Without some explicit expressions to discuss, I'm grasping at air here. But I find it difficult to credit that assuming a uniform distribution of n in (say) [1..100] would result in an estimated n substantially different from [1..10^10] , when --- in my present experiment --- I had k = 4 coupons at m = 7 trials, and k remains unchanged at m = 17 . The fact that classical probability cannot cope with a uniform denumerable distribution does not preclude the existence of a perfectly definite limiting answer as upper bound -> infinity. Fred Lunnon On 2/4/16, Mike Stay <metaweta@gmail.com> wrote:
On Wed, Feb 3, 2016 at 4:46 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I have no prior favouring any n ; all I know is that n is finite (rather than denumerably infinite). WFL
Well, there's no uniform distribution on a countably infinite set, so you have to choose some other one. May I suggest the universal prior from Kolmogorov complexity?
On 2/4/16, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 03/02/2016 23:46, Fred Lunnon wrote:
Ahem, I digress. Now I argue that if a trial reaches reaches k distinct coupons in m > c k log k trials, where (say) c = 2, or 10, then it's a damned good bet that n = k . Anybody disagree?
With a strong enough prior favouring a larger n, you will prefer the hypothesis that you just saw very improbable results.
(But you probably should never actually *have* a strong enough prior for that; if m and c are large enough, someone who thought they had such a prior would be saying "oops, I chose the wrong model", which means that in some sense their prior didn't favour large n so strongly after all.)
-- g
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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