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Suppose I have a binary outcome $Y$ and two binary covariates $X_1$ and $X_2$ with distribution $P(X_1, X_2)$ which I know. In addition to $P(X_1, X_2)$, I know $P(Y\mid X_1)$ and $P(Y\mid X_2)$. I would like to know how is $P(Y\mid X_1, X_2)$ bounded by the data I have at hand, and possibly know whether there is a natural visualisation of it.

I have checked previous questions here, here, and here. But unfortunately none of them provides an answer.

So far I have only been able to derive some trivial bounds on the probability. For example, if $P(y\mid x_1)\neq 0$ and $P(x_1, x_2)\neq 0$, then we can use $P(y\mid x_1) = \sum_{x_2} P(y\mid x_1, X_2=x_2)P(X_2=x_2\mid x_1)$ to conclude that $P(y\mid x_1, X_2=x_2)\neq 0$. But I haven't been able to come up with general formulas for the bounds on the full conditional.

The point of this question goes beyond three binary variables, I would even like to know whether there is a way to bound conditional probabilities based on arbitrary functions (say a conditional mean, instead of the conditional distribution) of marginalised distributions.

Sergio
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  • Could you clarify what you mean by "have access to"? It sounds like it means know, in which case you have full information about the joint probability and no bounds are needed. It also implies data are irrelevant, so references to data are difficult to reconcile with the rest of your question. – whuber Feb 08 '23 at 16:15
  • @whuber Yes, with "have access to" I mean I know it. I will edit accordingly. – Sergio Feb 08 '23 at 16:21
  • @whuber Can you please explain how do I have access to the full joint probability? – Sergio Feb 08 '23 at 16:22
  • As an additional question that arises from your comment: can't we call the knowledge of $P(Y\mid X_1)$ "data"? – Sergio Feb 08 '23 at 16:54
  • Data would consist of observations modeled by a probability distribution. The distribution, or any part or property of it, is a purely mathematical, hypothetical thing. You know the full probability because of the basic law of conditional probabilities: $\Pr(X,Y)=\Pr(Y\mid X)\Pr(X).$ This holds even when $X=(X_1,X_2)$ is multivariate. – whuber Feb 08 '23 at 17:44
  • @whuber thank you for taking the time. I think you might have misread my question. I don't know $P(Y\mid X_1, X_2)$ but only the joint of $X_1$ and $X_2$, and the conditionals $P(Y\mid X_1)$ and $P(Y\mid X_2)$. If your statement still holds, could you please elaborate further? – Sergio Feb 08 '23 at 18:31
  • Then you are correct that I misread it and you indeed do not have full information about the joint distribution. Concerning your data, would they be a random sample from the joint distribution or might they be a little different, such as observations of $Y$ based on values of $X$ determined by the experimenter (a typical regression situation)? – whuber Feb 08 '23 at 19:39

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