Question

If an individual has never had a previous automobile accident, then the probability he or she has an accident in the next \(h\) time units is \(\beta h+o(h) ;\) on the other hand, if he or she has ever had a previous accident, then the probability is \(\alpha h+o(h) .\) Find the expected number of accidents an individual has by time \(t\).

Short answer

The expected number of accidents an individual has by time \(t\) is given by the sum of the probabilities of having an accident in each interval until time \(t\), taking into account their probability of having a previous accident: \(E[t] = \sum_{i=1}^n P_i\). It depends on the values of \(\alpha\) and \(\beta\), which are not given.

Step-by-step solution

Define the problem

We are given two probability functions for the occurrence of an accident for an individual, with different coefficients for those who have never had a previous accident, and those who have had a previous accident. Let's denote the probability of having an accident in the next \(h\) time units is \(P(h)\), which is equal to \(\beta h + o(h)\) for those without a previous accident, and \(\alpha h + o(h)\) for those with a previous accident. Our goal is to find the expected number of accidents for an individual by time \(t\).

Calculate the expected number of accidents in the first interval

In the first interval occurring in the first \(h\) time units, we have: For individuals without a previous accident: \(P(h) = \beta h + o(h)\) For individuals with a previous accident: \(P(h) = \alpha h + o(h)\) We first find the probability that an individual has their first accident in the first interval. Since every individual starts without a previous accident, we can write \(P_1 = P(h) = \beta h + o(h)\) as the probability of having their first accident in the initial time unit.

Calculate the expected number of accidents in the second interval

In the second interval, which occurs in the next \(h\) time units, the probability of having an accident depends on whether the individual had an accident in the first interval. This means we need to calculate the conditional probability of having an accident in the second interval given their accident history in the first interval. If the individual had no accident in the first interval, the probability of having an accident in the second interval is still \(\beta h + o(h)\). The probability of having no accident in the first interval is \(1 - (\beta h + o(h))\). Thus, the probability of having the first accident in the second interval is \((1 - (\beta h + o(h))) (\beta h + o(h))\). On the other hand, if the individual had an accident in the first interval, the probability of having another accident in the second interval is \(\alpha h + o(h)\). Thus, the probability of having a second accident in the second interval is \((\beta h + o(h)) (\alpha h + o(h))\). Therefore, the total probability of having an accident in the second interval is: \(P_2 = (1 - (\beta h + o(h))) (\beta h + o(h)) + (\beta h + o(h)) (\alpha h + o(h))\)

Extend the calculation to time t

We can extend the process to find the expected number of accidents for any interval, as well as the cumulative expected number of accidents by time \(t\). Let \(n\) be the total number of intervals in time \(t\), and let \(P_n\) be the probability of having an accident in the nth interval. Then, the expected number of accidents by time \(t\) is given by the sum of the probabilities of having an accident in each interval until time \(t\): \(E[t] = \sum_{i=1}^n P_i\) The expected number of accidents will be the sum of probabilities for an individual having an accident in each interval, taking into account their probability of having a previous accident. Since the values of \(\alpha\) and \(\beta\) are not given, we cannot provide a specific numerical value for the expected number of accidents by time \(t\). Instead, the given sum formula, \(E[t] = \sum_{i=1}^n P_i\), will be the expected number of accidents by time \(t\).