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If the mean-value function of the renewal process \(\\{N(t), t \geqslant 0\\}\) is given by \(m(t)=\) \(t / 2, t \geqslant 0\), what is \(P[N(5)=0\\} ?\)

Short Answer

Expert verified
The probability \(P[N(5) = 0]\) is approximately \(0.0821\).

Step by step solution

01

Find the renewal function from the mean-value function

We are given the mean-value function, \(m(t) = \frac{t}{2}\), and we know that \(\frac{d}{dt} R(t) = m(t)\). We need to solve for the renewal function \(R(t)\). To do this, we will integrate \(m(t)\) with respect to \(t\). \[ R(t) = \int m(t) dt = \int \frac{t}{2} dt \]
02

Integrate

Now, we will integrate the function: \[ R(t) = \int \frac{t}{2} dt = \frac{1}{2} \int t dt = \frac{1}{2} \cdot \frac{t^2}{2} + C \] Here \(C\) represents the constant of integration. Since our given function is for \(m(t)\), we can assume that \(R(0) = 0\). Using this information, we can find out the value of C.
03

Determine the constant of integration

To find the constant of integration, use the fact that \(R(0) = 0\): \[ R(0) = 0 = \frac{1}{2} \cdot \frac{0^2}{2} + C \Rightarrow C = 0 \] Hence, our renewal function \(R(t)\) is given by: \[ R(t) = \frac{t^2}{4}, t\geq 0 \]
04

Find the probability \(P[N(5) = 0]\)

Now we can find the probability \(P[N(5) = 0]\) using the renewal function: \[ P[N(5) = 0] = e^{-R(5)} \] Substitute the value of \(R(5)\) in this equation: \[ P[N(5) = 0] = e^{-\frac{5^2}{4}} = e^{-\frac{25}{4}} \]
05

Calculate the final probability

Finally, we can use a calculator to get the numerical value of the probability: \[ P[N(5) = 0] = e^{-\frac{25}{4}} \approx 0.0821 \] So the probability \(P[N(5) = 0]\) is approximately \(0.0821\).

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Key Concepts

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

Renewal Process
A renewal process is a powerful concept in probability theory and stochastic processes that is used to model events that happen intermittently over time. It's like imagining a machine that undergoes repairs periodically—each repair 'renews' the machine, and the times between repairs are of interest.

The renewal process consists of a sequence of random variables representing the times at which events occur. These events could be anything from receiving emails to the arrival of buses at a stop. The central focus of the renewal process is to understand and predict the number of times an event occurs within a certain period. When a process starts anew after each event, it is called a renewal process because the process 'renews' itself.

One can predict various characteristics of a renewal process by analyzing the inter-renewal times—the periods between consecutive renewals. These inter-renewal times are often assumed to be independent and identically distributed (i.i.d.) random variables, leading to certain predictable long-term averages or 'regenerative' properties.
Mean-Value Function
Moving on to the mean-value function, this is a key concept that represents the expected number of times a renewal has occurred by time t. In simpler terms, it tells us on average how many times an event is expected to have happened by a certain point in time.

The mean-value function m(t) for a renewal process essentially gives us an average snapshot of the process at any given time t. To extract meaningful information from the mean-value function, you often need to understand its relation with the renewal function, which is derived by integrating the mean-value function. This relationship is crucial because it reflects how the average number of renewals grows over time.

The calculation process demonstrated in the exercise uses integration to obtain the renewal function from the given mean-value function. This step is pivotal as it sets the stage for predicting probabilities associated with the renewal process.
Integration in Probability
Integration plays a vital role in probability theory, especially when working with continuous random variables or processes like the renewal process. It helps you summarize entire distributions, calculate expected values, and, as seen in the exercise, derive functions that describe processes over time.

In our example, integrating the mean-value function m(t) with respect to time t gave us the renewal function R(t). Integration here allows us to accumulate all the infinitesimal contributions from the mean-value function up to time t, resulting in a formula that represents the overall behavior of the renewal process. In fact, without the tool of integration, it would be nearly impossible to make this leap from knowing the average rate of renewals at any time to knowing the total expected number of renewals over an interval.

Understanding integration within probability also means you can work out probabilities of certain events occurring, such as calculating the probability P[N(5) = 0] by integrating the mean-value function and then applying the known relation with the renewal function. Thus integration is essential for deriving meaningful insights from probabilistic models.

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Most popular questions from this chapter

For the renewal process whose interarrival times are uniformly distributed over \((0,1)\), determine the expected time from \(t=1\) until the next renewal.

Consider a renewal process \(\\{N(t), t \geqslant 0\\}\) having a gamma \((r, \lambda)\) interarrival distribution. That is, the interarrival density is $$ f(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{r-1}}{(r-1) !}, \quad x>0 $$ (a) Show that $$ P[N(t) \geqslant n]=\sum_{i=n r}^{\infty} \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ (b) Show that $$ m(t)=\sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ where \([i / r]\) is the largest integer less than or equal to \(i / r\). Hint: Use the relationship between the gamma \((r, \lambda)\) distribution and the sum of \(r\) independent exponentials with rate \(\lambda\) to define \(N(t)\) in terms of a Poisson process with rate \(\lambda\).

In Example \(7.6\), suppose that potential customers arrive in accordance with a renewal process having interarrival distribution \(F\). Would the number of events by time \(t\) constitute a (possibly delayed) renewal process if an event corresponds to a customer (a) entering the bank? (b) leaving the bank? What if \(F\) were exponential?

A taxi alternates between three different locations. Whenever it reaches location \(i\), it stops and spends a random time having mean \(t_{i}\) before obtaining another passenger, \(i=1,2,3 .\) A passenger entering the cab at location \(i\) will want to go to location \(j\) with probability \(P_{i j} .\) The time to travel from \(i\) to \(j\) is a random variable with mean \(m_{i j} .\) Suppose that \(t_{1}=1, t_{2}=2, t_{3}=4, P_{12}=1, P_{23}=1, P_{31}=\frac{2}{3}=1-P_{32}\) \(m_{12}=10, m_{23}=20, m_{31}=15, m_{32}=25 .\) Define an appropriate semi- Markov process and determine (a) the proportion of time the taxi is waiting at location \(i\), and (b) the proportion of time the taxi is on the road from \(i\) to \(j, i, j=1,2,3\).

Consider a semi-Markov process in which the amount of time that the process spends in each state before making a transition into a different state is exponentially distributed. What kind of process is this?

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