# If $X$ and $Y$ have the same marginal distribution, then do they have to have the same conditional distribution?

Suppose $$X$$ and $$Y$$ are two random variables that have the same distribution. Does

$$P[X \leq t \mid Y=a]$$

be necessarily equal to $$\;\ P[Y \leq t \mid X=a]?$$ Note that if $$X$$ and $$Y$$ are bivariate normal with correlation $$\rho$$ and each is marginally $$N(\mu,\sigma^2)$$, then it is necessarily true because both conditional distributions would be

$$N(\mu+\rho(a-\mu),(1-\rho^2)\sigma^2).$$

A simple counterexample:

$$P(X = 1, Y = 2) = \frac{1}{3}\\ P(X = 2, Y = 3) = \frac{1}{3}\\ P(X = 3, Y = 1) = \frac{1}{3}$$

Then $$P(X \le 1 | Y = 2) = 1$$, but $$P(Y \le 1 | X = 2) = 0$$.

How about $$X=-Y = \begin{cases} 0 \\ 1 \end{cases}\quad$$ each with probability $$1/2.$$

• In this case $X$ is either 0 or 1 and $Y$ is either 0 or -1, so $X$ and $Y$ don't have the same distribution. Dec 23, 2020 at 23:05

There are already answers with simple examples, so why one more? Because it is interesting to look at the general pattern. The wanted property is a kind of symmetry, so we should look for some asymmetrical joint distribution for $$X,Y)$$. Si if $$(X,Y)$$ has a permutable distribution in the sense that $$(X,Y)$$ and $$(Y,X)$$ have the same distribution, so that for the joint cumulative we have $$F(x,y)=F(y,x)$$ for all $$(x,y)$$, then the sought-after property will hold.

Let us use copulas. Let $$F$$ be the joint cdf (cumulative distribution function) and $$\DeclareMathOperator{\P}{\mathbb{P}} C(u,v)= \P(F(x) \le u, F(Y)\le v)$$ By the Fréchet–Hoeffding copula bounds (see linked wiki article above) we have $$W(u,v) \le C(u,v) \le M(u,v)$$ where $$W(u,v)= \max(u+v-1,0)$$ and $$M(u,v)=\min(u,v)$$. Both $$W,M$$ are copulas. $$W$$ describes the anti-monotonic case $$X=U, Y=1-U$$ for $$U$$ some uniform random variable. Now you can check that $$W$$ gives a counterexample. $$M$$ corresponds to $$X=U, Y=U$$ which is not a counterexample.

• If $X=U$ and $Y=1-U$, then $P[X\le t|Y=a]$ is the same as $P[Y \le t|X=a]$. Both conditional distributions are equal to the distribution of the random variable that takes the value $1-a$ with probability 1. Dec 27, 2020 at 21:11

Not necessarily true. Let $$X$$ and $$Y$$ be discrete random variables that take the values in {1, 2, 3} each with probability $$\frac{1}3$$; i.e. they have a discrete uniform distribution. Consider the joint probability mass function represented by the matrix below where the element in row $$i$$ and column $$j$$ is $$P[X=i, Y=j]$$:
$$\frac{1}{60}\begin{bmatrix} 3 & 6 & 11 \\ 3 & 12 & 5 \\ 14 & 2 & 4 \end{bmatrix}$$ Note that all rows and columns add up to $$\frac{1}3$$ and therefore the marginal distributions are the discrete uniform as stated. Now calculate the conditional probability $$P[X\leq2\mid Y=1]=\frac{\frac{3}{60}+\frac{3}{60}}{\frac{3}{60}+\frac{3}{60}+\frac{14}{60}}=\frac{3}{10}.$$ On the other hand, $$P[Y\leq2\mid X=1]=\frac{\frac{3}{60}+\frac{6}{60}}{\frac{3}{60}+\frac{6}{60}+\frac{11}{60}}=\frac{9}{20}.$$