Assume those $N$ Bernoulli random variables are i.i.d. with probability $p$, where $N \sim \operatorname{Poisson}(\lambda)$. The finite sum of them is Poisson distributed with $p\lambda$. I read a book claims that but without proof.
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$\begingroup$ Alternatively to the MGF proof, you can calculate the distribution directly from the law of total probability: $P(Y=j)=\sum_{n=0}^\infty P(Y=j|N=n)P(N=n)$, where $Y=\sum_1^N X_i$, $N$ is Poisson $\lambda$ and $X_i$ are iidrv Bernoulli. All one must know is the conditional distribution—which should be an extremely familiar distribution of a discrete RV related to sums of Bernoulli variables... $\endgroup$– Nap D. LoverOct 30, 2017 at 14:53
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2 Answers
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Here's a pedestrian way: Let $j\in \mathbb N$, $$\begin{align} P\left( \sum_{k=1}^N X_k = j \right) &= \sum_{l=j}^\infty P\left((N=l)\;\cap \left(\sum_{k=1}^l X_k = j\right)\right)\\ &=\sum_{l=j}^\infty P\left(N=l\right)P\left(\sum_{k=1}^l X_k = j\right) \quad \text{by independence of $N$ and $X_k$}\\ &=\sum_{l=j}^\infty e^{-\lambda}\frac{\lambda^l}{l!} \binom lj p^j(1-p)^{l-j}\\ &=\frac{e^{-\lambda}p^j\lambda^j}{j!} \sum_{l=j}^\infty \frac{\lambda^{l-j}(1-p)^{l-j}}{(l-j)!}\\ &= \frac{e^{-\lambda p}(\lambda p)^j}{j!} \end{align}$$
Hence $\sum_{k=1}^N X_k$ follows a Poisson distribution with parameter $\lambda p$.
answered Oct 30, 2017 at 12:19
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$\begingroup$ Could you please provide explanation to the last rewrite of equation? Where did the summation go? And how did u get p to the exponent of lambda? $\endgroup$– Smarty77Apr 28, 2018 at 11:29
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1$\begingroup$ @Smarty77 $$\sum_{l=j}^\infty \frac{\lambda^{l-j}(1-p)^{l-j}}{(l-j)!} = \sum_{l=0}^\infty \frac{\lambda^l(1-p)^l}{l!} = e^{\lambda (1-p)} $$ $\endgroup$ Apr 28, 2018 at 11:37
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$\begingroup$ ah, the Taylor Series! Thank you! And can we rewrite any j to 0 like that, because it is constant and it gets lost in infinite series, or is there some other reason? $\endgroup$– Smarty77Apr 28, 2018 at 11:47
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1$\begingroup$ @Smarty77 To be precise this is more related to the power series expansion of $\exp$: $e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$. I just reindexed the sum. If you wanted to be formal you could write $$\sum_{l=j}^\infty \frac{\lambda^{l-j}(1-p)^{l-j}}{(l-j)!} = \lim_{N\to \infty} \sum_{l=j}^{N+j} \frac{\lambda^{l-j}(1-p)^{l-j}}{(l-j)!} = \lim_{N\to \infty} \sum_{l=0}^N \frac{\lambda^l(1-p)^l}{l!}=\sum_{l=0}^\infty \frac{\lambda^l(1-p)^l}{l!}$$ where the reindexing is done on finite sums. $\endgroup$ Apr 28, 2018 at 12:03
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You can prove by showing that the moment generating function of $\sum_{i=1}^{N}{X_i}$ is that of a Poisson. \begin{align*} E\left[\exp\left(t \sum_{i=1}^{N}{X_i}\right)\right] &= E\left[E\left[\exp\left(t \sum_{i=1}^{N}{X_i}\right)\bigg|N\right]\right] \\ &= E\left[\prod_{i=1}^{N}E[\exp(t X_i)]\right] \\ &= E\left[(p\exp(t)+(1-p))^{N}\right] \\ &= \sum_{i=0}^{\infty} \frac{(p\exp(t)+(1-p))^{i} \exp(-\lambda)\lambda^{i}}{i!} \\ &= \exp(-\lambda) \sum_{i=0}^{\infty}\frac{(\lambda(p\exp(t)+(1-p)))^{i}}{i!} \\ &= \exp(\lambda(p\exp(t)+(1-p)-1)) \\ &= \exp(\lambda p(\exp(t)-1)) \end{align*} Note that $\exp(\lambda p(\exp(t)-1))$ is the moment generation function of a Poisson($\lambda p$).
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3$\begingroup$ This works only if you have a theorem that says a distribution with the same moment-generating function as a Poisson distribution has a Poisson distribution. $\endgroup$ Oct 30, 2017 at 16:15