Suppose $X$ is discrete and $Z,W$ are discrete or continuous, I am wondering if it is always the case (or at least non-trivially) that

$$ P(X=x\mid Z) \geq P(X=x\mid Z,W) $$

for all $x\in X$.

It appears to make intuitive sense to me because it appears the probability of $X$ given $Z$ and $W$ should be "subsetting" off the probability of $X$ given $Z$. That is, if there is some chance of $X$ given $Z$, then $X$ given $Z$ and $W$ is subdividing the occurrence of $X$ given $Z$ into many more chunks according to $W$, and thus the probability of a chunk occurring should be less than the whole.

In other words, it seems to hold in the finite-sampling perspective, where the above are empirical proportions over some collection of objects. However, the result doesn't seem to hold more generally. I am wondering why the finite-sampling intuition doesn't seem to extend. Is there a general result behind this?


closed as off-topic by Jochen Wengenroth, user44191, Mateusz Kwaśnicki, LSpice, Chris Godsil May 14 at 12:15

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Jochen Wengenroth, Mateusz Kwaśnicki, Chris Godsil
  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – user44191, LSpice
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  • $\begingroup$ This question is more suitable for MathStackExchange. Did you try the case $Z=0$ (constant random variable) and $W=X$? $\endgroup$ – Jochen Wengenroth May 14 at 6:54
  • $\begingroup$ You are comparing two probability distributions. They both sum to $1$, so it can't be the case that one of them is bigger everywhere, unless they are the same. $\endgroup$ – James Martin May 14 at 11:13

No. First of all, $Z$ is a red herring here -- just define a new probability space conditional on $Z$. Thus your question is really whether $P(X=x)\ge P(X=x|W=w)$ always holds. This is easily seen to be false: Let $(X,W)$ take on the values $(0,1)$ and $(1,0)$ with probability $1/2$. Then $P(X=1)=1/2$ but $P(X=1|W=1)=0$.


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