# Counting Events and Poisson Processes

Hi I'm having difficulty understanding this proof any help will be much appreciated. I'm sorry for the horrendous length of this post and the possible confusion that my lack of understanding might bring.

Let $N((0,t])$ be the number of events that occur in the interval (0,t].
We want to show that the following four postulates require $P_{k}(t) = Pr[N((0,t]) = k]$ is Poisson as $P_{k}(t) = \frac{(\lambda t)^{k}e^{-\lambda t}}{k!}$ for k=0,1,...

We have the four postulates:
1. The numbers of events that occur in disjoint intervals are independent random variables. For integers m = 2,3,... and time points $t_{0}<t_{1}<...<t_{m}$ the random variables $N((t_{0},t_{1}]),...,N((t_{m-1},t_{m}])$ are independent.

2. For $t \ge 0$ and $h > 0$ the probability distribution of $N((t,t+h])$ depends only on interval length h, not t.

3. There is a positive constant $\lambda$ for which the probability of at least one event happening in time interval h is $Pr[N(t,t+h] \ge 1] = \lambda h + o(h)$ as h approaches 0.

4. The probability of two or more events occurring in an interval of length h is o(h), namely $Pr[N((t,t+h]) \ge 2] = o(h)$ as h approaches 0.

Divide the interval (0,t] into n subintervals of length h = t/n and let
$\epsilon _{i} = 1$ if there is at least one event in the interval ((i-1)t/n,it/n]
$\epsilon _{i} = 0$ else

Then $S_{n}=\epsilon _{1}+...+\epsilon _{n}$ counts the total number of subintervals that contain at least one event and $p_{i} = Pr[\epsilon _{i} = 1] = \frac{\lambda t}{n} + o(\frac{t}{n})$.
Also we define $\mu = \sum_{i=1}{n}p_{i}=\lambda t + no(\frac{t}{n})$.
From the postulates we deduce:
$\lim_{n\to \infty} Pr[S_{n}=k]=\frac{\mu ^{k}e^{- \mu}}{k!}$ where $\mu = \lambda t$

We must now show $\lim_{n\to \infty} Pr[S_{n}=k] = Pr[N((0,t])=k] =P_{k}(t)$.
Note that $S_{n}$ differs from N((0,t]) only if at least one of the subintervals contains two or more events, and postulate 4 precludes this:

(Okay at this point I don't understand why the inequality below is always true.)
$| P_{k}(t)-Pr[S_{n}=k]| \le Pr[N((0,t]) \ne S_{n}]$
$\le \sum_{i=1}{n} Pr[N(( \frac{(i-1)t}{n} ,\frac{it}{n}]) \ge 2 ]$
$\le no(\frac{t}{n})$
$= t \frac{o(\frac{t}{n})}{\frac{t}{n}} \rightarrow 0$ as $n \rightarrow \infty$

So we conclude that the probabilities in the absolute value are equal as n grows infinitely large. I'm having difficulty with providing to myself a very convincing argument as to why the first inequality is always true. Thanks for all the help.

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I think it is because the 2 things inside the | | sign in the first expression only differ if $N((0,t])=k$ but $S_n\neq k$. The first inequality follows because we get it by summing over all $k$. Don't trust me on it. I spent 10 mins on this, this is the answer I came up with but I am not 100% convinced myself... –  Lost1 Mar 25 '13 at 23:52