P(y)=Integral( P(x|y)*P(y)*dx)
function is above if I didnt write wrongly. P(x|y) is conditional probability and it is known but P(y) is not known, thanks.
there may be an iterative solution but is there any analytical solution.
I am possibly confused here, but:
Integral( P(x|y)*P(y)*dx ) = P(y) * Integral( P(x|y)*dx ) # because P(y) is not a function of x, and
Integral( P(x|y)*dx ) = 1
so Integral( P(x|y)*P(y)*dx ) = P(y) * 1 = P(y)
Hi, in the paranthesis a bayes form of P(x, y) and if you take the integral according to dx you will get P(y) or if you take yhe integral according to dy you will get P(x). And I supposed that if you take the integral of P(x|y) *dx you will not get 1 everytime unless x and y are independent.
But I didnt ask that. I asked is there any name of integral which the output function is also in the integral, it may also be written such that
f(y) =Int(g(x, y) *f(y)*dx)
It seems recursive function but I a0m not sure anout it.
P(x|y) * P(y) = (P(x,y)/P(y))*P(y) = P(x,y) in which case
Integral( P(x,y)*dx)) is the marginal distribution of y. The integral you have written is an identity, which when P(x,y) is a density for jointly distributed random variable (X,Y) is by definition, the marginal distribution of Y.
Thanks you all for answers, I know that is bayes marginal distribution and I have rewritten the equation from bayes theorem, but here i want to find the P(y) while i assumed that i know p(x|y). In fact I want to ask the general name or solution of integral equation which the output is also in the integral. It has not to be probability equation.
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