buzzard at urubu.freeserve.co.uk
Wed Mar 14 02:25:42 CET 2007
Hendrik van Rooyen wrote:
> "Nick Craig-Wood" <nick at craig-wood.com> wrote:
>>Paul Rubin <http> wrote:
>>> The fencepost method still seems to be simplest:
>>> t = sorted(random.sample(xrange(1,50), 4))
>>> print [(j-i) for i,j in zip(+t, t+)]
>>Here is another effort which is easier to reason about the
>>distribution produced but not as efficient.
>>def real(N, M):
>> while 1:
>> t = [ random.random() for i in range(N) ]
>> factor = M / sum(t)
>> t = [ int(round(x * factor)) for x in t]
>> if sum(t) == M:
>> print "again"
>> assert len(t) == N
>> assert sum(t) == M
>> return t
>>It goes round the while loop on average 0.5 times.
>>If 0 isn't required then just test for it and go around the loop again
>>if found. That of course skews the distribution in difficult to
> I have been wondering about the following as this thread unrolled:
> Is it possible to devise a test that can distinguish between sets
> - five random numbers that add to 50, and
> - four random numbers and a fudge number that add to 50?
> My stats are way too small and rusty to attempt to answer
> the question, but it seems intuitively a very difficult thing.
> - Hendrik
Yes, if the generating processes yield numbers from different
probability mass functions. You could simply look at the likelihood
ratio. Otherwise, the likelihood ratio will be 1.
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