Re: Conditional dist' of a Gaussian dist' with exponentially dist' variance
From: Ian Jermyn (Ian.Jermyn_at_sophia.inria.fr)
Date: 07/28/04
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Date: Wed, 28 Jul 2004 10:29:08 +0200
Dear Hiu,
As I said, the second distribution is very counterintuitive, in fact I
believe it is impossible. Let's impose the condition Y = y rather than Y = 1
to keep track of the Y dependence. It is not true that
Pr(X = x | Y = y) = \int Dv Pr(X = x | Y = y, V = v) Pr(V = v ) .
It is true, however, that
Pr(X = x | Y = y) = \int Dv Pr(X = x | Y = y, V = v) Pr(V = v | Y = y) .
Since
Pr(V = v | Y = y) = Pr(Y = y | V = v) Pr(V = v) / Pr(Y = y) ,
and since
Pr(Y = y | V = v) = (2\pi U)^{-1/2} e^{-y^{2}/ 2U}
(where U is the sum of the variances U = \sum_{i} v_{i} ) is *determined* by
the form of the Gaussian distribution for X, you need a form of Pr(V) that
will generate an exponential form independent of y for Pr(V = v | Y = y),
which is what you want to use. This means that you are implicitly assuming
that
Pr(V = v) = A(y) U^{1/2) e^(y^{2}/2U) e^{- \sum_{i} \lambda_{i} v_{i}} , (*)
where A(y) is a normalizing constant. Unfortunately, A(y) is infinite as far
as I can see, and thus there is no choice that will lead to the behaviour
that you want. Even if this were not the case, A(y) would not cancel out the
exponential in equation (*), and thus Pr(V = v) would depend on the value of
Y, in defiance of logic.
Thus your choice is incorrect.
Ian.
-- -------------------------------------------------- Ian Jermyn INRIA (Ariana) 2004 route des Lucioles B. P. 93 06902 Sophia Antipolis Cedex France T: +33 (0)4 9238 7683 F: +33 (0)4 9238 7643 E: Ian.Jermyn@sophia.inria.fr W: http://www-sop.inria.fr/ariana/personnel/Ian.Jermyn "Hiu Chung Law" <antispam@antispam.org> a écrit dans le message de news:ce6b9j$1k70$1@msunews.cl.msu.edu... > Ian Jermyn <Ian.Jermyn@sophia.inria.fr> wrote: > > I am not answering your question I am afraid, but I do wonder why you > > would be considering the second of the two distributions. As you are > > aware, it is not consistent with the first, the reason being that you > > are implicitly using different prior distributions for V in the two > > cases: > > > Pr(X = x | Y = 1) = \int Dv Pr(X = x | Y = 1, V = v) Pr(V = v | Y = 1) > > . > > > The last term can be rewritten > > > Pr(V = v | Y = 1) = Pr(Y = 1 | V = v) Pr(V = v) / Pr(Y = 1) , > > > and it is clear that > > > Pr(Y = 1 | V = v) > > > depends on v, since Y is essentially the mean. Indeed, the > > distribution for Y is Gaussian, with variance \sum v_{i} I believe, > > and thus the sum of the v_{i} should not be too much smaller than 1 if > > Y = 1 is to have appreciable probability. The use of the exponential > > distribtions for the v_{i} as a substitute for Pr(V = v | Y = 1) > > therefore seems quite odd. Given your choice of the exponential form > > for Pr(V = v), the form for Pr(V = v | Y = 1) can be derived easily. > > In the case that all the lambda are equal and small, Pr(V = v | Y = 1) > > has a peak at \sum v_{i} = 1, as one would expect. > > > Ian. > > > > Just thank to say thank you for the input by David and Ian. > Your opinion has helped me to think more about the problem. > > Now I have settled into using the second distribution, because I > think it serves my overall modelling purpose better. > > Thank you again.
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