Is the joint distribution $P_{XY}(x,y)$ determined from the marginal $P_X(x)$ and the conditional $P_{X|Y}(x|y)$?

For simplicity assume that $$X,Y$$ are discrete, finite, random variables, with joint distribution $$P_{XY}(x,y) = \mathbb{P}(X=x\wedge Y=y)$$.

Now suppose that we do not know $$P_{XY}(x,y)$$, but are given the values of the marginal $$P_X(x)=\sum_y P_{XY}(x,y)$$ and the conditional $$P_{X|Y}(x|y)=P_{XY}(x,y)/P_Y(y)$$.

Is the knowledge of $$P_X(x)$$ and $$P_{X|Y}(x|y)$$ enough to recover the full joint distribution $$P_{XY}(x,y)$$?

Please note that this is different from Is the joint distribution $P_{XY}(x,y)$ determined from the conditionals $P_{X|Y}(x|y)$ and $P_{Y|X}(y|x)$?, because there I know the two conditionals, whereas here I know a conditional and a marginal.

• closely related: stats.stackexchange.com/q/427510/5536 – becko Sep 16 '19 at 18:29
• You have an incorrect formula for the marginal. When you use a correct one, it should become apparent that the answer is in the negative. Consider the case of independent variables, for instance: your conditions tell you nothing whatsoever about the distribution of $Y.$ – whuber Sep 16 '19 at 18:33
• Of course. That's an obvious counter-example. If you move your comment to an answer I'll accept it. Thanks. – becko Sep 16 '19 at 18:51
• @whuber Sorry, I don't see why my formula for the marginal is wrong? – becko Sep 17 '19 at 17:28
• To complete the two linked questions here is one example of the third flavour: Is it possible to derive joint probabilities from marginals with assumptions about the conditionals? – Sextus Empiricus Sep 18 '19 at 21:57

The marginal $$P_X(x)$$ can be found by summing (or integrating for continuous variables) the conditional $$P_{X|Y}(x|y)$$. Or in different words: the marginal probability of $$P_X(x)$$ is a sort of mixture of the conditional probabilities $$P_{X|Y}(x|y)$$ (at different values of $$Y$$) with the weights determined by the probability $$P_Y(y)$$.

$$P_X(x) = \sum_{\forall Y} P_{X|Y}(x|y)P_Y(y)$$

Since there may be multiple different $$P_Y(y)$$ that can lead to the same $$P_X(x)$$ for a given $$P_{X|Y}(x|y)$$ the information of $$P_X(x)$$ and $$P_{X|Y}(x|y)$$ can not be used to calculate backwards the $$P_Y(y)$$.

The simplest case is when $$P_{X|Y}(x|y) = P_X(x)$$ in which case any $$P_Y(y)$$ will be compatible.

For discrete variables (and you could extend the logic to continous variables), you could consider $$P_X(x) = \sum_{\forall Y} P_{X|Y}(x|y)P_Y(y)$$ as a matrix equation:

$$\begin{bmatrix} P_X(a_1) \\ P_X(a_2) \\ \vdots \\ P_X(a_n) \end{bmatrix} = \begin{bmatrix} P_{X|Y}(a_1|b_1) & P_{X|Y}(a_1|b_2) & \dots & P_{X|Y}(a_1|b_n) \\ P_{X|Y}(a_2|b_1) & P_{X|Y}(a_2|b_2) & \dots & P_{X|Y}(a_2|b_n) \\ \vdots & \vdots & & \vdots\\ P_{X|Y}(a_n|b_1) & P_{X|Y}(a_n|b_2) & \dots & P_{X|Y}(a_n|b_n) \\ \end{bmatrix} \cdot \begin{bmatrix} P_Y(b_1) \\ P_Y(b_2) \\ \vdots \\ P_Y(b_n) \end{bmatrix}$$

So when the $$P_{X|Y}(x|y)$$, when considered as vectors (a different one for each value of $$y$$) are linearly independent, then you can obtain $$P_Y(y)$$ from $$P_X(x)$$ and $$P_{X|Y}(x|y)$$.

This is a sufficient condition but not necessary. The additional restriction that all $$P_Y(y)>0$$ might make it possible that also linear dependent vectors might still result in a unique solution for $$P_Y(y)$$.

Example. When $$P_{X|Y}(x|y) \sim N(\mu = y,\sigma = 1)$$ then you can not write any of the conditional probabilities as a sum of the others and you should be able to recover $$P_Y(y)$$ when you know $$P_X(x)$$ (by some deconvolution).