# If $X$ is discrete and $Z,W$ are discrete or continuous, is it always the case that $P(X=x\mid Z) \geq P(X=x\mid Z,W)$? [closed]

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 GodsilMay 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
If this question can be reworded to fit the rules in the help center, please edit the question.

• This question is more suitable for MathStackExchange. Did you try the case $Z=0$ (constant random variable) and $W=X$? – Jochen Wengenroth May 14 at 6:54
• 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. – 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 $$P(X=1|W=1)=0$$.