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I'm am currently trying to show that the hypergeometric distribution converges to the Poisson distribution.

$$ \lim_{n,r,s \to \infty, \frac{n \cdot r}{r+s} \to \lambda} \frac{\binom{r}{k} \binom{s}{n-k}}{\binom{r+s}{n}} = \frac{\lambda^k}{k!}e^{-\lambda} $$

I know, how to show for specific values, that the hypergeometric distribution converges to the binomial distribution and from there we proved in our script that the binomial distribution converges to the Poisson distribution for specific values.

No the question is, can i show the approximation directly via the limit above? I came from the limit above to

$$ \lim_{n,r,s \to \infty, \frac{n \cdot r}{r+s} \to \lambda} \frac{\binom{r}{k} \binom{s}{n-k}}{\binom{r+s}{n}} = \cdots = \frac{\lambda^k}{k!}\frac{\frac{(r-1)!}{(r-k)!}\binom{s}{n-k}}{\binom{r+s-1}{n-1}}\left(\frac{1}{\lambda}\right)^{k-1} $$

But how to show that ? $$ \frac{\frac{(r-1)!}{(r-k)!}\binom{s}{n-k}}{\binom{r+s-1}{n-1}}\left(\frac{1}{\lambda}\right)^{k-1} = e^{-\lambda} $$

Davide Giraudo
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maskman10z
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1 Answers1

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This is the simplest proof I've been able to find.

Just by rearranging factorials, we can rewrite the hypergeometric probability function as
$$ \mathrm{Prob}(X=x) = \frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}} = \frac{1}{x!} \cdot \dfrac{M^{(x)} \, K^{(x)}}{N^{(x)}} \cdot \dfrac{(N-K)^{(M-x)}}{(N-x)^{(M-x)}}, $$ where $a^{(b)}$ is the falling power $a(a-1)\cdots(a-b+1)$. Since $x$ is fixed, \begin{align*} \dfrac{M^{(x)} \, K^{(x)}}{N^{(x)}} &= \prod_{j=0}^{x-1} \dfrac{(M-j) \cdot (K-j)}{(N-j)} \\ &= \prod_{j=0}^{x-1} \left( \dfrac{MK}{n} \right) \cdot \dfrac{(1-j/M) \cdot (1-j/K)}{(1-j/N)} \\ &= \left( \dfrac{MK}{N} \right) ^x \; \prod_{j=0}^{x-1} \dfrac{(1-j/M) \cdot (1-j/K)}{(1-j/N)}, \end{align*} which $\to \lambda^x$ as $N$, $K$ and $M$ $\to \infty$ with $\frac{MK}{N} = \lambda$.

Lets replace $N-x$, $K-x$ and $M-x$ by new variables $n$, $k$ and $m$ for simplicity. Since $x$ is fixed, as $N,K,M \to \infty$ with $KM/N \to \lambda$, so too $n,k,m \to \infty$ with $nk/m \to \lambda$. Next we write $$ A = \dfrac{(N-K)^{(M-x)}}{(N-x)^{(M-x)}} = \dfrac{(n-k)^{(m)} }{(n)^{(m)}} = \prod_{j=0}^{m-1} \left( \dfrac{n-j-k}{n-j} \right)= \prod_{j=0}^{m-1} \left( 1 - \dfrac{k}{n-j} \right)$$ and take logs: $$ \ln \, A = \sum_{j=0}^{m-1} \ln \left( 1 - \dfrac{k}{n-j} \right). $$ Since the bracketed quantity is an increasing function of $j$ we have $$ \sum_{j=0}^{m-1} \ln \left( 1 - \dfrac{k}{n} \right) \le \ln \, A \le \sum_{j=0}^{m-1} \ln \left( 1 - \dfrac{k}{n-m+1} \right), $$ or $$ m \, \ln \left( 1 - \dfrac{k}{n} \right) \le \ln \, A \le m \, \ln \left( 1 - \dfrac{k}{n-m+1} \right). $$ But $\ln (1-x) < -x$ for $0 < x < 1$, so $$ m \, \ln \left( 1 - \dfrac{k}{n} \right) \le \ln \, A < -m \, \left( \dfrac{k}{n-m+1} \right), $$ and dividing through by $km/n$ gives $$ \frac{n}{k} \, \ln \left( 1 - \dfrac{k}{n} \right) \le \dfrac{\ln \, A}{km/n} < - \, \left( \dfrac{n}{n-m+1} \right) = - \, \left( \dfrac{1}{1-m/n+1/n} \right). $$ Finally, we let $k$, $m$ and $n$ tend to infinity in such a way that $km/n \to \lambda$. Since both $k/n \to 0$ and $m/n \to 0$, both the left and right bounds $\to -1$. (The left bound follows from $\lim_{n \to \infty} (1-1/n)^n = e^{-1}$, which is a famous limit in calculus.) So by the Squeeze Theorem we have $\ln \, A \to -\lambda$, and thus $A \to e^{-\lambda}$. Putting all this together gives the result.

chri5
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