/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Suppose that people arrive at a ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that people arrive at a bus stop in accordance with a Poisson process with rate \(\lambda\). The bus departs at time \(t\). Let \(X\) denote the total amount of waiting time of all those who get on the bus at time \(t\). We want to determine \(\operatorname{Var}(X)\). Let \(N(t)\) denote the number of arrivals by time \(t .\) (a) What is \(E[X \mid N(t)] ?\) (b) Argue that \(\operatorname{Var}[X \mid N(t)]=N(t) t^{2} / 12\) (c) What is \(\operatorname{Var}(X) ?\)

Short Answer

Expert verified
The variance of the total waiting time of all those who get on the bus at time $t$ is given by: \(\operatorname{Var}(X) = (\lambda t * t^2/12) + (\lambda t * t^2/4)\).

Step by step solution

01

(Find E[X|N(t)])

First, if we have N(t) people arriving at the bus stop, they arrive uniformly at random on the interval [0, t]. Thus, conditioned on knowing N(t), the waiting time for a particular passenger at the bus stop follows a uniform distribution with parameters 0 and t. The expected waiting time for a single passenger is E[Waiting time of 1 passenger | N(t)] = t/2. Since E[X|N(t)] is the total waiting time of all passengers who get on the bus, we can multiply this single passenger's expected waiting time by N(t): E[X|N(t)] = N(t) * (t/2) Now, we have E[X|N(t)].
02

(Find Var[X|N(t)])

Now, to find the conditional variance Var[X|N(t)], we will use the fact that the waiting time of each passenger is uniformly distributed on the interval [0, t] and consider the variance of this uniform distribution: Var[Waiting time of 1 passenger | N(t)] = t^2/12. Since there are N(t) passengers and their waiting times are independent, the conditional variance Var[X|N(t)] can be determined as the sum of variances: Var[X|N(t)] = N(t) * (t^2/12)
03

(Find Var[X])

Finally, we need to find Var[X]. To do this, we will apply the Law of Total Variance: Var[X] = E[Var[X|N(t)]] + Var[E[X|N(t)]]. We already found that Var[X|N(t)] = N(t) * (t^2/12). Therefore: E[Var[X|N(t)]] = E[N(t) * (t^2/12)] We know that N(t) has a Poisson distribution with parameter λt, so its expected value is λt: E[N(t)] = λt E[Var[X|N(t)]] = λt * (t^2/12) We also found that E[X|N(t)] = N(t) * (t/2). Therefore: Var[E[X|N(t)]] = Var[N(t) * (t/2)], since (t/2) is constant, we get Var[N(t) * (t/2)] = (t^2/4)*Var[N(t)]. Since N(t) has a Poisson distribution, its variance equals its mean, which in this case is λt: Var[N(t)] = λt Var[E[X|N(t)]] = (t^2/4) * λt Now, combining everything using the Law of Total Variance: Var[X] = E[Var[X|N(t)]] + Var[E[X|N(t)]] Var[X] = λt * (t^2/12) + (t^2/4) * λt Var[X] = (λt * t^2/12) + (λt * t^2/4) The final answer is: Var[X] = (λt * t^2/12) + (λt * t^2/4)

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Key Concepts

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

Conditional Expectation
Understanding conditional expectation is essential in analyzing scenarios that involve a certain condition or piece of information influencing the outcome. When we talk about the conditional expectation of a random variable, we are trying to determine its expected value when the condition is satisfied.

In our bus stop scenario, the total amount of waiting time, denoted as the random variable X, can be thought of as a function of the number of people who arrive by time t, denoted by N(t). To calculate the conditional expectation E[X|N(t)], we imagine that we already know the number of people at the bus stop, and based on that, we seek the average waiting time.

For a single passenger, their waiting time, when conditioned on N(t), is uniformly distributed between 0 and t. Thus, the conditional expectation E[X|N(t)] is the sum of the individual expected waiting times. Since the waiting time of one passenger is t/2, when there are N(t) passengers, the total expected waiting time is N(t) × (t/2).

This step-wise approach simplifies the understanding and calculation of expected values in complex random processes.
Uniform Distribution
The uniform distribution is a probability distribution where all outcomes are equally likely. When we discuss waiting times at a bus stop in the context of the uniform distribution, we are assuming that at any point within the interval from when the bus stop starts operating to when the bus departs, the arrival of a passenger is equally likely.

Mathematically, the waiting time of a passenger uniformly distributed on the interval [0, t] has a constant probability density function, meaning the likelihood of arriving at any specific point is the same. When calculating the expected waiting time, the midpoint of the interval, or t/2, is used because it represents the average value over the entire interval.

For variance, the uniform distribution has a well-known formula: Var[Uniform(0, t)] = (t − 0)^2 / 12. This leads us to conclude that the conditional variance Var[X|N(t)] for the total waiting time at the bus stop, given the number of people N(t), is the sum of their individual variances, which is N(t) × (t^2/12).
Law of Total Variance
The Law of Total Variance, also known as the variance partitioning or Eve's Law, helps to decompose the overall variance of a random variable into its components. It expresses the total variance Var(X) as the sum of the expected variance of the conditional distribution E[Var[X|Y]] and the variance of the expected values of the conditional distributions Var[E[X|Y]].

This is powerful because it separates the randomness due to the condition Y from the inherent randomness of X. In our problem, we break down the total variance of the waiting time into two parts: (1) the expected variance of waiting times given N(t) arrivals and (2) the variance of the expected waiting times given these N(t) arrivals. By computing these separately and summing them up, we obtain the total variance of X. Through this method, we derive a more comprehensive understanding of how various sources of randomness contribute to the overall variability in our scenario.
Poisson Distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time, provided these events occur with a known constant mean rate and independently of the time since the last event. In our case, the arrival of people at the bus stop follows a Poisson process with rate λ.

The key properties of the Poisson distribution that we leverage in this problem are that the expected value and the variance of a Poisson-distributed random variable are both equal to its rate parameter times the length of the observed interval, λt. This simplifies the calculation of both the expected values and variances in our conditional settings.

By applying these properties, we can solve parts of our exercise based on the Poisson distribution characteristics, which help us eventually arrive at the variance of the total waiting time Var(X). Understanding the Poisson distribution's role in modeling and analyzing random arrival scenarios is crucial, providing clarity in various fields, including queuing theory, telecommunications, and traffic flow analysis.

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Most popular questions from this chapter

Consider a single server queueing 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.

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