MHB How Does the Poisson Process Model Customer Arrivals Over Time?

[Poirot1]

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

 

[CaptainBlack]

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

 

[Poirot1]

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

 

[CaptainBlack]

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

 

[Poirot1]

Ok I can easily do the rest now. Thanks