# conditional distribution determine joint distribution

$X,Y$ are two random variables and we know the distribution of $X|Y$ and $Y|X$, then I want a counterexample that $(X,Y)$ is not unique determined.

this question is motivated by Can conditional distributions determine the joint distribution? , but there isn't a counterexample. I followed the direction of @heropup and considered the discrete case.

I construct two random variable that both valued in $\{0,1\}$:$\mathbb P(X=0,Y=0)=\frac 1 2, \mathbb P(X=0,Y=1)=\frac 1 4, \mathbb P(X=1,Y=0)=\frac 1 8, \mathbb P(X=1,Y=1)=\frac 1 8$.

now given the conditional probability above $\mathbb P(X=i|Y=j)$ and $\mathbb P(Y=i|X=j)$, I want to construct two different random variables $\tilde{X},\tilde Y$ valued in $\{0,1\}$ s.t. $\mathbb P(X=i|Y=j)=\mathbb P(\tilde X=i|\tilde Y=j)$ and $\mathbb P(Y=i|X=j)= \mathbb P(\tilde Y=i|\tilde X=j)$

I used the method of undetermined coefficients, but unfortunately the $(\tilde X,\tilde Y)$ has the same distribution with $(X,Y)$.

I don't know how to come up with a counterexample now, any help will be appreciated!

• No counterexample can exist, even without independence hypothesis. When densities exist, note that $$f_{X\mid Y}(x\mid y)f_Y(y)=f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)f_X(x)$$ hence, $$f_Y(y)\int\frac{f_{X\mid Y}(x\mid y)}{f_{Y\mid X}(y\mid x)}dx=\int f_X(x)dx=1$$ hence $$f_Y(y)=\left(\int\frac{f_{X\mid Y}(x\mid y)}{f_{Y\mid X}(y\mid x)}dx\right)^{-1}$$ and $f_{X,Y}$ is fully determined by $f_{X\mid Y}$ and $f_{X\mid Y}$ through the formula $$f_{X,Y}(x,y)=f_{X\mid Y}(x\mid y)\left(\int\frac{f_{X\mid Y}(x\mid z)}{f_{Y\mid X}(z\mid x)}dz\right)^{-1}.$$ – Did Apr 30 '16 at 13:48
• @Did thanks, this is an exercise, I think I should focus on the general case, the conditional probability might be $0$. – Lookout May 1 '16 at 1:56

Let's say $X$ and $Y$ take values in finite sets $M$ and $N$, and that their joint mass function $\Bbb P[X=x,Y=y]$ is $>0$ for all $(x,y)\in M\times N$. Then $${\Bbb P[X=x|Y=y]\over \Bbb P[Y=y|X=x]}={\Bbb P[X=x]\over\Bbb P[Y=y]},$$ for all $(x,y)\in M\times N$. Summing over $x$ we find that $${1\over \Bbb P[Y=y]}=\sum_{x\in M}{\Bbb P[X=x|Y=y]\over \Bbb P[Y=y|X=x]}$$ is completely determined by the conditional distributions. Hence so is $\Bbb P[Y=y]$ determined. Finally, $\Bbb P[X=x,Y=y]=\Bbb P[X=x|Y=y]\cdot\Bbb P[Y=y]$ is also detemined by the conditional distributions.