negative hypergeometric distribution

There are $N$ elements, of which $K$ are defined as "successes" and the rest are "failures".

Elements are drawn one after the other, without replacements, until $r$ failures are encountered. Then, the drawing stops and the number $k$ of successes is counted. The negative hypergeometric distribution, $NHG\_\{N,K,r\}(k)$ is the discrete distribution of this $k$.

[https://www.encyclopediaofmath.org/index.php/Negative_hypergeometric_distribution Negative hypergeometric distribution] in Encyclopedia of Math.

The negative hypergeometric distribution is a special case of the beta-binomial distribution{{cite book |last1=Johnson |first1=Norman L. |last2=Kemp |first2=Adrienne W.|author2-link=Adrienne W. Kemp |last3=Kotz |first3=Samuel |date=2005 |title=Univariate Discrete Distributions |publisher=Wiley |isbn=0-471-27246-9}} §6.2.2 (p.253–254) with parameters $\backslash alpha\; =\; r$ and $\backslash beta\; =\; N-K-r+1$ both being integers (and $n\; =\; K$).

The outcome requires that we observe $k$ successes in $(k+r-1)$ draws and the $(k+r)\backslash text\{-th\}$ bit must be a failure. The probability of the former can be found by the direct application of the hypergeometric distribution $(HG\_\{N,K,k+r-1\}(k))$ and the probability of the latter is simply the number of failures remaining $(=N-K-(r-1))$ divided by the size of the remaining population $(=N-(k+r-1)$. The probability of having exactly $k$ successes up to the $r\backslash text\{-th\}$ failure (i.e. the drawing stops as soon as the sample includes the predefined number of $r$ failures) is then the product of these two probabilities:

:$\backslash frac\{\backslash binom\{K\}\{k\}\; \backslash binom\{N\; -\; K\}\{k+r-1-k\}\}\{\backslash binom\{N\}\{k+r-1\}\}\; \backslash cdot\; \backslash frac\{N-K-(r-1)\}\{N-(k+r-1)\}\; =\backslash frac\{\{\{k+r-1\}\backslash choose\{k\}\}\{\{N-r-k\}\backslash choose\{K-k\}\}\}\{N\; \backslash choose\; K\}.$

Therefore, a random variable $X$ follows the negative hypergeometric distribution if its probability mass function (pmf) is given by

:$f(k;\; N,\; K,\; r)\; \backslash equiv\; \backslash Pr(X\; =\; k)\; =\backslash frac\{\{\{k+r-1\}\backslash choose\{k\}\}\{\{N-r-k\}\backslash choose\{K-k\}\}\}\{N\; \backslash choose\; K\}\backslash quad\backslash text\{for\; \}k\; =\; 0,\; 1,\; 2,\; \backslash dotsc,\; K$

where

• $N$ is the population size,
• $K$ is the number of success states in the population,
• $r$ is the number of failures,
• $k$ is the number of observed successes,
• $a\; \backslash choose\; b$ is a binomial coefficient

By design the probabilities sum up to 1. However, in case we want show it explicitly we have:

:$\backslash sum\_\{k=0\}^\{K\}\; \backslash Pr(X=k)$

= \sum_{k=0}^{K} \frac{{{k+r-1}\choose{k}}{{N-r-k}\choose{K-k}}}{N \choose K}

= \frac{1}{N \choose K}\sum_{k=0}^{K} {{k+r-1}\choose{k}}{{N-r-k}\choose{K-k}}

= \frac{1}{N \choose K} {N \choose K}

= 1,

where we have used that,

:$\backslash begin\{align\}$

\sum_{j=0}^k \binom{j+m}{j} \binom{n-m-j}{k-j}

&=\sum_{j=0}^k (-1)^{j} \binom{-m-1}{j} (-1)^{k-j} \binom{m+1+k-n-2}{k-j}\\

&= (-1)^{k} \sum_{j=0}^k \binom{-m-1}{j} \binom{k-n-2-(-m-1)}{k-j}\\

&= (-1)^{k} \binom{k-n-2}{k}\\

&= (-1)^{k} \binom{k-(n+1)-1}{k}\\

&= \binom{n+1}{k},

\end{align}

which can be derived using the binomial identity,

:$\{\{n\; \backslash choose\; k\}\; =\; (-1)^k\; \{k-n-1\; \backslash choose\; k\}\},$

and the Chu–Vandermonde identity,

:$\backslash sum\_\{j=0\}^k\; \backslash binom\; m\; j\; \backslash binom\{n-m\}\{k-j\}\; =\; \backslash binom\; n\; k,$

which holds for any complex-values $m$ and $n$ and any non-negative integer $k$.

When counting the number $k$ of successes before $r$ failures, the expected number of successes is $\backslash frac\{r\; K\}\{N-K+1\}$ and can be derived as follows.

$\backslash begin\{align\}$

E[X] &= \sum_{k=0}^{K} k \Pr(X=k)

= \sum_{k=0}^{K} k \frac{{{k+r-1}\choose{k}}{{N-r-k}\choose{K-k}}}{N \choose K}

= \frac{r}{N \choose K}\left[\sum_{k=0}^{K} \frac{(k+r)}{r} {{k+r-1}\choose{r-1}}{{N-r-k}\choose{K-k}}\right]-r \\

&= \frac{r}{N \choose K}\left[\sum_{k=0}^{K} {{k+r}\choose{r}}{{N-r-k}\choose{K-k}}\right]-r

= \frac{r}{N \choose K}\left[\sum_{k=0}^{K} {{k+r}\choose{k}}{{N-r-k}\choose{K-k}}\right]-r\\

&= \frac{r}{N \choose K}\left[{{N+1} \choose K} \right]-r = \frac{rK}{N-K+1},

\end{align}

where we have used the relationship $\backslash sum\_\{j=0\}^k\; \backslash binom\{j+m\}\{j\}\; \backslash binom\{n-m-j\}\{k-j\}=\; \backslash binom\{n+1\}\{k\}$, that we derived above to show that the negative hypergeometric distribution was properly normalized.

The variance can be derived by the following calculation.

$\backslash begin\{align\}$

E[X^2] &= \sum_{k=0}^{K} k^2 \Pr(X=k)

= \left[\sum_{k=0}^{K} (k+r)(k+r+1) \Pr(X=k)\right]-(2r+1)E[X]-r^2-r \\

&=\frac{r(r+1)}{N \choose K}\left[\sum_{k=0}^{K} {{k+r+1}\choose{r+1}}{{N+1-(r+1)-k}\choose{K-k}}\right]-(2r+1)E[X]-r^2-r\\

&= \frac{r(r+1)}{N \choose K}\left[{{N+2} \choose K} \right]-(2r+1)E[X]-r^2-r

= \frac{r K (N - r + K r + 1)}{(N-K+1)(N-K+2)}

\end{align}

Then the variance is $\backslash textrm\{Var\}[X]=E[X^2]-\backslash left(E[X]\backslash right)^2\; =\; \backslash frac\{r\; K\; (N\; +\; 1)\; (N-K-r+1)\}\{(N\; -\; K\; +\; 1)^2\; (N\; -\; K\; +\; 2)\}$

If the drawing stops after a constant number $n$ of draws (regardless of the number of failures), then the number of successes has the hypergeometric distribution, $HG\_\{N,K,n\}(k)$. The two functions are related in the following way:

:$NHG\_\{N,K,r\}(k)\; =\; 1-HG\_\{N,N-K,k+r\}(r-1)$

Negative-hypergeometric distribution (like the hypergeometric distribution) deals with draws without replacement, so that the probability of success is different in each draw. In contrast, negative-binomial distribution (like the binomial distribution) deals with draws with replacement, so that the probability of success is the same and the trials are independent. The following table summarizes the four distributions related to drawing items:

class="wikitable"
	With replacements	No replacements
# of successes in constant # of draws	binomial distribution	hypergeometric distribution
# of successes in constant # of failures	negative binomial distribution	negative hypergeometric distribution

Some authorsRohatgi, Vijay K., and AK Md Ehsanes Saleh. An introduction to probability and statistics. John Wiley & Sons, 2015.Khan, RA (1994). A note on the generating function of a negative hypergeometric distribution. Sankhya: The Indian Journal of Statistics B, 56(3), 309-313. define the negative hypergeometric distribution to be the number of draws required to get the $r$th failure. If we let $Y$ denote this number then it is clear that $Y=X+r$ where $X$ is as defined above. Hence the PMF

:$\backslash Pr(Y=y)=\backslash binom\{y-1\}\{r-1\}\backslash frac\{\backslash binom\{N-y\}\{N-K-r\}\}\{\backslash binom\{N\}\{N-K\}\}.$

If we let the number of failures $N-K$ be denoted by $M$ means that we have

:$\backslash Pr(Y=y)=\backslash binom\{y-1\}\{r-1\}\backslash frac\{\backslash binom\{N-y\}\{M-r\}\}\{\backslash binom\{N\}\{M\}\}.$

The support of $Y$ is the set $\backslash \{r,r+1,\backslash dots,N-M+r\backslash \}$. It is clear that:

:$E[Y]=E[X]+r=\; \backslash frac\{r(N+1)\}\{M+1\}$

and $\backslash textrm\{Var\}[X]=\backslash textrm\{Var\}[Y]$.

{{reflist}}

{{ProbDistributions|discrete-finite}}

Category:Discrete distributions

Category:Factorial and binomial topics