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Chapter 5: Problem 14

Let \(X\) be an exponential random variable with rate \(\lambda .\) (a) Use the definition of conditional expectation to determine \(E[X \mid Xc] P[X>c \mid $$

Short Answer

Expert verified
The conditional expectation \(E[X \mid X<c]\) can be found using both the definition of conditional expectation and the given identity. After performing the computations for both methods, the conditional expectation is: $$ E[X \mid X<c] = \frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right] $$

Step by step solution

01

Define the Conditional PDF

Given that \(X\) follows exponential distribution with rate \(\lambda\), the probability density function of \(X\) is: $$ f_X(x) = \lambda e^{-\lambda x}, x \ge 0 $$ Since we are given \(X < c\) as the condition, we'll have: $$ f_{X\mid X<c}(x) = f_X(x), \text{ for } 0 \le x < c $$
02

Calculate \(P(X

To find the denominator \(P(X<c)\), we need to integrate the PDF over [0, c): $$ P(X<c) = \int_0^c \lambda e^{-\lambda x} dx = 1 - e^{-\lambda c} $$
03

Find the Conditional Expectation

Now plug the conditional PDF and \(P(X<c)\) into the formula for \(E[X\mid X<c]\) : $$ E[X \mid X<c] = \int_{0}^c x \frac{\lambda e^{-\lambda x}}{1 - e^{-\lambda c}} dx = \frac{1}{1 - e^{-\lambda c}} \int_{0}^c x \lambda e^{-\lambda x} dx $$ Using integration by parts, we find that: $$ E[X \mid X<c] = \frac{1}{1 - e^{-\lambda c}} \left[-xe^{-\lambda x} \Big|_{0}^{c} + \int_{0}^c e^{-\lambda x} dx\right] = \frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right] $$ So, the conditional expectation is: $$ E[X \mid X<c] = \frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right] $$ For Part (b),
04

Find Expected Value \(E[X]\)

Given that \(X\) follows an exponential distribution with rate \(\lambda\), we have: $$ E[X] = \frac{1}{\lambda} $$
05

Use Identity to Find Conditional Expectation

By substituting the known values into the given identity: $$ \frac{1}{\lambda} = E[X \mid Xc] P[X>c] $$ We already found \(P(Xc) = e^{-\lambda c}\), we have: $$ \frac{1}{\lambda} = \left(\frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right]\right)(1 - e^{-\lambda c}) + E[X \mid X>c] e^{-\lambda c} $$ Solving for \(E[X \mid X>c]\) and substituting back into the given identity, we find that: $$ E[X \mid X<c] = \frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right] $$ The results from both part (a) and part (b) match, thus verifying the computation.

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Key Concepts

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

Probability Density Function
Understanding the probability density function (PDF) is crucial when dealing with continuous random variables such as the time until a next event in a process, often modelled by an exponential distribution. The PDF describes the likelihood of the random variable taking on a specific value. In mathematical terms, the PDF of an exponential random variable with a rate parameter \(\lambda\) is written as \(f_X(x) = \lambda e^{-\lambda x}\), where \(x \) must be nonnegative since exponential variables cannot take negative values.

The exponential PDF decreases exponentially as \(x\) increases, which intuitively makes sense because in many real-world phenomena, the likelihood of waiting a long time for an event (like a bus arrival) decreases as the waiting time increases. When calculating the conditional expectation, we start by focusing on the interval \( [0, c)\) and within this interval, the PDF remains unchanged but is reinterpreted as the conditional PDF given \(X < c\).

By integrating the PDF, you can derive various probabilities, such as \(P(X
Random Variable
A random variable, often symbolized as \(X\), \(Y\), or \(Z\), is a numerical value resulting from some random process. In this exercise, \(X\) represents the time until an event occurs and follows an exponential distribution. This type of variable is continuous, meaning it can assume an infinite number of values within a given range.

When we talk about \(E[X \mid Xexpected value or the mean of the random variable \(X\) given that it is less than a certain threshold \(c\). This is known as the conditional expectation, which provides us with the average outcome we would expect under these constraints. It integrates the concept of a weighted average into the continuous domain, where the 鈥榳eights鈥� are the probabilities expressed by the PDF.

Random variables allow us to quantify uncertainty and calculate the likelihood of various outcomes in processes that are inherently unpredictable, like radioactive decay, or waiting times in queues. Being able to calculate with these variables, especially their expected values, gives us powerful insight into the behavior and characteristics of such stochastic processes.
Integration by Parts
When you're tackling integrals, especially those involving exponential functions and polynomials such as in our exercise, integration by parts is an effective technique. Putting the method to practice involves recognizing parts of the integrand that can be separately differentiated and integrated. The rule comes from the product rule in differentiation and is formally stated as \(\int u dv = uv - \int v du\).

In the given solution, to calculate \(E[X \mid X
The strength of integration by parts lies in its ability to transform a hard-to-solve integral into smaller, more manageable pieces. However, choosing \(u\) and \(dv\) wisely is key. A good rule of thumb is to pick \(u\) to be the function that gets simpler when differentiated, as we see with \(x\) becoming \(1\) in our example. This tactic can significantly streamline solving integrals, particularly in probability and statistical contexts.

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Most popular questions from this chapter

Teams 1 and 2 are playing a match. The teams score points according to independent Poisson processes with respective rates \(\lambda_{1}\) and \(\lambda_{2} .\) If the match ends when one of the teams has scored \(k\) more points than the other, find the probability that team 1 wins. Hint: Relate this to the gambler's ruin problem.

Events occur according 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 one's probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).

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.

The lifetimes of A's dog and cat are independent exponential random variables with respective rates \(\lambda_{d}\) and \(\lambda_{c} .\) One of them has just died. Find the expected additional lifetime of the other pet.

Let \(\left\\{M_{i}(t), t \geqslant 0\right\\}, i=1,2,3\) be independent Poisson processes with respective rates \(\lambda_{i}, i=1,2\), and set $$ N_{1}(t)=M_{1}(t)+M_{2}(t), \quad N_{2}(t)=M_{2}(t)+M_{3}(t) $$ The stochastic process \(\left\\{\left(N_{1}(t), N_{2}(t)\right), t \geqslant 0\right\\}\) is called a bivariate Poisson process. (a) Find \(P\left[N_{1}(t)=n, N_{2}(t)=m\right\\}\) (b) Find \(\operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right)\)
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