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| Science Forum Index » Statistics - Math Forum » Paradoxical (to me) Bayesian result... |
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| strnbrg... |
 Posted: Thu Oct 29, 2009 10:34 am |
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Suppose X is uniformly distributed on [0,A], and A is uniformly
distributed on [1,M] (M>1). I get this for the marginal density of X:
p(x) = Int{1,M}(p(x|a)p(a)da = Int{1,M}(1/(a(M-1)))da
= (1/(M-1))Int{1,M}(dlog(a)) = log(M)/(M-1)
So x is uniformly distributed not only conditionally but
unconditionally
as well -- uniform on [1,(M-1)/log(M)].
Could this be right? Something about it bothers me; saying X
is unconditionally uniform on [1,(M-1)/log(M)] implies there's zero
probability of X > (M-1)/log(M). And that can't be right! |
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| Jack Tomsky... |
| Posted: Thu Oct 29, 2009 11:35 am |
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[quote]Suppose X is uniformly distributed on [0,A], and A is
uniformly
distributed on [1,M] (M>1). I get this for the
marginal density of X:
p(x) = Int{1,M}(p(x|a)p(a)da =
da = Int{1,M}(1/(a(M-1)))da
= (1/(M-1))Int{1,M}(dlog(a)) = log(M)/(M-1)
So x is uniformly distributed not only conditionally
but
unconditionally
as well -- uniform on [1,(M-1)/log(M)].
Could this be right? Something about it bothers me;
saying X
is unconditionally uniform on [1,(M-1)/log(M)]
implies there's zero
probability of X > (M-1)/log(M). And that can't be
right!
[/quote]
The key is that the integration limits on A are from max(x,1) to M. (This is reflection of the fact that the density of x is zero when x > A.)
Break down the integration into two cases: according to whether x <= 1 or x > 1.
Jack
www.tomskystatistics.com |
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