Symmetry and Equal Marginal Distributions

Why Swapping the Variables Matters

A joint distribution describes two random variables together, including how their possible values occur in pairs. Sometimes the model treats the variables in exactly the same way: exchanging their labels leaves the joint law unchanged. In that case, neither variable has a different distribution from the other. This is useful when a model has no inherent “first” and “second” variable, but the conclusion requires symmetry of the joint law, not merely a support that looks symmetric.

For example, two randomly selected measurements might be governed by a law that assigns the same probability to the pair $(x,y)$ as to $(y,x)$. If that holds for every possible pair, the distribution of the first measurement must match the distribution of the second. The equality is about distributions; it does not say that the two random variables have equal values in every outcome.

Definition: The joint law of $(X,Y)$ is invariant under swapping its variables if $(X,Y)$ and $(Y,X)$ have the same probability distribution. Equivalently, for every pair of real numbers $x,y$, their joint CDF satisfies $$ F_{X,Y}(x,y)=F_{X,Y}(y,x). $$ For a discrete pair, this is equivalent to $P(X=a,Y=b)=P(X=b,Y=a)$ for all possible values $a,b$. For a pair with a joint density, it is equivalent to $f_{X,Y}(x,y)=f_{X,Y}(y,x)$ almost everywhere, with the density understood to be zero outside its support.

The Equal-Marginals Result

The marginal distribution of a random variable describes its probabilities without specifying the value of the other variable. The result below says that swap invariance is enough to make the two marginal distributions identical.

Theorem: If the joint law of $(X,Y)$ is invariant under swapping its variables, then $X$ and $Y$ have the same marginal distribution. In CDF notation, $$ F_X(t)=F_Y(t)\qquad\text{for every real }t. $$ Equivalently, $P(X\in A)=P(Y\in A)$ for every Borel set $A$.

The CDF proof is short, but it identifies exactly where the symmetry is used. The event $\{X\leq t\}$ can be written as the event that the first coordinate of the pair is at most $t$, with no restriction on the second coordinate.

\begin{aligned} F_X(t) &=P(X\leq t)\\ &=P(X\leq t,\ Y\in\mathbb{R})\\ &=P(Y\leq t,\ X\in\mathbb{R})\\ &=P(Y\leq t)\\ &=F_Y(t). \end{aligned}

The middle equality follows from swap invariance: swapping the coordinates turns the event $X\leq t,\ Y\in\mathbb{R}$ into $Y\leq t,\ X\in\mathbb{R}$, without changing its probability. Since the CDFs agree at every $t$, the marginal distributions agree. The same argument applies to any Borel set $A$: swap invariance gives $P(X\in A)=P(Y\in A)$.

In particular, if a discrete joint probability mass function satisfies $p_{X,Y}(a,b)=p_{X,Y}(b,a)$, summing over the other coordinate gives identical marginal masses. If a joint density is symmetric, integrating over the other coordinate gives identical marginal densities. These are two ways to verify the same distributional result.

Worked Example: Marginals from a Symmetric Discrete Table

A hypothetical model records two integer-valued measurements, $X$ and $Y$, each taking values in $\{0,1,2\}$. Their joint probability mass function is given by this table. Find both marginal distributions, check swap invariance, and calculate $P(X\leq 1)$ and $E[X]$.

$X\backslash Y$	0	1	2
0	$1/10$	$1/20$	$1/20$
1	$1/20$	$1/5$	$1/20$
2	$1/20$	$1/20$	$2/5$

Step 1: Check that the entries form a probability distribution. The diagonal entries sum to $1/10+1/5+2/5=7/10$. The six off-diagonal entries sum to $6(1/20)=3/10$. Thus the total is $7/10+3/10=1$.

Step 2: Check swap invariance. The table is symmetric across its diagonal: for example, the probabilities at $(0,2)$ and $(2,0)$ are both $1/20$, and the probabilities at $(0,1)$ and $(1,0)$ are both $1/20$. The diagonal entries are unchanged when the coordinates are swapped. Therefore $p_{X,Y}(a,b)=p_{X,Y}(b,a)$ for all possible pairs.

Step 3: Sum the rows to find the marginal distribution of $X$.

\begin{aligned} P(X=0)&=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}=\frac{1}{5},\\ P(X=1)&=\frac{1}{20}+\frac{1}{5}+\frac{1}{20}=\frac{3}{10},\\ P(X=2)&=\frac{1}{20}+\frac{1}{20}+\frac{2}{5}=\frac{1}{2}. \end{aligned}

Step 4: Sum the columns to find the marginal distribution of $Y$. The table is symmetric, so the column sums are the same as the row sums: $P(Y=0)=1/5$, $P(Y=1)=3/10$, and $P(Y=2)=1/2$. Each marginal totals $1/5+3/10+1/2=1$, as required.

Step 5: Calculate the requested probability and mean. The event $X\leq 1$ includes $X=0$ and $X=1$, so $P(X\leq 1)=1/5+3/10=1/2=0.5000$. Also,

E[X]=0\left(\frac{1}{5}\right)+1\left(\frac{3}{10}\right)+2\left(\frac{1}{2}\right) =\frac{3}{10}+1=\frac{13}{10}=1.3.

The same marginal distribution gives $E[Y]=1.3$. The check is consistent with the equal-marginals theorem. The method is to verify the entries under coordinate swapping, then sum over the other coordinate; it avoids assuming equality just because the two variables have similar names.

Continuous Models: Swap and Integrate

For a continuous pair with joint density $f_{X,Y}$, the marginal density of $X$ is obtained by integrating the joint density over all possible values of $Y$. The marginal density of $Y$ is obtained by integrating over all possible values of $X$. If the density and its support are unchanged by swapping coordinates, these two integrations give the same function. The bounds must cover the actual support; writing bounds for a convenient rectangle when the support is smaller can give an incorrect marginal.

Worked Example: A Symmetric Density on a Square

Let $(X,Y)$ have joint density $$ f_{X,Y}(x,y)=4(1-x)(1-y),\qquad 0<x<1,\quad 0<y<1, $$ and density zero elsewhere. Verify that this is a density, obtain both marginals, and calculate $P(Y\leq 1/2)$. Give the probability to four decimal places.

Step 1: Verify nonnegativity and normalization. On the unit square, both $1-x$ and $1-y$ are nonnegative. The integral is

\begin{aligned} \int_0^1\int_0^1 4(1-x)(1-y)\,dy\,dx &=4\left(\int_0^1(1-x)\,dx\right) \left(\int_0^1(1-y)\,dy\right)\\ &=4\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=1. \end{aligned}

Thus the stated function is a valid joint density. It is unchanged when $x$ and $y$ are exchanged, and the square support is also unchanged by that swap.

Step 2: Integrate out $Y$ to obtain the marginal density of $X$.

f_X(x)=\int_0^1 4(1-x)(1-y)\,dy =4(1-x)\left[\;y-\frac{y^2}{2}\;\right]_0^1 =2(1-x),\qquad 0<x<1.

Step 3: Integrate out $X$ to obtain the marginal density of $Y$. The same calculation gives $f_Y(y)=2(1-y)$ for $0<y<1$. The two marginal densities have the same formula, with the variable appropriate to each marginal.

Step 4: Calculate the requested probability.

\begin{aligned} P(Y\leq 1/2) &=\int_0^{1/2}2(1-y)\,dy\\ &=\left[2y-y^2\right]_0^{1/2}\\ &=1-\frac{1}{4}=\frac{3}{4}=0.7500. \end{aligned}

As a check, $P(X\leq 1/2)$ has the same integral and is also $0.7500$, as the theorem requires. The result comes from integrating the correctly derived marginal density; the number is not inferred merely from the visual symmetry of the square.

Symmetry of the Support Is Not Enough

The support is the set of points where a joint density can be positive. A symmetric support is necessary for swap invariance of a density, but it is not sufficient: the density values at swapped points must match as well. The following example shows the difference.

Worked Example: A Symmetric Support with Unequal Marginals

Suppose a continuous pair has density $f_{X,Y}(x,y)=2x$ on the unit square $0<x<1,\ 0<y<1$, and density zero elsewhere. Find both marginal densities and compare $P(X\leq 1/2)$ with $P(Y\leq 1/2)$. State why the equal-marginals theorem does not apply.

Step 1: Check normalization.

\int_0^1\int_0^1 2x\,dy\,dx =\int_0^1 2x(1)\,dx =\left[x^2\right]_0^1=1.

The density is nonnegative on the square and integrates to one.

Step 2: Calculate the marginal density of $X$. Integrating over $y$ gives

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

Step 3: Calculate the marginal density of $Y$. Integrating over $x$ gives

f_Y(y)=\int_0^1 2x\,dx=1,\qquad 0<y<1.

Step 4: Compare the probabilities.

\begin{aligned} P(X\leq 1/2)&=\int_0^{1/2}2x\,dx =\left[x^2\right]_0^{1/2}=\frac{1}{4}=0.2500,\\ P(Y\leq 1/2)&=\int_0^{1/2}1\,dy =\frac{1}{2}=0.5000. \end{aligned}

The marginals differ. Although the unit square is symmetric, the density is not: $f_{X,Y}(x,y)=2x$ while $f_{X,Y}(y,x)=2y$, which are not equal for general $x,y$. Therefore the joint law is not swap-invariant, and the theorem’s condition is not met. The different probabilities provide a final check on the marginal calculations.

What the Result Does—and Does Not—Say

Swap invariance is a statement about the entire joint law. It guarantees identical marginal distributions, so any marginal probability or marginal moment that exists is the same for $X$ and $Y$. For example, if the relevant expectations are finite, equality of the marginal distributions implies $E[X]=E[Y]$. It does not say that $X=Y$ for each outcome, and it does not by itself imply that $X$ and $Y$ are independent. A symmetric joint law can still describe dependent variables.

A useful diagnostic is to check the joint law directly. For a discrete model, compare each off-diagonal probability with its transposed entry. For a continuous model, check both the support and the density formula after exchanging its arguments. A symmetric-looking picture or equal-looking ranges alone are not a proof.

Common exam trap: Concluding “the marginals are equal because the support is symmetric” skips a required condition. In the square example above, the support is symmetric but the density $2x$ becomes $2y$ when its arguments are swapped; the correct marginal probabilities are $0.2500$ and $0.5000$, not equal. Check invariance of the joint probabilities or density, not just the set of possible pairs.

The central method is therefore: state the joint model and its support, verify that swapping the coordinates leaves the joint law unchanged, and then use the equal-marginals result or compute the marginals directly. This is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, verify the relevant joint-law condition before using the equal-marginals result.

A discrete pair takes values in $\{0,1\}^2$, with $P(0,0)=1/8$, $P(0,1)=3/8$, $P(1,0)=3/8$, and $P(1,1)=1/8$. Find $P(X=1)$ and $P(Y=1)$. (Answer: $P(X=1)=P(Y=1)=1/2$.)
A pair has density $f_{X,Y}(x,y)=4xy$ for $0<x<1,\ 0<y<1$, and zero elsewhere. Find the two marginal densities and $P(X\leq 1/2)$. (Answer: $f_X(x)=2x$ and $f_Y(y)=2y$ on $(0,1)$; $P(X\leq 1/2)=1/4=0.2500$.)
A pair has constant density $2$ on the triangle $x>0,\ y>0,\ x+y<1$, and zero elsewhere. Find the marginal density of $X$, and calculate $P(Y\leq 1/2)$. (Answer: $f_X(x)=2(1-x)$ for $0<x<1$; $P(Y\leq 1/2)=3/4=0.7500$.)
A pair has density $f_{X,Y}(x,y)=\frac{2}{3}(1+x)$ on the unit square and zero elsewhere. Find both marginal densities and compare the probabilities that each variable is at most $1/2$. (Answer: $f_X(x)=\frac{2}{3}(1+x)$, $f_Y(y)=1$ on $(0,1)$; $P(X\leq 1/2)=5/12=0.4167$ and $P(Y\leq 1/2)=1/2=0.5000$, so the joint law is not swap-invariant.)
A pair has density $f_{X,Y}(x,y)=\frac{1}{2}(1+x+y)$ on the unit square and zero elsewhere. Find the common marginal density and calculate $P(X\leq 1/2)$. (Answer: $f_X(t)=f_Y(t)=\frac{3}{4}+\frac{t}{2}$ for $0<t<1$; $P(X\leq 1/2)=7/16=0.4375$.)

\(X\backslash Y\)	0	1	2
0	\(1/10\)	\(1/20\)	\(1/20\)
1	\(1/20\)	\(1/5\)	\(1/20\)
2	\(1/20\)	\(1/20\)	\(2/5\)

What You'll Learn