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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?

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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 $x+y=(m+1)\lambda$.

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  • $\begingroup$ Oh yep, so close... but it works! $\endgroup$ – Christian Ivicevic Jun 25 '12 at 14:43
  • $\begingroup$ Using characteristic functions is also an option, since they are independent the char of the sum is just the multiplication of the chars, which are equal. $\endgroup$ – Jorge E. Cardona Nov 28 '13 at 18:04
  • $\begingroup$ @JorgeEduardoCardona: That would indeed be a very good way. From the wording of the question, it is likely not to be the intended way. Another way which I prefer is to show that the sum of independent Poisson satisfies the "axioms" under which one concludes that a random variable is Poisson. $\endgroup$ – André Nicolas Nov 28 '13 at 18:11

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