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I'm trying to solve the probability that a cumulative discrete distribution (specifically, the negative hypergeometric distribution) equals 0.5 for a particular parameter $\in \mathbb{N}$. Since it's discrete, there is no exact solution, but I was wondering if Mathematica had a way to find the nearest solution. At this point I'm reduced to plugging in and guessing, and NSolve is not working (probably because its not just checking the naturals).

So I need a approximate solution $\in \mathbb{N}$ that may not be very close.


Edit:

Here's the code:

Solve[.5 ==
Sum[
    (Binomial[k + 1 - 1, k]*Binomial[540000 - 1 - k, 539000 - k])
    / 
    (Binomial[540000, 539000]
    ),
    {k, 1, n}
   ],
    n, Integers]

The (approximate) solution is 375.

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  • $\begingroup$ FindMinimum of the distribution minus 0.5 could work. $\endgroup$ – Marius Ladegård Meyer Aug 15 '17 at 7:26
  • $\begingroup$ Welcome to Mathematica Stack-Exchange! In general people get much better answers when they copy and paste (as text) the actual Mathematica they've tried. A non-obvious advantage of doing so: not everyone on MSE is a native english speaker, but everyone here speaks Mathematica, so you'll reach a much wider audience. It takes a little longer but it ends up making both the question and subsequent answers much more valuable. $\endgroup$ – John Joseph M. Carrasco Aug 15 '17 at 8:42
  • $\begingroup$ InverseCDF[] or Quantile[] are usable if your distribution is expressible in terms of built-ins. $\endgroup$ – J. M.'s discontentment Aug 15 '17 at 14:51
  • $\begingroup$ For some reason the negative hypergeometric isn't built in. $\endgroup$ – ketchup Aug 15 '17 at 23:40
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As noted in the docs, the negative hypergeometric is expressible in terms of BetaBinomialDistribution[]:

NegativeHypergeometricDistribution[r_, nsucc_, ntot_] :=
BetaBinomialDistribution[r, ntot - nsucc - r + 1, nsucc]

Check:

FullSimplify[(Binomial[k + 1 - 1, k] Binomial[540000 - 1 - k, 539000 - k])/
             (Binomial[540000, 539000]) == 
             PDF[NegativeHypergeometricDistribution[1, 539000, 540000], k], 
             0 <= k <= 539000]
   True

Thus, your problem is solved by either of

Quantile[NegativeHypergeometricDistribution[1, 539000, 540000], 1/2]
   373

or

InverseCDF[NegativeHypergeometricDistribution[1, 539000, 540000], 1/2]
   373
| improve this answer | |
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