As $N\to\infty$ Hypergeometric distribution reduces to Poisson distribution

Question

Let $X_N\sim Hg(N,\lambda N^2,N^3),\quad N=1,2,\ldots$

$$P(X_N=m)=\frac{\binom{\lambda N^2}{m}\binom{N^3}{N-m}}{\binom{\lambda N^2+N^3}{N}}$$

Now I have to show that for fixed $m=0,1,\ldots,$

$$P(X_N=m)=\frac{\lambda^m}{m!}\exp[-\lambda]$$ as $N\to\infty$

Answer 1

The case $m=0$ holds:

By definition, $$ P(X_N=0)=\frac{(N^3)!\cdot(N^3+\lambda N^2-N)!}{(N^3-N)!\cdot(N^3+\lambda N^2)!}. $$ The argument of each factorial goes to infinity and, when $n\to\infty$, Stirling's formula says that $$ n!\sim\sqrt{2\pi n}\frac{n^n}{\mathrm e^n}. $$ In the ratio of the Stirling's equivalents of the four factorials we are interested in, the $\sqrt{2\pi}$ factors cancel out, as well as the $\mathrm e^n$ factors. Each of the $\sqrt{n}$ factors is equivalent to $\sqrt{N^3}$ hence they also cancel out. We are left with the ratios of four powers $(N^3+N^3a_N)^{N^3+N^3a_N}$, where $a_N\to0$. Each of these is $(N^3)^{N^3+N^3a_N}\cdot(1+a_N)^{N^3+N^3a_N}$ hence the four powers $(N^3)^{N^3+N^3a_N}$ cancel out. All this leads to $$ P(X_N=0)\sim\frac{(1+\lambda/N-1/N^2)^{N^3+\lambda N^2-N}}{(1-1/N^2)^{N^3-N}\cdot(1+\lambda/N)^{N^3+\lambda N^2}}. $$ Equivalently, $$ P(X_N=0)\sim(1+\varepsilon_N)^{N^3+\lambda N^2-N}\cdot\frac{(1-1/N^2)^{\lambda N^2}}{(1+\lambda/N)^{N}}, $$ where $$ \varepsilon_N=\frac{1+\lambda/N-1/N^2}{(1-1/N^2)\cdot(1+\lambda/N)}-1=\frac{\lambda/N^3}{(1-1/N^2)\cdot(1+\lambda/N)}\sim\frac{\lambda}{N^3}. $$ Using thrice the asymptotics $\left(1+x_n\right)^{y_n}\to\mathrm e^{z}$ for every $x_n\to0$ and $y_n\to\infty$ such that $x_ny_n\to z$, one sees that $$ P(X_N=0)\to\mathrm e^\lambda\cdot\frac{\mathrm e^{-\lambda}}{\mathrm e^\lambda}=\mathrm e^{-\lambda}. $$ The case $m=0$ implies every case $m\geqslant0$:

For every $m\geqslant0$, $$ P(X_N=m)=P(X_N=0)\cdot\frac1{m!}\cdot\frac{(\lambda N^2)!\cdot N!\cdot(N^3-N)!}{(\lambda N^2-m)!\cdot(N-m)!\cdot(N^3-N+m)!}. $$ Using the fact that, when $n\to\infty$, $$ \frac{n!}{(n\pm m)!}\sim \frac1{n^{\pm m}}, $$ one gets $$ P(X_N=m)\sim P(X_N=0)\cdot\frac1{m!}\cdot\frac{(\lambda N^2)^m\cdot N^m}{(N^3)^m}= P(X_N=0)\cdot\frac{\lambda^m}{m!}, $$ which completes the proof.