I try to understand the proof of the poisson aproximation theorem: Let $\alpha>0$ and $p_n >0$ with $\lim_{n \to \infty} n*p_n=\alpha$. Then for all $k \in \mathbb{N}$
$$\lim_{n \to \infty} \text{Bin}_{n,p_n}(k)=\text{Poi}_{\alpha}(k).$$ (where Bin is the binomial distribution and Poi is the Poisson distribution with the respective parameters)
Proof: For $n \geq k$
(where $a_n$ ~ $b_n$ means $\lim_{n \to \infty} \frac{a_n}{b_n}=1$). The right side then converges to $\text{Poi}_{\alpha}(k).$
I do not understand this equality:
I get that $(1-p_n)= (1-\frac{\alpha}{n}+\mathcal{o}(\frac{1}{n}))$ since $\frac{(1-p_n-1+\frac{\alpha}{n})}{\frac{1}{n}}$ converges to $(-\alpha+\alpha)=0$. From this we can conclude that $(1-p_n)^{n-k}=(1-\frac{\alpha}{n}+\mathcal{o}(\frac{1}{n}))^{n-k} $ but the exponential on the right side is now $n$ and not $n-k$. Which argument allows this?
Best Answer
Let me give you a different but faster proof. Invoking Lévy-Cramér theorem we know that
$$F_{X_n}\xrightarrow{\mathcal{L}}F_X \leftrightarrow M_{X_n}(t) \rightarrow M_X(t)$$
Where $M_X(t)$ is the Moment Generating Function.
Thus if we take $X_n\sim Bin (n;p)$ we have that
$$M_{X_n}(t)=[1-p+pe^t]^n$$
now let's set $p=\frac{\theta}{n}$ obtaining
$$M_{X_n}(t)=\left[1-\frac{\theta}{n}+\frac{\theta}{n}e^t\right]^n=\left[1+\frac{\theta(e^t-1)}{n}\right]^n$$
Thus
$$\lim\limits_{n\to +\infty}M_{X_n}(t)=e^{\theta(e^t-1)}$$
And immediately we recognize in this limit result the MGF of a Poisson distribution with parameter $\theta$
as per your question is concerned:
Just substitute $p_n=\frac{\alpha}{n}$ and consider $n$ large
In fact
$$\left(1-\frac{\theta}{n}\right)^{n-k}=\left(1+\frac{-\theta}{n}\right)^{n}\cdot\left(1-\frac{\theta}{n}\right)^{-k}$$
and passing to the limit you get that
$$\left(1-\frac{\theta}{n}\right)^{n-k}=\underbrace{\left(1+\frac{-\theta}{n}\right)^{n}}_{\to e^{-\theta}}\cdot\underbrace{\left(1-\frac{\theta}{n}\right)^{-k}}_{\to 1}$$