PGF of a Poisson process using differential equations - telephone answering service

119 Views Asked by At

Could really do with help on this particular problem. I am comfortable with finding PGFs for most process but have no idea where to start for this question which begins with difference equations. Thanks.

A telephone answering service receives calls whose frequency varies with time but independently of other calls perhaps with a daily pattern – more during the day than the night. The rate $\lambda(t)\geq 0$ becomes a function of the time $t.$ The probability that a call arrives in the small time interval $(t,t+\delta t)$ when $n$ calls have been received at time $t$ satisfies

$$p_{n+1}(t+\delta t) = p_{n-1}(t)(\lambda(t)\delta t+o(t\delta))+p_{n}(t)(1-\lambda(t)\delta t+o(\delta t))$$

with

$$p_o(t+\delta t) = (1-\lambda(t)\delta t+o(\delta t))p_o(t)$$

It is assumed that the probability of two or mote calls arriving in the time interval $(t,t+\delta t)$ is negligible. Find the set of differential-difference equations for $p_n(t)$ Hence show that the probability generating function for the process is given by $$G(s,t)= \exp\left[(s - 1)\int_0^t\lambda(u) \, du\right].$$

1

There are 1 best solutions below

0
On BEST ANSWER

To start with since we have been asked to generate differential difference equations, we should be looking at how derivate might come up here. We know by definition that $$ \frac{dp_n}{dt} = lim_{\delta \rightarrow 0} \frac{p_n(t+\delta t)-p_n(t)}{\delta t}$$ Then substituting in what you know $p_n(t)$ and $p_n(t+\delta t)$ to be, we see that $$ \frac{dp_n}{dt} = lim_{\delta \rightarrow 0} (p_{n-1}\lambda - p_n\lambda + O(\delta t))=p_{n-1}\lambda - p_n\lambda$$ And using the definition for $p_0$, $$ \frac{dp_0}{dt} = - p_0\lambda$$ This is our set of differential difference equations. Then we have: $$ G(s,t)=\sum\limits_{n=0}^{\infty} p_n(t)s^n$$ So that, assuming we can pass the derivative through the sum (which is true for $|S|<1$?): $$ \frac{dG(s,t)}{dt} = \frac{d}{dt} \sum\limits_{n=0}^{\infty} p_n(t)s^n = \sum\limits_{n=0}^{\infty} \frac{dp_n(t)}{dt}s^n \\ = \sum\limits_{n=0}^{\infty} (p_{n-1}\lambda - p_n\lambda)s^n \\ $$ We then recognise the summand as being similar to that in the definition of G, so that we can write: $$ \frac{dG(s,t)}{dt} = \lambda s G(s,t) - \lambda G(s,t) = \lambda (s-1) G(s,t) $$ Then you can solve that (partial) differential equation in G using an integrating factor to give the desired result.