Question 7.22: The rate a certain insurance company charges its policyholde......
The rate a certain insurance company charges its policyholders alternates between r_{1} and r_{0}. A new policyholder is initially charged at a rate of r_{1} per unit time. When a policyholder paying at rate r_{1} has made no claims for the most recent s time units, then the rate charged becomes r_{0} per unit time. The rate charged remains at r_{0} until a claim is made, at which time it reverts to r_{1}. Suppose that a given policyholder lives forever and makes claims at times chosen according to a Poisson process with rate λ, and find
(a) P_{i} , the proportion of time that the policyholder pays at rate r_{i} , i = 0, 1;
(b) the long-run average amount paid per unit time.
If we say that the system is “on” when the policyholder pays at rate r_{1} and “off” when she pays at rate r_{0}, then this on–off system is an alternating renewal process with a new cycle starting each time a claim is made. If X is the time between successive claims, then the on time in the cycle is the smaller of s and X. (Note that if X < s, then the off time in the cycle is 0 .) Since X is exponential with rate \lambda, the preceding yields that
\begin{aligned}E[\text { on time in cycle }] & =E[\min (X, s)] \\& =\int_0^s x \lambda e^{-\lambda x} d x+s e^{-\lambda s} \\& =\frac{1}{\lambda}\left(1-e^{-\lambda s}\right)\end{aligned}
Since E[X]=1 / \lambda, we see that
P_1=\frac{E[\text { on time in cycle }]}{E[X]}=1-e^{-\lambda s}
and
P_0=1-P_1=e^{-\lambda s}
The long-run average amount paid per unit time is
r_0 P_0+r_1 P_1=r_1-\left(r_1-r_0\right) e^{-\lambda s}