Bayesian Analysis!

It reminds me of this
P(A | B)=

\sum_{i}P(A|C_{i})P(C_{i}|B)[br]

but I can't understand what is this this thing in the denominator. Please help me...

I know what is the posterior and the conjugate and I can solve easily these integrals, but I can't understand the formula I highlighted (sorry for the bad photo too)

(edited 9 years ago)

Reply 1

Smaug123

15

Original post by ted25

It reminds me of this
P(A | B)=

\sum_{i}P(A|C_{i})P(C_{i}|B)[br]

but I can't understand what is this this thing in the denominator. Please help me...

I know what is the posterior and the conjugate and I can solve easily these integrals, but I can't understand the formula I highlighted (sorry for the bad photo too)

I imagine it's a normalising constant. That is, it's there to ensure that the posterior probability sums to 1.

Reply 2

ted25

OP

thanks, but still can't find how from this

[br] f(\Theta |x_{1},...,x_{n})\frac{ f(x_{1},...,x_{2}|\Theta)u(\Theta)}{\int f(x_{1},...,x_{n}|\Theta )u(\Theta)d\Theta}[br]

or the formula above we get to that of the picture. (Which one is which?)

(edited 9 years ago)

Reply 3

ted25

OP

Found it. The denominator equals one. All he's doing is the P(A|B) formula I wrote above.

The reason is doing it is for computational reasons. If the numerator is left untouched it gives us some Gamma functions that cannot be eliminated and which are similar to the Gammas of the posterior ( g(θ) ).

So first he eliminates all (same) Gammas, then he transforms the integrals into slightly different Gammas which can be eliminated easily, using the known fact that

\int

PDF = (Gamma part)* (number part) dx = 1

\int

(number part) dx = 1/(Gamma part)

(edited 9 years ago)

Bayesian Analysis!

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