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Chapter 5: Problem 51

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

Expert verified
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

01

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 \(t\).
02

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.
03

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: \(P_2 = (1 - (\beta h + o(h))) (\beta h + o(h)) + (\beta h + o(h)) (\alpha h + o(h))\)
04

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 \(t\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Functions
Probability functions are central to understanding the likelihood of events happening within a given timeframe. In this exercise, the focus is on accidents occurring over a set period. We have two essential probability functions here:- For individuals without any preceding accidents, the probability function is represented by \( \beta h + o(h) \).- For those who have experienced an accident before, the probability function is expressed as \( \alpha h + o(h) \).The notation \( o(h) \) indicates terms that become negligible as \( h \) approaches zero, allowing for simple calculations of probabilities over small intervals. Essentially, these functions help us quantify the chance of an accident occurring, taking into account past history.
Conditional Probability
Conditional probability refers to determining the likelihood of an event occurring under a given condition or past event. In this exercise, we explore how the probabilities of subsequent accidents depend on the occurrence of initial accidents within earlier intervals:- For someone who had no accident in the first time interval, the chance of having an accident in the subsequent interval remains as \( \beta h + o(h) \).- Conversely, if an accident already occurred in the first interval, the next interval's probability is modified to align with \( \alpha h + o(h) \).This distinction is crucial as it mirrors real-world scenarios where past experiences influence future probabilities, such as an individual's behavior affecting accident likelihood.
Cumulative Probability
Understanding cumulative probability involves looking at the probability of events accumulating over several intervals, up to a set time \( t \). This concept allows us to evaluate the overall likelihood of a series of accidents happening as time progresses.The expected number of accidents by a certain time \( t \), as noted in the steps, can be summed up using:\[ E[t] = \sum_{i=1}^n P_i \]Here, each \( P_i \) represents the probability of an accident occurring in a specific interval, adjusted for whether there was prior history of an accident. This cumulative approach helps in forecasting the total number of incidents over time, providing a broad perspective.
Time Interval Analysis
Time interval analysis breaks down the study of events, such as accidents, within small, manageable units. Each interval \( h \) is a sliver of time within the larger timeframe, \( t \), allowing precise probability calculations.- By examining individual \( h \) intervals, we can systematically calculate the probability for each period and adjust based on prior event occurrences.- Over multiple intervals, this method enables the compilation of an overarching probability, capturing the entire duration \( t \).Time interval analysis is invaluable for accurately modeling scenarios over elongated timelines. This granular approach is pivotal in predicting and understanding complex events through cumulative analysis.

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Most popular questions from this chapter

Suppose that customers arrive to a system according to a Poisson process with rate \(\lambda\). There are an infinite number of servers in this system so a customer begins service upon arrival. The service times of the arrivals are independent exponential random variables with rate \(\mu\), and are independent of the arrival process. Customers depart the system when their service ends. Let \(N\) be the number of arrivals before the first departure. (a) Find \(P(N=1)\). (b) Find \(P(N=2)\) (c) Find \(P(N=j)\). (d) Find the probability that the first to arrive is the first to depart. (e) Find the expected time of the first departure.

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu .\) Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \text { and } X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(E\left[X_{(1)}\right]\) (b) \(\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[X_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

Consider a single server queuing system where customers arrive according to a Poisson process with rate \(\lambda\), service times are exponential with rate \(\mu\), and customers are served in the order of their arrival. Suppose that a customer arrives and finds \(n-1\) others in the system. Let \(X\) denote the number in the system at the moment that customer departs. Find the probability mass function of \(X\). Hint: Relate this to a negative binomial random variable.

Shocks occur according to a Poisson process with rate \(\lambda\), and each shock independently causes a certain system to fail with probability \(p .\) Let \(T\) denote the time at which the system fails and let \(N\) denote the number of shocks that it takes. (a) Find the conditional distribution of \(T\) given that \(N=n\). (b) Calculate the conditional distribution of \(N\), given that \(T=t\), and notice that it is distributed as 1 plus a Poisson random variable with mean \(\lambda(1-p) t .\) (c) Explain how the result in part (b) could have been obtained without any calculations.

Let \(X_{1}, X_{2}, \ldots\) be independent positive continuous random variables with a common density function \(f\), and suppose this sequence is independent of \(N, a\) Poisson random variable with mean \(\lambda\). Define $$ N(t)=\text { number of } i \leqslant N: X_{i} \leqslant t $$ Show that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t)=\lambda f(t)\).
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