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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.

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2 Answers 2

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

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ 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
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$X_N\sim\;hypergeom(N, r, x)\\ with\;x=0, 1, 2,..., n.$

When $N$ and $N-r$ are large and $n$ is small, $X_N$ has a binomial distribution because the draws are 'almost' independents.

Fixing $n$ and letting $r$ dependent of $N$ in order to $\frac{r}{N} \rightarrow p$, as $N \rightarrow \infty$, with $0 < p < 1$.

$P(X_N = x) = \frac{r!}{[x!](r-x)!}\frac{(N-r)!}{[(n-x)!](N-r-n+x)!}\frac{[n!](N-n)!}{N!}$

$=\binom{n}{x}\frac{r(r-1)...(r-k+1)(N-r)(N-r-1)...(N-r-n+k+1)}{N(N-1)...(N-n+1)}$

$=\binom{n}{x}\frac{\frac{r}{N}(\frac{r}{N}-\frac{1}{N})...(\frac{r}{N}-\frac{x-1}{N})(1-\frac{r}{N})(1-\frac{r+1}{N})...(1-\frac{r+n-x-1}{N})}{1(1-\frac{1}{N})...(1-\frac{n-1}{N})}$

Then, with $\frac{r}{N}\rightarrow p$ and fixed $n$.

$P(X_n = x)\xrightarrow{\text{$N\rightarrow\infty$}} \binom{n}{x}p^k(1-p)^{n-x},\;\;x=0, 1,...,\;n$

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