I know that

$$P(X\,|\,Y)\le P(X)\quad\Leftrightarrow\quad \frac{P(X , Y)}{P(Y)}\le P(X)\quad\Leftrightarrow\quad P(X , Y)\le P(Y)P(X)$$

is a proof for $P(X=x|Y=y)\le P(X=x)$ but i don't get the intuition.

If we just consider marginal probability of $X$ and probability of $X=x$ be $p$ , it isn't possible that probability of $X=x$ become greater than $p$ after revealing that $Y=y$?

  • $\begingroup$ The inequality $P(X|Y)\le P(X)$ just happen in some circumstances, but it is not generally true (consider the case of $X=Y$ and some event such that $P(X)<1$). I dont understand what you mean by "intuition" here. Intuition about what? About why an inequality holds for some particular choices of $X$ and $Y$? $\endgroup$ – Masacroso Nov 8 '18 at 20:53
  • $\begingroup$ If we considere this $$P(B\cap A)=P(B , A) $$ so intersection of two sets will be less than sum of two set. then can't we say $$P(B\cap A)=P(B , A)≤P(A)P(B) $$ ? $\endgroup$ – alireza Nov 8 '18 at 20:57
  • $\begingroup$ but probabilities are numbers between zero and one, so the inequality it is not clear. We will have $P(A,B)\le \min\{P(A),P(B)\}$, but we still dont know if the inequality holds. Suppose that $P(A,B)=0.1$ and $P(A)=P(B)=0.2$, then you will had that $P(A,B)>P(A)P(B)$. $\endgroup$ – Masacroso Nov 8 '18 at 21:05
  • $\begingroup$ and it can also be the case that $P(A,B)=0$ (events are mutually exclusive) but $P(A)=P(B)=0.2$, in this case we will have $P(A,B)\le P(A)P(B)$, etc... $\endgroup$ – Masacroso Nov 8 '18 at 21:10

You should read the inequality $P(X,Y) \leq P(X)P(Y)$ as: 'the events $X$ and $Y$ are negatively correlated'.

If $X$ was independent from $Y$, then $P(X)P(Y)$ would have been the probability of the two events happening together. What the inequality tells you is that there is some dependence between the events in such a way that they have a lower chance to happen together.

Now the intuition should be clear. If you know that $Y$ happened then $X$ now has a smaller chance of happening as well, so $P(X|Y) \leq P(X)$.

  • $\begingroup$ So you say that $P(X,Y)≤P(X)P(Y)$ is just an assumption and in general case we can have $P(X,Y)>P(X)P(Y)$ ? $\endgroup$ – alireza Nov 8 '18 at 20:46
  • $\begingroup$ @alireza "in general" there is not order relation between $P(X|Y)$ and $P(X)$. The term "in general" in mathematics means "for all cases". $\endgroup$ – Masacroso Nov 8 '18 at 20:54

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