Talk:Negative hypergeometric distribution

The PDF formula refers to the parameter n, which is undefined. This is probably a typo. --Erel Segal (talk) 15:15, 23 March 2015 (CET)

Typo, indeed. It should be $N$, not $n$. Thank you. Would you like to fix it yourself? --Boris Tsirelson (talk) 18:08, 23 March 2015 (CET)
The equation was an image I could not edit, so I replaced it with a latex equation - I hope I did this right. --Erel Segal (talk) 20:31, 24 March 2015 (CET)
Yes, this is one of our problems; you could look at Help:HowTo EoM, or just see some new articles made in TeX. Anyway, I did it a bit nicer. Boris Tsirelson (talk) 08:17, 25 March 2015 (CET)
Yes, this looks much nicer. While we are at it: I did not understand why this formula is correct. Can you please add an intuitive explanation? Also, I didn't understand the formula at the bottom, connecting the negative-hypergeometric with the hypergeometric. Can you explain this too? --Erel Segal (talk) 08:25, 25 March 2015 (CET)
Wow... not now, maybe later. Boris Tsirelson (talk) 08:30, 25 March 2015 (CET)

The PMF formula (*) is suspicious. If m<N, then k+m-N<k, so the top-left factor is zero! --Erel Segal (talk) 19:23, 25 March 2015 (CET)

Some sources available:
Boris Tsirelson (talk) 20:45, 26 March 2015 (CET)

Hmmm... After rewriting the three formulas in the last three sources in terms of EoM I got, respectively, $\frac{ {k+m-1 \choose k}{N-m-k \choose M-m} } { {N \choose M} } \, ;$ $\frac{ {M \choose k-1}{N-M \choose m} } { {N \choose m+k-1} } \cdot \frac{ M-k+1 }{ N-m-k+1 } \; ,$ $\frac{ {k+m-1 \choose k-1}{N-m-k \choose M-m} } { {N \choose M} } \, .$ I am puzzled. Boris Tsirelson (talk) 21:43, 26 March 2015 (CET)

In PlanetMath, the meaning of the parameters is only partially explained. But probably their $B$ and $W$ are our $M$ and $N-M$, and their $b$ and $x$ are our $m$ and $k$. If so, then their formula is identical to the formula from Balakrishnan and Nevzorov. At least, a hope to see two identical formulas!.. Boris Tsirelson (talk) 21:55, 26 March 2015 (CET)

Good news! Two more sources give the same as Balakrishnan and Nevzorov:

• G.K. Miller, S.L. Fridell, "A forgotten discrete distribution? Reviving the negative hypergeometric model", The American Statistician 2007 61:4, 347-350.
• E.F. Schuster, W.R. Sype, "On the negative hypergeometric distribution", Int. J. Math. Educ. Sci. Technol. 1987 18:3, 453-459.

Boris Tsirelson (talk) 22:36, 26 March 2015 (CET)

So, now I feel sure enough for correcting the article (even though I did not check the calculations myself). Thanks to Erel Segal for the good question. Boris Tsirelson (talk) 22:42, 26 March 2015 (CET)

Thanks for sorting this out! --Erel Segal (talk) 05:46, 27 March 2015 (CET)

Proof of the formula for the probability

The proof, given by Miller and Fridell, is easy. The outcome $X=k$ requires that, just before, we have $m-1$ marked and $k$ unmarked in a sample of size $m+k-1$, which has the probability $\frac{ {M \choose m-1}{N-M \choose k} } { {N \choose m+k-1} }$ according to Hypergeometric distribution. This is multiplied by the probability of getting one of the remaining $M-m+1$ marked elements among $N-m-k+1$ remaining elements; \begin{multline*} \frac{ {M \choose m-1}{N-M \choose k} } { {N \choose m+k-1} } \cdot \frac{ M-m+1 }{ N-m-k+1 } = \\ \frac{M!}{(m-1)!(M-m+1)!} \frac{(N-M)!}{k!(N-M-k)!} \frac{(m+k-1)!(N-m-k+1)!}{N!} \cdot \frac{ M-m+1 }{ N-m-k+1 } = \\ \frac{M!}{(m-1)!(M-m)!} \frac{(N-M)!}{k!(N-M-k)!} \frac{(m+k-1)!(N-m-k)!}{N!} = \\ \frac{M!(N-M)!}{N!} \cdot \frac{(m+k-1)!}{(m-1)!k!} \cdot \frac{(N-m-k)!}{(M-m)!(N-M-k)!} = \frac{ {k+m-1 \choose k}{N-m-k \choose M-m} } { {N \choose M} } \, . \end{multline*} Enjoy! Boris Tsirelson (talk) 09:31, 27 March 2015 (CET)

Thanks! --Erel Segal (talk) 11:13, 27 March 2015 (CET)
How to Cite This Entry:
Negative hypergeometric distribution. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Negative_hypergeometric_distribution&oldid=36382