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As application of convergence theorem in our probability lecture we want to show the generating function of sequence of binomially distributed random variables converges to the generating function of the Poisson distribution (poisson limit theorem):

Let $(p_n)_{n \in \mathbb{N}_0}$ be a sequence of success parameters such that $n p_n \xrightarrow{n \to \infty} \lambda > 0$ and $(G_n(s) := (s p_n + 1 - p_n )^n )_{n \in \mathbb{N}_0}$ the generating functions of the binomial distribution Bin$(n,p)$. Let $\lambda_n := n p_n$. Then we have \begin{equation*} G_n(s) = (s p_n + 1 - p_n )^n = \left( 1 - (1 - s) \frac{\lambda_n}{n} \right)^n \xrightarrow{n \to \infty} e^{(s - 1)\lambda}, \end{equation*} because $\left(1 - (1 - s)\frac{\lambda}{n}\right)^n \xrightarrow{n \to \infty} e^{(s - 1)\lambda}$ for all $\lambda \in \mathbb{R}$ and

\begin{equation*} \lim_{n \to \infty} \left| \left( 1 - (1 - s) \frac{\lambda_n}{n}\right)^n - \left( 1 - (1 - s) \frac{\lambda}{n}\right)^n \right| \le \lim_{n \to \infty} | 1 - s | | \lambda_n - \lambda | = 0. \end{equation*} How does one see the last inequality? Does it hold without the limit, too?

I tried to use the binomial theorem and obtained \begin{align*} \left| \left( 1 - (1 - s) \frac{\lambda_n}{n}\right)^n - \left( 1 - (1 - s) \frac{\lambda}{n}\right)^n \right| & = \left| \sum_{k = 0}^{n} \binom{n}{k} \left( (s - 1) \frac{\lambda_n}{n} \right)^k - \sum_{k = 0}^{n} \binom{n}{k} \left( (s - 1) \frac{\lambda}{n} \right)^k \right| \\ & \overset{\triangle \ne}{\le} \sum_{k = 0}^{n} \binom{n}{k} \left|\frac{s - 1}{n}\right|^k \left| \lambda_n^k - \lambda^k \right| \\ & \le \sum_{k = 0}^{n} \binom{n}{k} \left|\frac{s - 1}{n}\right|^k \left| \lambda_n - \lambda \right|^k \\ & = \left( 1+ \frac{|1 - s||\lambda_n - \lambda|}{n} \right)^n. \end{align*} Another approach could be \begin{align*} \left| \left(1 - (1 - s) \frac{\lambda_n}{n}\right)^n - \left( 1 - (1 - s) \frac{\lambda}{n}\right)^n \right| & \overset{\triangle \ne}{\le} \left| 1 - (1 - s) \frac{\lambda_n}{n} \right|^n + \left| 1 - (1 - s) \frac{\lambda}{n} \right|^n \\ & \le e^{1 - s}\left( e^{\lambda_n} + e^{\lambda} \right) \end{align*} since the limit $\left(1 + \frac{x}{n}\right)^n \xrightarrow{n \to \infty} e^x$ is monotone.

Another approach could be the use the mean value theorem for $f(x) := \left(1 - (1 - s) \frac{x}{n} \right)^n$. Then we know that there exists a $\xi \in (\lambda_n, \lambda)$ such that \begin{align*} \left|\left(1 - (1 - s) \frac{\lambda_n}{n}\right)^n - \left( 1 - (1 - s) \frac{\lambda}{n}\right)^n \right| & = | f(\lambda_n) - f(\lambda) | \overset{\textrm{MVT}}{=} | f'(\xi) | | \lambda_n - \lambda | \\ & = | 1 - s| \left( 1 - \frac{(1 - s) \xi}{n} \right)^{n - 1} | \lambda_n - \lambda | \end{align*} Therefore we have $$ \lim_{n \to \infty} \left|\left(1 - (1 - s) \frac{\lambda_n}{n}\right)^n - \left( 1 - (1 - s) \frac{\lambda}{n}\right)^n \right| \le \left(\lim_{n \to \infty} | 1 - s| | \lambda_n - \lambda |\right) \cdot \left(\lim_{n \to \infty} \left( 1 - \frac{(1 - s) \xi}{n} \right)^{n - 1}\right) = \left(\lim_{n \to \infty} | 1 - s| | \lambda_n - \lambda |\right) \cdot \exp(-(1 - s) \lambda), $$ because $\lambda_n \to \lambda$ as $n \to \infty$ implying $\xi \to \lambda$ and the exponential function is continuous (does it really work that way or is this just a different formulation for the above problem?)

Probably related question.

Here's an answer how this can be shown a different way.

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The inequality holds without the limit at least for large enough $n$ and it is quite easy to prove. Let $a=1-(1-s)\frac {\lambda_n} n$ and $b=1-(1-s)\frac {\lambda} n$. For large $n$ and $s \in [0,1]$ note that $a,b \in [0,1]$. Hence $|a^{n}-b^{n}|=|a-b| |a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}| \leq n|a-b|$ and this gives the inequality.

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  • $\begingroup$ Do you see how any of my approaches could be brought to fruition? $\endgroup$
    – Ramanujan
    Jun 16 '19 at 14:16

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