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Chapter 7: Problem 3

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

Short Answer

Expert verified
Given the mean-value function \(m(t) = \frac{t}{2}\), the probability of zero events occurring at \(t=5\) is: \[P(N(5)=0) = e^{-\frac{5}{2}}\]

Step by step solution

01

The mean-value function, denoted by \(m(t)\), is the expected number of events that have occurred up to time \(t\). It is given by: \[m(t) = E[N(t)]\] In our case, it is given by the function: \[m(t) = \frac{t}{2}\] #Step 2: Determine the expected number of events at \(t=5\)#

Using the provided mean-value function, we can determine the expected number of events at \(t=5\), which is given by: \[m(5) = \frac{5}{2}\] #Step 3: Use the Poisson distribution to estimate the probability#
02

Since we are dealing with a renewal process, we can model the number of events occurring at any given time using the Poisson distribution. The probability mass function of a Poisson distribution is given by: \[P(N(t)=k) = \frac{(\lambda t)^{k}e^{-\lambda t}}{k!}\] Where \(k\) is the number of events, \(\lambda\) is the rate parameter, and \(t\) is the time. #Step 4: Determine the rate parameter of the Poisson distribution#

For the Poisson distribution we are using, the mean is directly proportional to the rate parameter \(\lambda\) and time \(t\). \[E[N(t)] = \lambda t\] Using the expected number of events from Step 2, and \(t=5\), we can calculate the rate parameter: \[\frac{5}{2} = \lambda \cdot 5\] \[\Rightarrow \lambda = \frac{1}{2}\] #Step 5: Compute the probability of zero events occurring at \(t=5\)#
03

Using the Poisson distribution and the rate parameter we found in Step 4, we can now compute the probability of zero events occurring at \(t=5\): \[P(N(5)=0) = \frac{[\frac{1}{2}\cdot(5)]^{0}e^{-\frac{1}{2}\cdot(5)}}{0!}\] #Step 6: Calculate and simplify the expression#

We can now simplify the expression from Step 5: \[P(N(5)=0) = \frac{1^{0}e^{-\frac{5}{2}}}{1}\] \[P(N(5)=0) = e^{-\frac{5}{2}}\] #Solution# Given the mean-value function \(m(t) = \frac{t}{2}\), the probability of zero events occurring at \(t=5\) is: \[P(N(5)=0) = e^{-\frac{5}{2}}\]

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Key Concepts

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

Mean-Value Function
The mean-value function, often symbolized as \( m(t) \), plays a crucial role in understanding renewal processes. It represents the expected number of events occurring up to a specific time \( t \). To simplify, think of it as the average number of events expected to happen by the time \( t \) has passed.
In the context of the given problem, we see \( m(t) = \frac{t}{2} \). This tells us that for every unit of time, we expect half of an event. Put simply, if \( t = 2 \), you typically anticipate one event to have occurred on average.

  • It's a linear function, implying a constant rate of event occurrence over time.
  • The slope \( \frac{1}{2} \) indicates the rate at which events occur per unit time.
  • This continuous and predictable pattern aids in planning and analysis of event occurrences.
In this way, the mean-value function simplifies understanding the cumulative aspect of events in a renewal process.
Poisson Distribution
The Poisson distribution is a pivotal concept when examining the behavior of events happening independently within a fixed interval of time or space. It's most often applied when you deal with rare events.
The probability mass function (PMF) for the Poisson distribution is defined as:\[P(N(t) = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}\]where:
  • \( k \) is the number of events or happenings you are calculating the probability for.
  • \( \lambda \) is the rate at which these events occur.
  • \( t \) is the time interval.
In this scenario, since the problem is rooted in a renewal process, using a Poisson distribution is fitting. As events could vary in time, having a formulation like this assists in computing the chances of observing a specific number of events \( k \). Its flexibility for handling diverse scenarios makes it indispensable in statistical modeling.
Probability Mass Function
A Probability Mass Function (PMF) is used specifically for discrete random variables. It allows us to calculate the probability that a discrete random variable is exactly equal to some value.
In relation to the Poisson distribution, the PMF is key to determining the likelihood that a certain number of events occur within a given time period. Here's how it works:
  • The PMF provides the probability for every possible value of a random variable \( k \).
  • The sum of these probabilities across all potential values of \( k \) will equal 1, ensuring all conceivable outcomes are accounted for.
  • This function is particularly useful in scenarios where the average rate of occurrence is known but occurrences are random.
By using the PMF with a Poisson distribution, it is straightforward to understand how likely it is to experience exactly \( k \) events. This extends our ability to not only anticipate event outcomes but quantify them, such as finding \( P(N(5)=0) \), the probability of zero events by time \( t=5 \). Such use of PMF makes predictive tasks in probability transparent and manageable.

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Most popular questions from this chapter

Write a program to approximate \(m(t)\) for the interarrival distribution \(F+G\), where \(F\) is exponential with mean 1 and \(G\) is exponential with mean 3.

In Example 7.7, 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 (possible delayed) renewal process if an cvent corresponds to a customer (a) entering the bank? (b) leaving the bank? What if \(F\) were exponential?

Consider a renewal process with mean interarrival time \(\mu .\) Suppose that cach event of this process is independently "counted" with probability \(p\). Let \(N_{c}(t)\) denote the number of counted events by time \(t, t>0\). (a) Is \(N_{c}(t), t \geq 0\) a rencwal process? (b) What is \(\lim _{t \rightarrow \infty} N_{c}(t) / t ?\)

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 a process is this?

For an interarrival distribution \(F\) having mean \(\mu\), we define the equilibrium distribution of \(F\), denoted \(F_{e}\), by $$ F_{\mathrm{e}}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y $$ (a) Show that if \(F\) is an exponential distribution, then \(F=F_{t^{*}}\) (b) If for some constant \(c\), $$ F(x)=\left\\{\begin{array}{ll} 0, & x
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