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Mathematically, $P(Y|X) = \frac{P(X,Y)}{P(X)}$ and so $P(Y|X)$ must depend on $P(X)$. Since $P(Y|X)$ will change when $P(X)$ changes.

However, consider this scenario:

  • X = amount of red meat consumed per week.
  • Y = probability of colon cancer.

Clearly, the probability of colon cancer given the amount of red meat consumed per week, $P(Y|X)$, does not depend on the distribution of red meat consumption in the population, $P(X)$.

The way I interpreted this situation is by treating $P(Y|X)$ and $P(X)$ as "independent variables" that can vary on their own. While $P(X,Y)$ is a "dependent variable" that cannot vary on its own and is purely calculated from $P(Y|X)$ and $P(X)$.

However, this feels like something I made up instead of a formal mathematical justification. Is there a canonical way of expressing this scenario in terms of mathematical probability?

Legendre
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    You can easily construct examples where any two of $P(X),P(Y \mid X), P(X,Y)$ vary and the third does not change (rather like pressure, volume and temperature) so I have difficulty understanding the question – Henry Jul 30 '23 at 02:44
  • The question is unclear: $P(X)$ is a random variable as a transform of the rv $X$, and the same for $P(Y|X)$, hence one can consider whether or not the two rv's are independent. However, it seems that the OP considers the densities $p(y|x)$ and $p(x)$ for which "independence" has no specific meaning. – Xi'an Jul 30 '23 at 06:54

1 Answers1

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  1. $P(Y|X)$ won't change when $P(X)$ changes because $X$ is fixed in $ P(Y|X)$.

  2. Mathematically, $P(Y,X)=P(Y|X)P(X)$, so if you change $P(X)$, both sides change proportionately to the change in $P(X)$ and $P(Y|X)$ is unchanged.

jbowman
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