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Chapter 5: Problem 45

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\), that is independent of the nonnegative random variable \(T\) with mean \(\mu\) and variance \(\sigma^{2}\). Find (a) \(\operatorname{Cov}(T, N(T))\) (b) \(\operatorname{Var}(N(T))\)

Short Answer

Expert verified
(a) \(\operatorname{Cov}(T, N(T)) = 0\) (b) \(\operatorname{Var}(N(T)) = \lambda * \mu\)

Step by step solution

01

Given that \(T\) has a mean \(\mu\), we can write \(E[T] = \mu\). #Step 2: Find E[N(T)]#

As N(T) follows a Poisson process with rate parameters \(\lambda\), we have \(E[N(T)] = \lambda * E[T] = \lambda * \mu\). #Step 3: Find E[T * N(T)]#
02

Since \(T\) and \(N(T)\) are independent, we can write \(E[T * N(T)] = E[T] * E[N(T)] = (\mu) * (\lambda * \mu) = \lambda * \mu^2\). #Step 4: Calculate Cov(T, N(T))#

Now, we can calculate the covariance using the formula \(\operatorname{Cov}(T, N(T)) = E[T * N(T)] - E[T] * E[N(T)]\). Plugging in the values obtained in steps 1, 2, and 3, we get: \(\operatorname{Cov}(T, N(T)) = (\lambda * \mu^2) - (\mu) * (\lambda * \mu) = 0\). #Step 5: Find E[(N(T))^2]#
03

For a Poisson random variable (with rate \(\lambda\)), we have \(E[(N(T))^2] = \operatorname{Var}(N(T)) + (E[N(T)])^2 = \lambda * E[T] + (\lambda * E[T])^2\). Substituting \(E[T]=\mu\), we get \(E[(N(T))^2] = \lambda * \mu + (\lambda * \mu)^2\). #Step 6: Calculate Var(N(T))#

Now, we calculate the variance using the formula \(\operatorname{Var}(N(T)) = E[(N(T))^2] - E[N(T)]^2\). Plugging in the values from Step 2 and Step 5, we get \(\operatorname{Var}(N(T)) = (\lambda * \mu + (\lambda * \mu)^2) - (\lambda * \mu)^2 = \lambda * \mu\). So, we have derived the required covariance and variance: (a) \(\operatorname{Cov}(T, N(T)) = 0\) (b) \(\operatorname{Var}(N(T)) = \lambda * \mu\)

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Key Concepts

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

Covariance
Covariance is a measure of how much two random variables change together. It is used to determine the relationship between these variables; if they increase and decrease together, the covariance is positive, while if one increases when the other decreases, the covariance is negative. In our Poisson process example, we evaluated the covariance between the time variable, represented by the random variable T, and the number of events occurring in that timeframe, N(T). From the solution, we found the covariance to be zero, indicating that T and N(T) do not linearly relate to one another. This means that knowing the occurrence of events up to time T does not provide any information regarding the numerical value of T and vice versa, suggesting independence in their variations.
Variance
Variance is a statistical measure that describes the spread of a set of data points. It measures how far each number in the set is from the mean and thus from every other number in the set. In terms of a random variable, variance gauges how much the values of the variable are expected to deviate from the expected value. When we solved for the variance of N(T) in the exercise, we determined that it equals \( \lambda * \mu \), where \( \lambda \) is the rate of the Poisson process, and \( \mu \) is the mean of T. This result encapsulates the expected amount by which the number of events in time T differs from its mean value, squared to keep the measure non-negative.
Random variable
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. There are two types of random variables: discrete, like the number of events in a Poisson process, and continuous, such as time. The concept of a random variable is central to probability theory and statistics because it provides a quantitative description of the outcomes of random events. In the case of our Poisson process, N(T), the number of events that occur by time T, is a discrete random variable, and its properties鈥攕uch as expected value and variance鈥攖ell us much about the characteristics of the random Poisson process we are examining.
Expected value
The expected value, often denoted by E[X] for a random variable X, represents the mean, or average, value the variable would take if the random process could be repeated many times. It's the long-term average result of the random variable and is weighted by probabilities in the case of discrete data or density functions for continuous data. The expected value of a random variable gives a sense of the 'center' of its distribution. In our solution for E[N(T)], we saw that it equals \( \lambda * \mu \), which tells us that the average number of events by time T in our Poisson process is proportional to both the process's rate and the expected amount of time T.

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Most popular questions from this chapter

Each entering customer must be served first by server 1 , then by server 2 , and finally by server \(3 .\) The amount of time it takes to be served by server \(i\) is an exponential random variable with rate \(\mu_{i}, i=1,2,3 .\) Suppose you enter the system when it contains a single customer who is being served by server \(3 .\) (a) Find the probability that server 3 will still be busy when you move over to server 2 . (b) Find the probability that server 3 will still be busy when you move over to server \(3 .\) (c) Find the expected amount of time that you spend in the system. (Whenever you encounter a busy server, you must wait for the service in progress to end before you can enter service.) (d) Suppose that you enter the system when it contains a single customer who is being served by server \(2 .\) Find the expected amount of time that you spend in the system.

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

Let \(S(t)\) denote the price of a security at time \(t .\) A popular model for the process \(\\{S(t), t \geqslant 0\\}\) supposes that the price remains unchanged until a "shock" occurs, at which time the price is multiplied by a random factor. If we let \(N(t)\) denote the number of shocks by time \(t\), and let \(X_{i}\) denote the \(i\) th multiplicative factor, then this model supposes that $$ S(t)=S(0) \prod_{i=1}^{N(t)} X_{i} $$ where \(\prod_{i=1}^{N(t)} X_{i}\) is equal to 1 when \(N(t)=0\). Suppose that the \(X_{i}\) are independent exponential random variables with rate \(\mu\); that \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda\); that \(\\{N(t), t \geqslant 0\\}\) is independent of the \(X_{i}\); and that \(S(0)=s\). (a) Find \(E[S(t)]\). (b) Find \(E\left[S^{2}(t)\right]\)

A two-dimensional Poisson process is a process of randomly occurring events in the plane such that (i) for any region of area \(A\) the number of events in that region has a Poisson distribution with mean \(\lambda A\), and (ii) the number of events in nonoverlapping regions are independent. For such a process, consider an arbitrary point in the plane and let \(X\) denote its distance from its nearest event (where distance is measured in the usual Euclidean manner). Show that (a) \(P\\{X>t\\}=e^{-\lambda \pi t^{2}}\), (b) \(E[X]=\frac{1}{2 \sqrt{\lambda}}\).

Customers arrive at a bank at a Poisson rate \(\lambda\). Suppose two customers arrived during the first hour. What is the probability that (a) both arrived during the first 20 minutes? (b) at least one arrived during the first 20 minutes?
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