The rate a certain insurance company charges its policyholders alternates between r1 and r0. A new policyholder is initially charged at a rate of r1 per unit time. When a policyholder paying at rate r1 has made no claims for the most recent s time units, then the rate charged becomes r0 per unit

Question

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.

(a) [latex]P_{i}[/latex] , the proportion of time that the policyholder pays at rate [latex]r_{i}[/latex] , i = 0, 1;
(b) the long-run average amount paid per unit time.

Accepted Answer

If we say that the system is “on” when the policyholder pays at rate [latex]r_{1}[/latex] and “off” when she pays at rate [latex]r_{0}[/latex], 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 [latex]\lambda[/latex], the preceding yields that
[latex]\begin{aligned}E[ ext { 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} ight)\end{aligned}[/latex]
Since [latex]E[X]=1 / \lambda[/latex], we see that
[latex]P_1=\frac{E[ ext { on time in cycle }]}{E[X]}=1-e^{-\lambda s}[/latex]
and
[latex]P_0=1-P_1=e^{-\lambda s}[/latex]
The long-run average amount paid per unit time is
[latex]r_0 P_0+r_1 P_1=r_1-\left(r_1-r_0 ight) e^{-\lambda s}[/latex]

Question 7.22: The rate a certain insurance company charges its policyholde......

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