The infinitesimal definition of the Poisson process is given as follows
$$P(N(t+h)=n|N(t)=m)= \begin{cases}
\lambda h +o(h) & \text{if } n=m+1\\
1- \lambda h +o(h) &\text{if } n=m \\
o(h) & \text{if } n \ge m+2 \\
0 & \text{otherwise}
\end{cases} $$
Here $N(t)$ counts the number of arrivals in the interval $[0,t]$. Also it has rate parameter $\lambda$.
The problem we have is as follows: "Customers arrive at a ticket office as a Poisson process of rate $λ.$ Suppose that
each customer buys a ticket for $£2$. Let $X(t)$ be the amount of money (in
pounds) collected by the ticket office up to time $t$.
Write down an expression for $P(X(t + h) = n | X(t) = m)$ analogous to the
infinitesimal description of the Poisson process. "
I am finding this quite tricky as I don't really like working with the infinitesimal definition. My thought process is as follows. $X(t)$ must have parameter $2 \lambda$ as for every arrival the customer spends $£2$, then I think the expression would look like this:
$$P(X(t + h) = n | X(t) = m) = \begin{cases}
2\lambda h +o(h) & \text{if } n=m+1\\
1- 2\lambda h +o(h) &\text{if } n=m \\
o(h) & \text{if } n \ge m+2 \\
0 & \text{otherwise}
\end{cases} $$
This seems somehow "too simple" does anyone know how I should proceed?
Best Answer
$$P(X(t + h) = n | X(t) = m) = \begin{cases} \lambda h +o(h) & \text{if } n=m+2\\ 1- \lambda h +o(h) &\text{if } n=m \\ o(h) & \text{if } n \ge m+2 \\ 0 & \text{otherwise} \end{cases} $$
Note that $X(t)$ characterizes the state of the Markov chain at time $t$.
The state increments by 2 at each transition. The rates remain the same.
The ingredients for a Markov process are
you have only one state variable, namely, the cost. You have infinite states, and the transition rates are defined above. From each state, you have only one possible transition, namely, to increment the state variable by 2 units.