# Simple Derivation of Marginalization

So P where P is Probability so P(X) = Prob(X) = P(X|I) = The Probability of Event X occurring:

how can you derive this from P(X) = P(X,Y) $\cap$ P(X,$\bar{Y}$)?

Thank you,

Note: sorry it should be P(X) in the first one, it's the same X. X and Y are events, so it is events $Y_1, Y_2,Y_3 ... Y_m = \{Y_k\}$

This ideally is the beginning of the marginalization equation to get $\int_{-\infty}^{+\infty} prob(Y|X)\ dY = 1$.

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I take it $P(X,Y)$ is a number, but I don't know what you mean by the intersection of two numbers. –  Gerry Myerson May 16 '12 at 4:12
Que? Probabilities are numbers. $Y$ may be an event, but $P(Y)$ is a number, and so, I expect, is $P(X,Y)$, although you haven't told us what $P(X,Y)$ means, so it's hard to be sure. Take some time to think your question through, and try again when you understand what you are actually trying to ask. –  Gerry Myerson May 16 '12 at 4:21
@Eiyrioü von Kauyf : I think you must have meant $P((X\cap Y)\cup(X\cap \bar{Y}))$. Notice that in that expression the letter $P$ appears only ONCE. Notice also which ones say $\cap$ and which ones say $\cup$. One may write $P(A\cup B)=P(A)+P(B)$, but one should never write $P(A)\cup P(B)$ or the like---that is nonsense. –  Michael Hardy May 16 '12 at 4:28
Satisfied with my answer? –  Did Sep 19 '12 at 18:17