Chapter 3, Problem 212SE

To determine

Show that $\underset{N\to \infty }{\lim }\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}=\left(\begin{array}{l}n\\ y\end{array}\right)p^y{q}^{n-y}$.

Explanation of Solution

Calculation:

Consider the expression $\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}$.

$\begin{array}{c}\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}=\frac{\left(\frac{r!}{y!\left(r-y\right)!}\right)\left(\frac{\left(N-r\right)!}{\left(n-y\right)!\left(N-r-\left(n-y\right)\right)!}\right)}{\left(\frac{N!}{n!\left(N-n\right)!}\right)}\\ =\left(\frac{r!}{y!\left(r-y\right)!}\right)\left(\frac{\left(N-r\right)!}{\left(n-y\right)!\left(N-r-\left(n-y\right)\right)!}\right)\left(\frac{n!\left(N-n\right)!}{N!}\right)\\ =\left(\frac{n!}{y!\left(n-y\right)!}\right)\left(\frac{r!}{\left(r-y\right)!}\right)\left(\frac{\left(N-r\right)!\left(N-n\right)!}{N!\left(N-r-\left(n-y\right)\right)!}\right)\\ =\left(\begin{array}{l}n\\ y\end{array}\right)\left(\frac{r!}{\left(r-y\right)!}\right)\left(\frac{\left(N-r\right)!\left(N-n\right)!}{N!\left(N-r-\left(n-y\right)\right)!}\right)\left(\frac{\left(N-y\right)!}{\left(N-y\right)!}\right)\end{array}$

                    $\begin{array}{l}=\left(\begin{array}{l}n\\ y\end{array}\right)\times \frac{\left(\frac{r!}{\left(r-y\right)!}\right)}{\left(\frac{N!}{\left(N-y\right)!}\right)}\times \left(\frac{\left(N-r\right)!\left(N-n\right)!}{\left(N-y\right)!\left(N-r-\left(n-y\right)\right)!}\right)\\ =\left(\begin{array}{l}n\\ y\end{array}\right)\times \frac{\left(\frac{r!}{\left(r-y\right)!}\right)}{\left(\frac{N!}{\left(N-y\right)!}\right)}\times \left(\frac{\left(N-r\right)!\left(N-n\right)!}{\left(N-n+\left(n-y\right)\right)!\left(N-r-\left(n-y\right)\right)!}\right)\\ =\left(\begin{array}{l}n\\ y\end{array}\right)\times \frac{\left(\frac{r!}{\left(r-y\right)!}\right)}{\left(\frac{N!}{\left(N-y\right)!}\right)}\times \frac{\left(\frac{\left(N-r\right)!}{\left(N-r-\left(n-y\right)\right)!}\right)}{\left(\frac{\left(N-n\right)!}{\left(N-n-\left(n-y\right)\right)!}\right)}\end{array}$

                    $\begin{array}{l}=\left[\begin{array}{l}\left(\begin{array}{l}n\\ y\end{array}\right)\times \frac{\left(\frac{r\left(r-1\right)\left(r-2\right)\mathrm{...}\left(r-y+1\right)\left(r-y\right)!}{\left(r-y\right)!}\right)}{\left(\frac{N\left(N-1\right)\left(N-2\right)\mathrm{...}\left(N-y+1\right)\left(N-y\right)!}{\left(N-y\right)!}\right)}\times \\ \frac{\left(\frac{\left(N-r\right)\left(N-r-1\right)\mathrm{...}\left(N-r-\left(n-y\right)+1\right)\left(N-r-\left(n-y\right)\right)!}{\left(N-r-\left(n-y\right)\right)!}\right)}{\left(\frac{\left(N-n\right)\left(N-n-1\right)\mathrm{...}\left(N-n-\left(n-y\right)+1\right)\left(N-n-\left(n-y\right)\right)!}{\left(N-n-\left(n-y\right)\right)!}\right)}\end{array}\right]\\ =\left[\begin{array}{l}\left(\begin{array}{l}n\\ y\end{array}\right)\times \left(\frac{r\left(r-1\right)\left(r-2\right)\mathrm{...}\left(r-y+1\right)}{N\left(N-1\right)\left(N-2\right)\mathrm{...}\left(N-y+1\right)}\right)\times \\ \left(\frac{\left(N-r\right)\left(N-r-1\right)\mathrm{...}\left(N-r-\left(n-y\right)+1\right)}{\left(N-n\right)\left(N-n-1\right)\mathrm{...}\left(N-n-\left(n-y\right)+1\right)}\right)\end{array}\right]\\ =\left(\begin{array}{l}n\\ y\end{array}\right)\times {\displaystyle \prod _{a=1}^y\frac{r-y+a}{N-y+a}\times }{\displaystyle \prod _{b=1}^{n-y}\frac{N-r-n+y+b}{N-n+b}}\end{array}$

For value of y being fixed and a belongs to 1,…,y, take the limit with $N\to \infty$ for the expression. Also, $p=\frac{r}{N}$ is a constant.

$\begin{array}{c}\underset{N\to \infty }{\lim }\frac{r-y+a}{N-y+a}=\underset{N\to \infty }{\lim }\frac{r-y+a}{N-y+a}\left(\frac{N}{N}\right)\\ =\underset{N\to \infty }{\lim }\frac{\left(\frac{r-y+a}{N}\right)}{\left(\frac{N-y+a}{N}\right)}\\ =\underset{N\to \infty }{\lim }\frac{\left(\frac{r}{N}-\frac{y}{N}+\frac{a}{N}\right)}{\left(\frac{N}{N}-\frac{y}{N}+\frac{a}{N}\right)}\\ =\frac{\left(\frac{r}{N}-0+0\right)}{\left(1-0+0\right)}\left(\text{Since }\frac{r}{N}\text{is constant}\right)\\ =\frac{r}{N}\\ =p\end{array}$

For value of y being fixed and b belongs to $1,\mathrm{...},\left(n-y\right)$, take the limit with $N\to \infty$ for the expression. If $p=\frac{r}{N}$ is a constant, then $q=1-p$ is also a constant.

$\begin{array}{c}\underset{N\to \infty }{\lim }\frac{N-r-n+y+b}{N-n+b}=\underset{N\to \infty }{\lim }\frac{N-r-n+y+b}{N-n+b}\left(\frac{N}{N}\right)\\ =\underset{N\to \infty }{\lim }\frac{\left(\frac{N-r-n+y+b}{N}\right)}{\left(\frac{N-n+b}{N}\right)}\\ =\underset{N\to \infty }{\lim }\frac{\left(\frac{N}{N}-\frac{r}{N}-\frac{n}{N}+\frac{y}{N}+\frac{b}{N}\right)}{\left(\frac{N}{N}-\frac{n}{N}+\frac{b}{N}\right)}\\ =\frac{\left(1-\frac{r}{N}-0+0+0\right)}{\left(1-0+0\right)}\left(\text{Since }\frac{r}{N}\text{is constant}\right)\\ =1-\frac{r}{N}\\ =q\end{array}$

The value for $\underset{N\to \infty }{\lim }\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}$ is,

$\begin{array}{c}\underset{N\to \infty }{\lim }\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}=\underset{N\to \infty }{\lim }\left(\begin{array}{l}n\\ y\end{array}\right)\times {\displaystyle \prod _{a=1}^y\frac{r-y+a}{N-y+a}\times }{\displaystyle \prod _{b=1}^{n-y}\frac{N-r-n+y+b}{N-n+b}}\\ =\underset{N\to \infty }{\lim }\left(\begin{array}{l}n\\ y\end{array}\right)\times \underset{N\to \infty }{\lim }{\displaystyle \prod _{a=1}^{y}p\times }\underset{N\to \infty }{\lim }{\displaystyle \prod _{b=1}^{n-y}q}\\ =\underset{N\to \infty }{\lim }\left(\begin{array}{l}n\\ y\end{array}\right)p^y{q}^{n-y}\end{array}$

Hence, it is showed that $\underset{N\to \infty }{\lim }\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}=\left(\begin{array}{l}n\\ y\end{array}\right)p^y{q}^{n-y}$ as $N\to \infty$.