I have problems with this probability exercise

Let $f : \mathbb{R} \to \mathbb{R}$ a continuous and bounded function. Prove that for every $\lambda > 0$

$$\lim_{n \to \infty}\sum_{k=1}^{\infty} f \left(\frac{k}{n} \right)\frac{(n\lambda)^k}{k!}e^{-n\lambda}=f(\lambda)$$

I think I could prove this with epsilon-delta definition. Then, I tried the next but I am not sure.

\begin{align*} \left|\sum_{k=1}^{\infty} f \left(\frac{k}{n} \right)\frac{(n\lambda)^k}{k!}e^{-n\lambda}-f(\lambda)\right| &= \left|\sum_{k=1}^{\infty} f \left(\frac{k}{n} \right)\frac{(n\lambda)^k}{k!}e^{-n\lambda}-f(\lambda)\sum_{k=1}^{\infty}\frac{(n\lambda)^k}{k!}e^{-n\lambda}\right|\\ &=\left|\sum_{k=1}^{\infty} \left(f \left(\frac{k}{n} \right)-f(\lambda)\right) \frac{(n\lambda)^k}{k!}e^{-n\lambda}\right|\\ \end{align*}

I guess the next step is to use the fact that function is continuous and bounded but I can argue exactly why. Any help?


A probabilistic proof of the result is probably easier than using the epsilon-delta definition.

First, observe that $$\mathbb{E}\left[f\left(\frac{N_{n\lambda}}{n}\right)\right] = \sum_{k = 0}^\infty f\left(\frac{k}{n}\right)\frac{(n\lambda)^k}{k!}e^{-n\lambda} = \sum_{k = 1}^\infty f\left(\frac{k}{n}\right)\frac{(n\lambda)^k}{k!}e^{-n\lambda} + e^{-n \lambda},$$ where $N_{n\lambda}$ is a Poisson random variable with parameter $n\lambda $. Thus the quantity inside the limit is $\mathbb{E}\left[f\left(\frac{N_{n\lambda}}{n}\right)\right] - e^{-n\lambda}$.

It is a well know fact about the sum of Poisson random variables that if $P_1, \dots, P_N$ are independent Poisson random variables with parameters $\lambda_1, \dots, \lambda_N$, then their sum is also Poisson with parameter $\sum_{i =1}^N \lambda_i$. Therefore, $N_{n\lambda} \overset{law}{=} \sum_{i =1}^N N_{\lambda}^i$ with the $N_{\lambda}^i$'s being independent. As a consequence we have, by the strong law of large numbers,
$$ \frac{N_{n\lambda}}{n} \to \mathbb{E}\left[N_\lambda\right] = \lambda \quad \text{almost surely}.$$ Then by dominated convergence we have that since $f$ is bounded $$\lim_{n\to \infty}\mathbb{E}\left[f\left(\frac{N_{n\lambda}}{n}\right)\right] = f\left(\lambda\right).$$

Injecting this result into the limit we want to solve gives that $$\lim_{n \to \infty}\sum_{k=1}^{\infty} f \left(\frac{k}{n} \right)\frac{(n\lambda)^k}{k!}e^{-n\lambda} = \lim_{n\to \infty} \left[\mathbb{E}\left[f\left(\frac{N_{n\lambda}}{n}\right)\right] - e^{-n\lambda}\right] = f\left(\lambda\right).$$

  • $\begingroup$ Wow! I see clearer now. Only one thing... we can suppose that Poisson r.v. are independent? $\endgroup$ Jan 14 at 23:51
  • 1
    $\begingroup$ They are independent by construction as we had chosen them to be independent $\endgroup$
    – Shiva
    Jan 15 at 0:26
  • $\begingroup$ Yes, you right! Thanks! $\endgroup$ Jan 15 at 0:46

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