# Poisson process derivation

#### Poirot

##### Banned
Let customers arrive according to a poisson process with parameter st and let $X_{t}$ denote number of customers in the system by time t. Consider an interval [t,t+h] with h small.

Show that P(1 arrival)= sh + o[h], P(more than one arrival)=o[h] and P(no arrival)=1-sh+o[h].

I know P(1 arrival)=$she^{-sh}$ but how to get further?

Thanks

Last edited:

#### CaptainBlack

##### Well-known member
Let customers arrive according to a poisson process with parameter st and let $X_{t}$ denote number of customers in the system by time t. Consider an interval [t,t+h] with h small.

Show that P(1 arrival)= sh + o[h], P(more than one arrival)=o[h] and P(no arrival)=1-sh+o[h].

I know P(1 arrival)=$she^{-sh}$ but how to get further?

Thanks
We have a Poisson process with mean arrival rate $$s$$, then the number of arrivals in an interval of length $$h$$ is $$N$$ and $$N \sim P(sh)$$.

Then:

$$\displaystyle p(N=1)=sh\; e^{-sh} =sh \left( 1-sh + \frac{s^2h^2}{2}-... \right) = sh(1 + R(sh))=sh + shR(sh)$$

where the $$|R(sh)|$$ for small positive $$h$$ is bounded by $$sh$$ (the truncation error for an alternating series of terms of decreasing absolute value is bounded by the absolute value of the first neglected term), so:

$$\displaystyle |sh \;R(sh)| < s^2h^2 = o ( h)$$

since $$\lim_{h \to 0} (s^2 h^2)/h = 0$$.

Now do the no arrivals case, then $$p(N>1)=1-(p(N=0)+p(N=1))$$

CB

#### Poirot

##### Banned
Thanks, but can you explain what R(sh) is ?

#### CaptainBlack

##### Well-known member
Thanks, but can you explain what R(sh) is ?
It is the remainder when the series is truncated after the first term, or the sum of the tail of the infinite series (they are the same thing).

CB

#### Poirot

##### Banned
Ok I can easily do the rest now. Thanks