/*! 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 29 Let \(X\) and \(Y\) be independe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\), where \(\lambda>\mu .\) Let \(c>0\) (a) Show that the conditional density function of \(X\), given that \(X+Y=c\), is $$ f_{X \mid X+Y}(x \mid c)=\frac{(\lambda-\mu) e^{-(\lambda-\mu) x}}{1-e^{-(\lambda-\mu) c}}, \quad 0

Short Answer

Expert verified
The conditional density function of \(X\) given that \(X+Y=c\) is: $$ f_{X|X+Y}(x|c) = \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}}, \quad 0

Step by step solution

01

Write the individual PDFs for \(X\) and \(Y\)

Since \(X\) and \(Y\) are exponentially distributed, they have the following probability density functions: $$ f_X(x) = \lambda e^{-\lambda x}, \quad x > 0 \\ f_Y(y) = \mu e^{-\mu y}, \quad y > 0 $$
02

Write the joint PDF for \(X\) and \(Y\)

Since \(X\) and \(Y\) are independent, their joint probability density function is the product of their individual PDFs: $$ f_{X,Y}(x,y) = f_X(x)f_Y(y) = \lambda \mu e^{-\lambda x} e^{-\mu y}, \quad x > 0, y > 0 $$
03

Write the conditional density function for \(X\) given \(X+Y=c\)

To find the conditional density function \(f_{X|X+Y}(x|c)\), we need to find the joint probability density function of \(X\) and \(Y\) constrained to the condition \(X+Y=c\). We can do this by finding the Jacobian of the transformation and then integrating out the \(Y\) component. The transformation is: $$ u = x \\ v = x + y $$ Taking the inverse transformation, we get: $$ x = u \\ y = v - u $$ Computing the Jacobian: $$ J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 $$ Now, we can write the constrained joint density function: $$ f_{U,V}(u,v) = f_{X,Y}(u,v-u) |J| = \lambda \mu e^{-\lambda u} e^{-\mu (v-u)} $$ Now, to find the conditional density function \(f_{X|X+Y}(x|c)\), we need to integrate out the \(V\) component, setting \(v=c\): $$ f_{X|X+Y}(x|c) = \frac{f_{U,V}(x,c)}{\int_{0}^{c} f_{U,V}(x,v) dv} $$ Let's compute the integral and simplify the expression: $$ \int_{0}^{c} f_{U,V}(x,v) dv = \lambda \mu e^{-\lambda x} \int_{0}^{c} e^{-\mu (c-x)} dv = \frac{\lambda \mu e^{-(\lambda-\mu)x}(1-e^{-(\lambda-\mu)c})}{\lambda - \mu} $$ Now, we can substitute this back into the conditional density function: $$ f_{X|X+Y}(x|c) = \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}}, \quad 0
04

Find the conditional mean

To find the conditional mean \(E[X|X+Y=c]\), we need to take the expectation with respect to the conditional density function \(f_{X|X+Y}(x|c)\): $$ E[X|X+Y=c] = \int_{0}^{c} x f_{X|X+Y}(x|c) dx $$
05

Evaluate the integral

Let's compute the integral: $$ E[X|X+Y=c] = \int_{0}^{c} x \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}} dx $$ Substitute \(u = (\lambda - \mu)x\) and \(du = (\lambda - \mu)dx\), we get: $$ E[X|X+Y=c] = \frac{1}{(\lambda - \mu)(1-e^{-(\lambda-\mu)c})} \int_{0}^{\lambda c - \mu c} \frac{u}{\lambda-\mu} e^{-u} du $$ Integrating by parts, we have: $$ E[X|X+Y=c] = \frac{1}{(\lambda - \mu)(1-e^{-(\lambda-\mu)c})} \left[-\left(\frac{u}{\lambda-\mu}+1\right)e^{-u}\right]_0^{\lambda c - \mu c} = \frac{c}{\lambda + \mu} $$ Thus, \(E[X|X+Y=c] = \frac{c}{\lambda + \mu}\). #c) Find \(E[Y|X+Y=c]\)
06

Apply conditional expectation

From part (b) we found the conditional expectation of \(X\) given \(X + Y = c\): \(E[X|X+Y=c] = \frac{c}{\lambda + \mu}\). Now, we will use this result to find \(E[Y|X+Y=c]\): $$ E[Y|X+Y=c] = E[c-X|X+Y=c] = c - E[X|X+Y=c] $$
07

Substitute and simplify

Substitute the result from part (b) and simplify: $$ E[Y|X+Y=c] = c-\frac{c}{\lambda+\mu} = \frac{c\mu}{\lambda+\mu} $$ Thus, \(E[Y|X+Y=c] = \frac{c\mu}{\lambda+\mu}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Random Variables
Exponential random variables play a critical role in statistical analysis, particularly in modeling the time until an event occurs, such as the lifespan of an electronic component or the time between customer arrivals at a service center. An exponential random variable, denoted by the symbol 'X' or 'Y' in this context, is defined by a single parameter, often referred to as the rate parameter (denoted as \(\lambda\) or \(\mu\) respectively). This parameter captures the intensity or frequency of the event occurring.

The probability density function (PDF) for an exponential random variable is expressed mathematically as \(f(x) = \lambda e^{-\lambda x}\), with x being the time and \(\lambda > 0\). One characteristic feature of exponential random variables is their memoryless property, meaning the probability of an event occurring in the future is independent of how much time has already elapsed.

When dealing with scenarios involving two independent exponential random variables, such as 'X' and 'Y' in our problem, we consider the cumulative interaction of these two variables. For example, when summing two independent exponential random variables, we may be interested in the new random variable 'Z = X + Y', representing the total time for both events to occur. Understanding the behavior of 'Z', including its conditional density function given 'X + Y = c', where 'c' is constant, provides valuable insights into the combined process' dynamics.
Probability Density Function
The probability density function (PDF) is a fundamental concept used to specify the probability of a continuous random variable falling within a particular range of values. For an exponential random variable with rate parameter \(\lambda\), the PDF is given by \(f_X(x) = \lambda e^{-\lambda x}\) for \(x > 0\), which dictates the likelihood of the variable 'X' being at or near a specific value.

The key purpose of the PDF is to serve as a tool for finding probabilities for continuous variables. For example, to find the probability that 'X' is between two values 'a' and 'b', we would integrate the PDF from 'a' to 'b'. It's important to note that the integral of the PDF across the entire range of the random variable is always equal to 1, representing the certainty that the variable will take on a value within its domain.

In practical terms, when we have two independent random variables with their respective PDFs, as in the case of 'X' and 'Y', we can determine their joint PDF. Since 'X' and 'Y' are independent, the joint PDF is simply the product of their individual PDFs. This mathematical detail is crucial when deriving the conditional density function for 'X' given 'X + Y = c'.
Expectation Conditional
Conditional expectation extends the concept of expected value into the realm of conditioned probabilities. It represents the expected value (or mean) of a random variable given that a certain condition or event has occurred. In the context of our problem, we are interested in \(E[X | X+Y=c]\), which is the expected value of 'X' given that the sum of 'X' and 'Y' is equal to some constant 'c'.

Computing the conditional expectation involves integrating over the conditional density function, which provides a weighted average of all possible values that 'X' can take, given the conditioning event. The expression \(E[X | X+Y=c]\) can be calculated by integrating the product of 'x' with the conditional density function of 'X' over the interval from 0 to 'c'. The power of conditional expectations lies in their ability to provide insights into one variable's behavior when another variable's behavior is known.

As part of the solution improvement advice, a clear understanding of how to apply the process of integration by parts—an essential technique in calculating expectations—is pivotal. When properly applied, this method simplifies the computation of expected values for complex probability distributions, such as those resulting from conditioning one random variable on another.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the times between successive arrivals of customers at a single- server station are independent random variables having a common distribution \(F .\) Suppose that when a customer arrives, he or she either immediately enters service if the server is free or else joins the end of the waiting line if the server is busy with another customer. When the server completes work on a customer, that customer leaves the system and the next waiting customer, if there are any, enters service. Let \(X_{n}\) denote the number of customers in the system immediately before the \(n\) th arrival, and let \(Y_{n}\) denote the number of customers that remain in the system when the \(n\) th customer departs. The successive service times of customers are independent random variables (which are also independent of the interarrival times) having a common distribution \(G\). (a) If \(F\) is the exponential distribution with rate \(\lambda\), which, if any, of the processes \(\left\\{X_{n}\right\\},\left[Y_{n}\right\\}\) is a Markov chain? (b) If \(G\) is the exponential distribution with rate \(\mu\), which, if any, of the processes \(\left\\{X_{n}\right\\},\left\\{Y_{n}\right\\}\) is a Markov chain? (c) Give the transition probabilities of any Markov chains in parts (a) and (b).

A certain scientific theory supposes that mistakes in cell division occur according to a Poisson process with rate \(2.5\) per year, and that an individual dies when 196 such mistakes have occurred. Assuming this theory, find (a) the mean lifetime of an individual, (b) the variance of the lifetime of an individual. Also approximate (c) the probability that an individual dies before age \(67.2\), (d) the probability that an individual reaches age 90 ,

Consider a single server queuing 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.

If an individual has never had a previous automobile accident, then the probability he or she has an accident in the next \(h\) time units is \(\beta h+o(h) ;\) on the other hand, if he or she has ever had a previous accident, then the probability is \(\alpha h+o(h) .\) Find the expected number of accidents an individual has by time \(t\).

Suppose that customers arrive to a system according to a Poisson process with rate \(\lambda\). There are an infinite number of servers in this system so a customer begins service upon arrival. The service times of the arrivals are independent exponential random variables with rate \(\mu\), and are independent of the arrival process. Customers depart the system when their service ends. Let \(N\) be the number of arrivals before the first departure. (a) Find \(P(N=1)\). (b) Find \(P(N=2)\) (c) Find \(P(N=j)\). (d) Find the probability that the first to arrive is the first to depart. (e) Find the expected time of the first departure.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.