25
$\begingroup$

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.

$\endgroup$
1

1 Answer 1

30
$\begingroup$

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 $$ \lim_{N \to \infty} \frac{(N-r-(n-x)+m)}{(N-n+m) } = \lim_{N \to \infty} \frac{N-r}{N} = 1-p $$

$\endgroup$
5
  • 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, 2013 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, 2013 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, 2013 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, 2013 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, 2013 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.