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(Feller Volume 1 p.194 Q.10) Normal approximation to the hypergeometric distribution. Let $n, m , k$ be positive integers and suppose that they tend to infinity in such a way that $\frac{r}{n+m} \to t, \frac{n}{n+m} \to p, \frac{m}{n+m} \to q, h(k-rp) \to x$ where $h = 1/\sqrt{(n+m)pqt(1-t)}$. Prove that $${n \choose k}{m \choose r-k}/{n+m \choose r} \sim h\phi(x).$$ Hint: Use the normal approximation to the binomial distribution rather than Stirling's formula.

$\phi(x)$ is a normal pdf and $\sim$ denotes that the ratio converges to 1 for large sample size.

I know that the hypergeometric distribution can be approximated by the binomial distribution (for example, see here). Also, by the Demoivre-Laplace theorem, the binomial distribution can be approximated by the normal distribution. However, I am struggling to prove this question using the conditions given. Can you give me some hint?

shk910
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1 Answers1

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Hint: One version of the normal approximation to the binomial is: Let $p_n+q_n=1$ with $p_n\to p$ and $q_n\to q$. If $n$ and $k$ tend to infinity in such a way that $(k-np_n)/\sqrt{np_nq_n}\to y$, then $$ {n\choose k}p_n^kq_n^{n-k}\sim \frac1{\sqrt{npq}}\phi(y)\tag N $$ as $n\to\infty$. To apply this to the hypergeometric, the trick is to write $$ \frac{\displaystyle{n\choose k}{m\choose r-k}}{\displaystyle{n+m\choose r}}= \frac{\displaystyle{r\choose k}{n+m-r\choose n-k}}{\displaystyle{n+m\choose n}}= \frac{\displaystyle{r\choose k}p_n^kq_n^{r-k}{n+m-r\choose n-k}p_n^{n-k}q_n^{m+k-r}}{\displaystyle{n+m\choose n}p_n^nq_n^m} $$ where $p_n:=\frac n{n+m}$, $q_n:=\frac m{n+m}$. Now apply the normal approximation (N) three times to the RHS.

grand_chat
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