Re: Conditional dist' of a Gaussian dist' with exponentially dist' variance
From: Ian Jermyn (Ian.Jermyn_at_sophia.inria.fr)
Date: 07/24/04
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Date: 24 Jul 2004 03:01:17 -0700
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.
Hiu Chung Law <lawhiu@cse.msu.edu> wrote in message news:<cdr5aq$15op$1@msunews.cl.msu.edu>...
> ( This article is a bit long... Thank you for your time of reading it.)
>
> Let X[i] be a random variable that follows the Gaussian distribution
> N(0, V[i]), with i from 1 to n.
> V[i] is itself a random variable that follows the
> exponential distribution with mean 2/lambda[i]. Different X[i]s are
> independent. Different V[i]s are independent. In other words,
>
> p( X[i] | V[i] ) = exp( -0.5 X[i]^2 / V[i] ) / sqrt( 2 Pi V[i] )
> p( V[i] ) = exp( -lambda[i]/2 V[i] ) * lambda[i] / 2
>
> It is well-known that the distribution of X[i], after marginalizing
> out V[i], is Laplacian.
>
> p( X[i] ) = sqrt(lambda[i]) / 2 * exp( - sqrt(lambda[i]) * abs(X[i]) )
>
> I am interested in the conditional distribution of X[i]s when their
> sum is one. I have tried to use Fourier transform to find the pdf
> of the random variable Y = X[1] + ... + X[n] for small values of n,
> but the form is cumbersome and I failed to see any pattern
> for general n. This makes me suspect that
> the pdf p(X[1], ..., X[n] | Y=1) probably does not have a nice
> form.
>
> Because the conditional distribution of Gaussian random variables is nice,
> I instead consider the following distribution.
>
>
> q(X[1], ..., X[n] | Y=1)
> = int N(X[1], ..., X[n] | Y=1, V[1], ..., V[n]) p(V[1]) ... p(V[n])
> d V[1] ... d V[n]
>
> Recall N(.|.) is the Gaussian distribution, and, in this case, it has
> the covariance matrix diag(V[1], ..., V[n]), and p(V[i]) is the
> exponential distribution.
>
> In general, q(X[1], ..., X[n] | Y=1) is different from
> p(X[1], ..., X[n]|Y=1), because
>
> p(a|b) = ( int p(a,b|theta) p(theta) ) / ( int p(b|theta) p(theta) )
> <> int ( p(a,b|theta) / p(b|theta) p(theta) )
>
> However, when I studied the case when n=2 and n=3 numerically, I was
> surprised to find out that q(.) and p(.) are qualitative very
> similar. They look like to be in the same family of distribution,
> but with different parameters.
>
> Is this finding purely accidental? Or would it be possible to prove
> some relationship between p(X[1], ..., X[n] | Y=1) and
> q(X[1], ..., X[n] | Y=1)? I have tried to perform the integration
> of q(X[1], ..., X[n] | Y=1) analytically, but I am going nowhere.
>
> Any input is highly appreciated. Thank you!
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