23
$\begingroup$

I have this problem on a textbook that doesn't have a solution. It is:

Let $$f(x)=\frac{\binom{r}{x} \binom{N-r}{n-x}}{\binom{N}{n}}\;,$$ and keep $p=\dfrac{r}{N}$ fixed. Prove that $$\lim_{N \rightarrow \infty} f(x)=\binom{n}{x} p^x (1-p)^{n-x}\;.$$

Although I can find lots of examples using the binomial to approximate the hypergeometric for very large values of $N$, I couldn't find a full proof of this online.

|cite|improve this question
$\endgroup$
21
$\begingroup$

Write the pmf of the hypergeometric distribution in terms of factorials: $$\begin{eqnarray} \frac{\binom{r}{x} \binom{N-r}{n-x}}{\binom{N}{n}} &=& \frac{r!}{\color\green{x!} \cdot (r-x)!} \frac{(N-r)!}{\color\green{(n-x)!} \cdot (N-n -(r-x))!} \cdot \frac{\color\green{n!} \cdot (N-n)!}{N!} \\ &=& \color\green{\binom{n}{x}} \cdot \frac{r!/(r-x)!}{N!/(N-x)!} \cdot \frac{(N-r)! \cdot (N-n)!}{(N-x)! \cdot (N-r-(n-x))!} \\ &=& \binom{n}{x} \cdot \frac{r!/(r-x)!}{N!/(N-x)!} \cdot \frac{(N-r)!/(N-r-(n-x))!}{(N-n+(n-x))!/(N-n)! } \\ &=& \binom{n}{x} \cdot \prod_{k=1}^x \frac{(r-x+k)}{(N-x+k)} \cdot \prod_{m=1}^{n-x}\frac{(N-r-(n-x)+m)}{(N-n+m) } \end{eqnarray} $$ Now taking the large $N$ limit for fixed $r/N$, $n$ and $x$ we get the binomial pmf, since $$ \lim_{N \to \infty} \frac{(r-x+k)}{(N-x+k)} = \lim_{N \to \infty} \frac{r}{N} = p $$ and $$ \lim_{N \to \infty} \frac{(N-r-(n-x)+m)}{(N-n+m) } = \lim_{N \to \infty} \frac{N-r}{N} = 1-p $$

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ Alternatively, one can look at the ratio of hypergeometric PMF for $X=x$ and for $X=x+1$ which equals $\frac{(n-x)(r-x)}{(x+1)(1+N-n-r+x}$. In the limit the ratio becomes $\frac{n-x}{x+1} \frac{p}{1-x}$. Compounded with $\mathbb{P}(X=0) = \frac{\binom{N-r}{n}}{\binom{N}{n}} \to (1-p)^n$ we recover the pmf of the limiting distribution as $(1-p)^n \prod_{m=1}^x \frac{n-m}{m+1} \frac{p}{1-p} = \binom{n}{x} p^x (1-p)^{n-x}$ $\endgroup$ – user40314 Mar 14 '13 at 19:41
  • $\begingroup$ @Sasha, Sorry, I didn't get the part about $$\lim_{x\rightarrow \infty} \frac{(r-x+k)}{(N-x+k)}$$. How does knowing that help anything? I'm not familiar at all with computing big pi expressions... $\endgroup$ – ithisa Mar 17 '13 at 18:35
  • 1
    $\begingroup$ @EricDong Remember that $0\leq k \leq x$ and these remain fixed as $N$ increases. On the other hand $r$ increases with $N$. Thus $$ \lim_{N \to \infty} \frac{r-x+k}{N-x+k} = \lim_{N \to \infty} \frac{\frac{r}{N}+\frac{k-x}{N} }{1+\frac{k-x}{N}}$$ Since $k-x$ is fixed, $\frac{k-x}{N}$ will become smaller as $N$ grows and equals zero in the limit, therefore $$ \lim_{N \to \infty} \frac{r-x+k}{N-x+k} = \lim_{N \to \infty} \frac{r}{N}$$ The latter ratio equals $p$ as given in the problem. $\endgroup$ – Sasha Mar 17 '13 at 18:40
  • 1
    $\begingroup$ I do understand why the expression equals what it does, but how does that help us evaluate the big pi expression above? $\endgroup$ – ithisa Mar 17 '13 at 22:42
  • 1
    $\begingroup$ It does becaused $$\lim_{N \to \infty} \prod_{k=1}^x \frac{r-x+k}{N-x+k} = \prod_{k=1}^x \lim_{N \to \infty} \frac{r-x+k}{N-x+k} = \prod_{k=1}^x p = p^x$$ Likewise $$ \lim_{N \to \infty} \prod_{m=1}^{n-x}\frac{(N-r-(n-x)+m)}{(N-n+m) } = \prod_{m=1}^{n-x} (1-p) = (1-p)^{n-x}$$ $\endgroup$ – Sasha Mar 18 '13 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.