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Consider a lattice $\mathbb{Z_+}$ and immagine that on each site $i \in \mathbb{Z}_+$ there is a number of particles $X_i$, where $X_i$ are i.i.d. Poisson random variables having expectation $\mu$.

Define $Y_n =\sum_{k=1}^{n} X_k$, the total number of particles in the interval $\{0, 1, 2, \ldots n\}$. Now let $W_n$ be the position of the $n$-th particle, where particles are labaled with natural numbers from the origin to infinity, e.g. the particle $n=0$ is the closest to the origin and $W_0$ is the site where it is located, the particle $n=1$ is the one which is immediatly on its right or which eventually shares the same site, and so on...

How can I show that the expectation of $W_s$ is $\mathbb{E}[W_s] = s / \mu $ ? It is clear that the expectation of $Y_n$ is $n \mu$.

What else can I say on the distribution of $W_s$? Is it a sum of Poisson random variables?

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up vote 2 down vote accepted

I can give you an exact distribution for $W_s$; it isn't pretty, though.

The big thing to remember here is that sums of independent Poisson-distributed random variables simplify nicely: in particular, we have $$ Y_n=X_1+\cdots+X_n\overset{\mathcal{D}}{\equiv}\text{Pois}(n\mu). $$ So, the event that $W_s>n$, which is precisely the event $Y_n<s$, has $$ P(W_s>n)=\sum_{d=0}^{s-1}P(Y_n=d)=\sum_{d=0}^{s-1}\frac{e^{-n\mu}(n\mu)^{d}}{d!}. $$ But, since $W_s$ takes values in $\mathbb{N}$, we have $$ P(W_s=n)=P(W_s>n-1)-P(W_s>n)=\sum_{d=0}^{s-1}\left(\frac{e^{-(n-1)\mu}((n-1)\mu)^d}{d!}-\frac{e^{-n\mu}(n\mu)^d}{d!}\right). $$ Another way you could do this: $$ \begin{align*} P(W_s=n)&=P(Y_{n-1}<s, Y_n\geq s)\\ &=\sum_{d=0}^{s-1}\sum_{t=s-d}^{\infty}P(Y_{n-1}=d, X_n=t)\\ &=\sum_{d=0}^{s-1}\sum_{t=s-d}^{\infty}\frac{e^{-(n-1)\mu}((n-1)\mu)^d}{d!}\cdot\frac{e^{-\mu}\mu^t}{t!} \end{align*} $$ These do not, unfortunately, simplify in any nice sort of way.

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