Given the identically distributed and independant random variables $X_1,X_2,\ldots\sim\operatorname{Po}(\lambda)$ and $S_n=X_1+\ldots+X_n$ show with induction that
$$\Pr[S_n=k]=\frac{(n\lambda)^k}{k!}e^{-n\lambda}.$$
So far for
$n=1$ via the definition of the density function for Poisson:$$\Pr[S_1=k]=\Pr[X_1=k]=\frac{e^{-\lambda}\lambda^k}{k!}$$
With $n=2$ and independency:
$$\Pr[X_1+X_2=k]=\sum\limits_{n=0}^k\Pr[X_1=n]\Pr[X_2=k-n]=\sum\limits_{n=0}^k\frac{\lambda^n}{n!}e^{-\lambda}\frac{\lambda^{k-n}}{(k-n)!}e^{-\lambda}$$$$=\frac{1}{k!}e^{-2\lambda}\sum\limits_{n=0}^k\binom{k}{n}\lambda^n\lambda^{k-n}=\frac{(2\lambda)^k}{k!}e^{-2\lambda}$$
I assume now that with $S_n\sim\operatorname{Po}(\lambda)$ every $X_i\sim\operatorname{Po}(\lambda/n)$ and every Poisson variable can obviously split up, but how can I prove this?
Best Answer
Assume that the sum of $m$ independent Poisson with parameter $\lambda$ is Poisson with parameter $m\lambda$. We show that the sum of $m+1$ independent Poisson with parameter $\lambda$ is Poisson parameter $(m+1)\lambda$.
You post basically does it: exactly the same reasoning that got you from $1$ to $2$ gets you from $m$ to $m+1$.
Using almost exactly your notation, we find that $$\Pr(S_{m+1}=k)=\sum\limits_{n=0}^k\Pr(S_m=n)\Pr(X_{m+1}=k-n)=\sum\limits_{n=0}^k\frac{(m\lambda)^n}{n!}e^{-m\lambda}\frac{\lambda^{k-n}}{(k-n)!}e^{-\lambda}.$$ It follows that $$\Pr(S_{m+1}=k)=\frac{1}{k!}e^{-(m+1)\lambda}\sum\limits_{n=0}^k\binom{k}{n}(m\lambda)^n\lambda^{k-n}=\frac{((m+1)\lambda)^k}{k!}e^{-(m+1)\lambda}$$
The last equation is by the Binomial Theorem. For $$\sum_{n=0}^k \binom{k}{n}x^ny^{k-n}=(x+y)^k.$$ We are putting $x=m\lambda$ and $y=\lambda$, so $x+y=(m+1)\lambda$.