/*! 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 36 Let \(S(t)\) denote the price of... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 36

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 \(i^{\text {th }}\) multiplicative factor, then this model supposes that $$ S(t)=S(0) \prod_{l=1}^{N(t)} X_{i} $$ where \(\prod_{l=1}^{N(t)} X_{i}\) i? equal to 1 when \(N(t)=0\). Suppose that the \(X_{l}\) 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 \(E\left[S^{2}(t)\right]\)

Short Answer

Expert verified
For part (a), the expectation of the price of the security at time \(t\), \(E[S(t)]\), can be found as: $$ E[S(t)] = s e^{-\lambda t (1 - \frac{1}{\mu})} $$ For part (b), the expectation of the square of the price at time \(t\), \(E[S^2(t)]\), can be found as: $$ E\left[S^{2}(t)\right] = s^2 e^{-\lambda t} e^{\frac{\lambda t}{\mu^2}} $$

Step by step solution

01

Write down the expression for E[S(t)]

We start by finding the expectation of the price of the security at time \(t\), \(E[S(t)]\). Using the given model and the law of total expectation: $$ E[S(t)] = E\left[s \prod_{i=1}^{N(t)} X_i \right] $$
02

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

In order to find the expectation of \(S(t)\), we need to find the expectation of S(t) conditioning on N(t), \(E[S(t) | N(t)]\). Note that when conditioning on N(t), the summation over \(X_i\) becomes fixed, so we can rewrite the expression as: $$ E[S(t) | N(t)] = s \prod_{i=1}^{N(t)} E[X_i] $$ Since the \(X_i\) are independent exponential random variables with rate \(\mu\), we know that \(E[X_i]=\frac{1}{\mu}\), so the expression becomes: $$ E[S(t) | N(t)] = s \prod_{i=1}^{N(t)} \frac{1}{\mu} = s\left( \frac{1}{\mu} \right)^{N(t)} $$
03

Use the law of total expectation to find E[S(t)]

Using the law of total expectation, we have: $$ E[S(t)] = E\left[ E[S(t) | N(t)] \right] = E\left[ s\left( \frac{1}{\mu} \right)^{N(t)} \right] $$ Since \(N(t)\) is a Poisson process with rate \(\lambda\), we can calculate this expectation by summing over all possible values of \(N(t)\) with their respective probabilities: $$ E[S(t)] = \sum_{n=0}^{\infty} P(N(t) = n) s\left( \frac{1}{\mu} \right)^n = s \sum_{n=0}^{\infty} \left( \frac{1}{\mu} \right)^n P(N(t) = n) $$ Recall that the probability mass function of a Poisson distribution with rate \(\lambda\) is \(P(N(t) = n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}\). Plugging this in, we get: $$ E[S(t)] = s \sum_{n=0}^{\infty} \left( \frac{1}{\mu} \right)^n \frac{(\lambda t)^n e^{-\lambda t}}{n!} = s e^{-\lambda t} \sum_{n=0}^{\infty} \frac{\left(\frac{\lambda t}{\mu}\right)^n}{n!} $$ The last sum is an expansion of the exponential function, so we have: $$ E[S(t)] = s e^{-\lambda t} e^{\frac{\lambda t}{\mu}} = s e^{\frac{\lambda t}{\mu} - \lambda t} = s e^{-\lambda t (1 - \frac{1}{\mu})} $$ For part (a), the answer is: $$ E[S(t)] = s e^{-\lambda t (1 - \frac{1}{\mu})} $$
04

Find E[S^2(t)]

To find \(E\left[S^{2}(t)\right]\), we can follow a similar approach as above: $$ E\left[S^{2}(t)\right] = E\left[ \left( s \prod_{i=1}^{N(t)} X_i \right)^2 \right] = E[S(t)S(t)] $$ Using the law of total expectation for the joint event, we will condition on N(t): $$ E\left[S^{2}(t)\right] = E\left[ E\left[S(t) S(t) | N(t)\right] \right] = E\left[ \left(s\left(\frac{1}{\mu}\right)^{N(t)}\right)^2 \right] $$ Summing over all possible values of \(N(t)\) with their respective probabilities, just like before: $$ E\left[S^{2}(t)\right] = \sum_{n=0}^{\infty} P(N(t) = n) \left(s\left(\frac{1}{\mu}\right)^n \right)^2 = s^2 \sum_{n=0}^{\infty} \left(\frac{1}{\mu^2}\right)^n P(N(t) = n) $$ Plugging in the Poisson probabilities, we get: $$ E\left[S^{2}(t)\right] = s^2 e^{-\lambda t} \sum_{n=0}^{\infty} \frac{\left(\frac{\lambda t}{\mu^2}\right)^n}{n!} = s^2 e^{-\lambda t} e^{\frac{\lambda t}{\mu^2}} $$ For part (b), the answer is: $$ E\left[S^{2}(t)\right] = s^2 e^{-\lambda t} e^{\frac{\lambda t}{\mu^2}} $$

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Key Concepts

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

Exponential random variables
Exponential random variables are a cornerstone of probability theory and play a pivotal role in modeling the times between successive events in a Poisson process. This concept is crucial to understand for students who encounter models describing time-to-event data.

An exponential random variable, let's call it X, is characterized by a single parameter, the rate \( \mu \), which defines the average number of events per unit time. The probability density function (pdf) for an exponential random variable is given by \( f(x;\mu) = \mu e^{-\mu x} \) for \( x \geq 0 \) and \( f(x;\mu) = 0 \) for \( x < 0 \). From this, it has a memoryless property, meaning that the probability of an event occurring in the next instant is independent of how much time has already elapsed without the event.

Crucially, the expected value or mean of an exponential random variable is given by \( E[X] = \frac{1}{\mu} \) and the variance by \( Var(X) = \frac{1}{\mu^2} \). Knowing these properties helps us to solve complex problems, such as determining the expected price of a security over time if price changes are determined by exponential factors.
Law of total expectation
A powerful tool in probability theory is the law of total expectation, which can be used to simplify the calculation of expected values by breaking them down into conditional expectations. This law states that if we have a random variable X and another random variable Y, then the expected value of X can be found by taking the expectation of the conditional expectation of X given Y. Mathematically, it is written as \( E[X] = E[E[X|Y]] \).

When applied to complex scenarios such as the calculation of the expected price of a security, conditioned on the number of shocks it receives, the law enables us to first find the expectation assuming a fixed number of shocks and then averaging these conditional expectations with respect to the distribution of shocks.

This law is particularly beneficial when dealing with a sequence of random steps, such as the price changes in the exercise, which are contingent on the occurrence and magnitude of random events. By breaking the problem into smaller, more manageable parts, it can considerably streamline the problem-solving process.
Probability Models
Probability models are mathematical representations of real-world random processes. They come in different forms and serve as the backbone for statistical analysis and prediction. A probability model defines the structure of a process and the rules governing the occurrence of events within that process.

In the case of the security price model presented in the exercise, we combine the concepts of exponential random variables and a Poisson process to model the changes in security prices. A Poisson process is used to model the rate at which 'shocks' happen (events causing price changes), while the exponential random variables represent the multiplicative factors that change the security price when these shocks occur.

Understanding these models allows students to make inferences about the likelihood of certain outcomes, predict future events, and make informed decisions based on probabilistic forecasts鈥攌ey skills in fields like finance, engineering, and economics.

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Most popular questions from this chapter

Customers can be served by any of three servers, where the service times of server \(i\) are exponentially distributed with rate \(\mu_{i}, i=1,2,3\). Whenever a server becomes free, the customer who has been waiting the longest begins service with that server. (a) If you arrive to find all three servers busy and no one waiting, find the expected time until you depart the system. (b) If you arrive to find all three servers busy and one person waiting, find the expected time until you depart the system.

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) \(E[X]=\frac{1}{2 \sqrt{\lambda}}\)

Events occur according to a nonhomogeneous Poisson process whose mean value function is given by $$ m(t)=t^{2}+2 t, \quad t \geqslant 0 $$ What is the probability that \(n\) events occur between times \(t=4\) and \(t=5 ?\)

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 \(\operatorname{Var}(X) ?\)

Events occur acc贸rding to a Poisson process with rate \(\lambda\). Each time an event occurs, we must decide whether or not to stop, with our objective being to stop at the last event to occur prior to some specified time \(T\), where \(T>1 / \lambda\). That is, if an event occurs at time \(t, 0 \leqslant t \leqslant T\), and we decide to stop, then we win if there are no additional events by time \(T\), and we lose otherwise. If we do not stop when an event occurs and no. additional events occur by time \(T\), then we lose. Also, if no events occur by time \(T\), then we lose. Consider the strategy that stops at the first event to occur after some fixed time \(s, 0 \leqslant s \leqslant T\). (a) Using this strategy, what is the probability of winning? (b) What value of \(s\) maximizes the probability of winning? (c) Show that on?'s probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).
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