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I try to solve the following:

Let $X_1,X_2,\dots$ be independant random variables. We set $S_k=\sum\limits_{i=1}^kX_i$ for $k\geq 1$.

a) For $i\geq 1$ the random variables $X_i$ have a Poisson distribution with parameter $\lambda_i> 0$. What is the law of $S_k$

b) For $i\geq 1$, let $X_i$ have the binomial distribution with parameter $(n_i,p)$, where $n_i\in\mathbb{N}$ and $p\in[0,1]$. What is the law of $S_k$?


What I thought:

a) $$P(X_i=n)=e^{-\lambda_i}\frac{\lambda_i^n}{n!}$$

Since $X_1,X_2,\dots$ are independant we have: $$P(X_1=n,\dots,X_k=n)=P(X_1=n)\cdots P(X_k=n)=e^{-\lambda_1\dots-\lambda_k}\frac{(\lambda_1 \cdots \lambda_k)^n}{(n!)^k}$$

So the distribution is $$P(S_k\leq n)=\sum\limits_{l=0}^n e^{-\lambda_1\dots-\lambda_k}\frac{(\lambda_1 \cdots \lambda_k)^n}{(l!)^k}$$

b) $$P(X_i=l)=\binom{n_i}{l}p^l(1-p)^{n_i-l}$$

Since $X_1,X_2,\dots$ are independant we have: $$P(X_1=l,\dots,X_k=l)=P(X_1=l)\cdots P(X_k=l)=P(X_i=l)=\prod_{i=1}^k\binom{n_i}{l}p^l(1-p)^{n_i-l}$$

So the distribution is: $$P(S_k\leq n)=\sum\limits_{l=0}^n\prod_{i=1}^k\binom{n_i}{l}p^l(1-p)^{n_i-l}$$

But unfortunately I have a feeling that this is not correct.

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  • $\begingroup$ The equations you have written are correct, but they do not address the distribution of the sum $S_k$. $\endgroup$ – André Nicolas Mar 24 '16 at 21:02
  • $\begingroup$ @AndréNicolas Is it now better? I really don't know how to solve this task. $\endgroup$ – Matriz Mar 24 '16 at 21:26
  • $\begingroup$ Do you know already that the sum of two independent Poisson is Poisson? Then the result for $k$ Poisson follows fairly quickly. As to the binomial, there is an informal way and a more formal way. If $X_1$ to $X_k$ are the numbers of successes in $n_1$ to $n_k$ trials, then $S_k$ is the number of successes in $n_1+\cdots+n_k$ trials. $\endgroup$ – André Nicolas Mar 24 '16 at 21:30
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First, in Poisson case, $S_k \sim Pois(\sum_{i=1}^k \lambda_i)$ and in the binomial case, $S_k \sim Binom(\sum_{i=1}^k n_i, p).$

By intuition. Intuitively, these are sensible answers.

In the Poisson case, suppose you have $k$ radioactive sources, individually emitting particles at rates $\lambda_1, \lambda_2, \dots, \lambda_k.$ The number of particles $X_i$ emitted by each source is well modeled by a Poisson distribution, so that $X_i \sim Pois(\lambda_k).$ Then if we put all of the sources together, in combination it makes sense that the number of particles $S_k$ emitted by all sources will be Poisson with the sum of the rates of the individual sources: $S_k \sim Pois(\sum_{i=1}^k \lambda_i).$

In the binomial case, suppose the $i$th of $k$ students rolls a fair die $n_i$ times and counts the number of $X_i$ of 'sixes' seen. Then the distribution for each student will be $X_i \sim Binom(n_i, p = 1/6).$ Altogether, the $k$ students will have rolled a die $\sum_{i=1}^k$ times and the total number of 'sixes' seen will be $S_k \sim binom($\sum_{i=1}^k n_i, p = 1/6).$

Formally with MGFs. I don't know if you are familiar with moment generating functions. Both results follow from MGFs.

Poisson: The MGF for $X_i \sim Pois(\lambda_i)$ is $M_{X_i}(t) = \exp(\lambda_i(e^t - 1)).$ The MGF of the sum $S_k = \sum_{i=1}^k X_i$ is the product of the MGFs $M_{S_k}(t) = \prod_{i=1}^k M_{X_i}(t) = \exp(\Lambda(e^t - 1)),$ where $\Lambda = \sum_{i=1}^k \lambda_i.$ We recognize $M_{S_k}$ as the MGF of $Pois(\Lambda).$

Binomial: The MGF for $X_i \sim Binom(n_i, p)$ is $M_{X_i}(t) = (pe^t + (1-p))^{n_i},$ and the argument for the MGF of $S_k$ proceeds similarly.

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I would suggest that the easiest way is to look at the case when $k=2$ and you can generalize it to the case when for $k=n$.

So, in general, If $X_i \sim Poisson(\lambda_i)$, then $X_1+\cdot\cdot\cdot+X_k=Poisson(\lambda_1+\cdot\cdot\cdot+\lambda_k)$

So it would instead be $$P(S_k=n)=e^{-(\lambda_1+\cdot\cdot\cdot+\lambda_k)}{(\lambda_1+\cdot\cdot\cdot+\lambda_k)}^n/n!$$

To start this proof, I would look at $$P(X_1+X_2=n)=\sum_{m=0}^nP(X_1=m)P(X_2=n-m)$$

For the binomial distribution,

$$X_1+\cdot\cdot\cdot+X_k \sim Bin(n_1+\cdot\cdot\cdot+n_k,p)$$

This can also be generalized from the case when $k=2$.

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