinate itself is shared.

Worked Example: A Marginal Probability and a Joint Rectangle

Let the atoms have weights $0.25$ at $(-1,2)$ and $0.15$ at $(1,0)$. Let the density be $$ g(x,y)=\frac{3x}{5}\,\mathbf{1}_{[0,1]\times[-1,1]}(x,y). $$ The density integrates to $0.6$, because its $y$-interval has length $2$ and $2\int_0^1(3x/5)\,dx=0.6$. Together with atomic mass $0.4$, this is a probability law.

For $X$, integrating over $y$ yields $h_X(x)=6x/5$ on $[0,1]$. The atom at $X=-1$ has weight $0.25$, and the atom at $X=1$ has weight $0.15$. Therefore

P(X\leq 0.5)=0.25+\int_0^{0.5}\frac{6x}{5}\,dx =0.25+\frac{3}{5}(0.5)^2 =0.25+0.15=0.40.

The joint probability $P(X\leq0.5,\ Y\leq0)$ includes neither atom: $(-1,2)$ fails the $Y$ condition, and $(1,0)$ fails the $X$ condition. The relevant density region has $0\leq x\leq0.5$ and $-1\leq y\leq0$, a $y$-interval of length $1$. Thus $$ P(X\leq0.5,\ Y\leq0)=\int_0^{0.5}\int_{-1}^{0}\frac{3x}{5}\,dy\,dx =\int_0^{0.5}\frac{3x}{5}\,dx =\frac{3}{40}=0.075. $$

The marginal calculation sums all mass with $X\leq0.5$, including the atom at $(-1,2)$. The rectangle calculation must also enforce the restriction on $Y$, which excludes that atom.

A Reliable Calculation Strategy

Separate the components.
List the joint atom locations and weights, and write the density on its support. Check that their total mass is one.

Project the atoms.
For the $X$ marginal, keep each atom's $x$-coordinate; for the $Y$ marginal, keep its $y$-coordinate. Add weights when projected coordinates coincide.

Integrate the density.
Integrate $g(x,y)$ over the other coordinate to obtain $h_X(x)$ or $h_Y(y)$, respecting the support.

Report and check the marginal.
State the atomic probabilities separately from the density component. Confirm their masses sum to one and use them separately when forming a CDF or event probability.

A frequent mistake is to treat the joint density as if it were a table of probabilities, or to integrate it and then omit the atoms. The correction is to use the representation component by component: sums for the point masses, integrals for the density. Another error is to count every atom separately after projection, even when several have the same coordinate; marginal atom probabilities are grouped by coordinate.

Key takeaway: A mixed marginal is generally not described by one density alone. It consists of projected atom weights together with the density obtained by integrating the joint density over the other variable.

Check Your Understanding

For each question, keep the atomic and density contributions separate.

In the marginal formula for $X$, why does integrating $g(x,y)$ over $y$ produce a density component rather than a point probability?
If two distinct joint atoms have the same $x$-coordinate, what is the marginal probability at that coordinate?
A joint density is positive at $x=a$. Does its continuous marginal component assign positive probability to $X=a$? Explain.
For a mixed joint law, what two checks verify that a proposed marginal is normalized?
When calculating $P(X\in B,\ Y\in C)$, why might summing all atom weights with $x_r\in B$ give the wrong answer?

Marginals of Mixed Discrete-Continuous Laws

What You'll Learn