/*! 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 39 A pizza place has two phones. On... [FREE SOLUTION] | 91Ó°ÊÓ

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A pizza place has two phones. On each phone the waiting time until the first call is exponentially distributed with mean one minute. Each phone is not influenced by the other. Let \(X\) be the shorter of the two waiting times and let \(Y\) be the longer. It can be shown that the joint pdf of \(X\) and \(Y\) is \(f(x, y)=2 e^{-(x+y)}, 0

Short Answer

Expert verified
a. \( f_X(x) = 2e^{-2x} \); b. \( e^{-y+2x} \); c. \( 1/e \); d. No. e. \( x+1 \), yes; f. 1.

Step by step solution

01

Determine marginal density of X

The marginal density of \( X \) can be found by integrating the joint probability density function (pdf) \( f(x, y) = 2 e^{-(x+y)} \) over all values of \( y \) from \( x \) to \( \infty \).\[ f_X(x) = \int_x^\infty 2 e^{-(x+y)} \, dy \]Compute the integral:\[ f_X(x) = 2 e^{-x} \int_x^\infty e^{-y} \, dy \]Using the standard integral formula \( \int e^{-y} \, dy = -e^{-y} \), we have:\[ f_X(x) = 2 e^{-x} [-e^{-y}]_x^\infty = 2 e^{-x} (0 - (-e^{-x})) = 2 e^{-2x} \]Thus, the marginal density of \( X \) is \( f_X(x) = 2 e^{-2x} \).
02

Determine the conditional density of Y given X=x

The conditional density \( f_{Y|X}(y|x) \) is given by the joint density \( f(x, y) \) divided by the marginal density \( f_X(x) \):\[ f_{Y|X}(y|x) = \frac{f(x, y)}{f_X(x)} = \frac{2 e^{-(x+y)}}{2 e^{-2x}} \]Simplifying, we find:\[ f_{Y|X}(y|x) = e^{-y + 2x} \]for \( y > x \).
03

Determine the probability that Y > 2 given X = 1

Use the conditional density \( f_{Y|X}(y|1) \) to find the probability:\[ P(Y > 2 | X = 1) = \int_2^\infty e^{-(y-2)} \, dy \]This simplifies and evaluates to:\[ \int_2^\infty e^{-y+2} \, dy = e^2 \int_2^\infty e^{-y} \, dy = e^2 [-e^{-y}]_2^\infty = e^2 (0 - (-e^{-2})) = e^2 e^{-2} = e^0 = 1 \]Thus, \( P(Y > 2 | X = 1) = 1/e \).
04

Evaluate independence of X and Y

Two variables \( X \) and \( Y \) are independent if the joint density \( f(x, y) \) factorizes into a product of the marginal densities \( f_X(x) \) and \( f_Y(y) \). From Step 1, we have the marginal density \( f_X(x) = 2e^{-2x} \), but we know \( f_Y(y) \) depends on \( x \) due to the constraint \( 0 < x < y \).Since we can't express \( f(x,y) = f_X(x)f_Y(y) \), \( X \) and \( Y \) are not independent.
05

Determine the conditional mean of Y given X=x

Given the conditional density \( f_{Y|X}(y|x) = e^{-(y-x)} \), the conditional expected value is:\[ E(Y|X=x) = \int_x^\infty y e^{-(y-x)} \, dy \]Perform the substitution \( u = y - x \), \( du = dy \):\[ E(Y|X=x) = \int_0^\infty (u + x) e^{-u} \, du = x \int_0^\infty e^{-u} \, du + \int_0^\infty u e^{-u} \, du \]Using standard integrals \( \int_0^\infty e^{-u} \, du = 1 \) and \( \int_0^\infty u e^{-u} \, du = 1 \) (mean of exponential distribution):\[ E(Y|X=x) = x + 1 \]So \( E(Y|X=x) = x + 1 \). This is a linear function of \( x \).
06

Determine the conditional variance of Y given X=x

The conditional variance \( Var(Y|X=x) \) is calculated as:\[ Var(Y|X=x) = E(Y^2|X=x) - (E(Y|X=x))^2 \]Compute \( E(Y^2|X=x) = \int_x^\infty y^2 e^{-(y-x)} \, dy \) by using substitution \( u = y - x \):\[ E(Y^2|X=x) = \int_0^\infty (u + x)^2 e^{-u} \, du = \int_0^\infty (u^2 + 2ux + x^2) e^{-u} \, du \]Calculate separately:- \( \int_0^\infty u^2 e^{-u} \, du = 2 \)- \( \int_0^\infty 2ux e^{-u} \, du = 2x \cdot 1 = 2x \)- \( \int_0^\infty x^2 e^{-u} \, du = x^2 \)Thus, \( E(Y^2|X=x) = 2 + 2x + x^2 = x^2 + 2x + 2 \).The conditional variance is:\[ Var(Y|X=x) = x^2 + 2x + 2 - (x+1)^2 = x^2 + 2x + 2 - (x^2 + 2x + 1) = 1 \]So \( Var(Y|X=x) = 1 \).

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Key Concepts

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

Exponential Distribution
An exponential distribution is often used to model the time between events in a Poisson process. It has a constant failure rate and is characterized by the parameter \(\lambda\), which is the rate parameter. In practice, this means if \(\lambda\) is larger, the waiting time for an event is shorter. When the mean waiting time is one, \(\lambda = 1\). For the exponential distribution, the probability density function (pdf) is given by: \[ f(x) = \lambda e^{-\lambda x}, \quad x \geq 0 \] The exponential distribution is memoryless, meaning that the probability of an event occurring in the future is independent of the past. This property is unique for exponential distributions and simplifies many calculations.
  • The pdf decreases exponentially, illustrating the rapid decline in probability as the waiting time increases.
  • It is a continuous distribution, supporting that the time variable is continuous rather than discrete.
Marginal Density
Marginal density refers to the probability distribution of a subset of a collection of random variables. In other words, it is what you get when you look at one variable in isolation. To find the marginal density of a variable, you integrate out the other variables from the joint probability density function (pdf). For example: To find the marginal density of \(X\), we integrate the joint pdf \(f(x,y)\) over all values of \(y\) from \(x\) to infinity: \[ f_X(x) = \int_x^\infty f(x, y) \, dy \] This process simplifies the consideration from two variables \(X\) and \(Y\) to just one, \(X\).
  • Makes calculations more manageable by focusing on less complex distributions.
  • Essential for deriving properties and probabilities involving individual variables from joint behavior.
Conditional Probability
Conditional probability quantifies the probability of an event given that another event has occurred. It is particularly helpful in deriving probabilities when additional information is available. In the context of two random variables, \(X\) and \(Y\), the conditional probability density function \(f_{Y|X}(y|x)\) is calculated using: \[ f_{Y|X}(y|x) = \frac{f(x, y)}{f_X(x)} \] This expression shows that the conditional density is derived by dividing the joint pdf by the marginal pdf of \(X\). By doing so, it adjusts the probability distribution of \(Y\) to reflect the new "condition" that \(X = x\).
  • Enables understanding of the relationship between \(X\) and \(Y\) beyond joint properties.
  • Essential for calculating probabilities under given conditions.
This probability is dynamic as it often involves changes with different conditions that make problem-solving targeted and efficient.
Independence of Random Variables
Two random variables \(X\) and \(Y\) are independent if the occurrence of one does not affect the probability of the other. In terms of joint probability densities, \(X\) and \(Y\) are independent if the joint density \(f(x, y)\) equals the product of their marginal densities \(f_X(x)\) and \(f_Y(y)\): \[ f(x, y) = f_X(x) \cdot f_Y(y) \] If this condition holds, it implies that knowing the outcome of one variable provides no information about the other. Independence greatly simplifies calculations because it allows the separation of variables in probability expressions.
  • Leads to simplified analysis and computation in probability problems.
  • Aids in understanding the interaction (or lack thereof) between variables by comparing joint behavior against combined individual behavior.
In this exercise, \(X\) and \(Y\) are found to not be independent, because their joint pdf does not factor into the product of individual marginal pdfs. This means that one variable influences the distribution of the other, at least partially.

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Most popular questions from this chapter

The result of the previous exercise suggests how observed values of two independent standard normal variables can be generated by first generating their polar coordinates with an exponential rv with \(\lambda=\frac{1}{2}\) and an independent uniform \((0,2 \pi)\) rv: Let \(U_{1}\) and \(U_{2}\) be independent uniform \((0,1)\) rv's, and then let $$ \begin{gathered} Y_{1}=-2 \ln \left(U_{1}\right) \quad Y_{2}=2 \pi U_{2} \\ Z_{1}=\sqrt{Y_{1}} \cos \left(Y_{2}\right) \quad Z_{2}=\sqrt{Y_{1}} \sin \left(Y_{2}\right) \end{gathered} $$ Show that the \(Z_{\mathrm{i}}\) 's are independent standard normal. [Note: This is called the Box-Muller transformation after the two individuals who discovered it. Now that statistical software packages will generate almost instantaneously observations from a normal distribution with any mean and variance, it is thankfully no longer necessary for people like you and us to carry out the transformations just described - let the software do it!]

Let \(X_{1}\) denote the time (hr) it takes to perform a first task and \(X_{2}\) denote the time it takes to perform a second one. The second take always takes at least as long to perform as the first task. The joint pdf of these variables is \(f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{cl}2\left(x_{1}+x_{2}\right) & 0 \leq x_{1} \leq x_{2} \leq 1 \\ 0 & \text { otherwise }\end{array}\right.\) a. Obtain the pdf of the total completion time for the two tasks. b. Obtain the pdf of the difference \(X_{2}-X_{1}\) between the longer completion time and the shorter time.

A class has 10 mathematics majors, 6 computer science majors, and 4 statistics majors. A committee of two is selected at random to work on a problem. Let \(X\) be the number of mathematics majors and let \(Y\) be the number of computer science majors chosen. a. Determine the joint probability mass function \(p(x, y)\). This generalizes the hypergeometric distribution studied in Section 3.6. Give the joint probability table showing all nine values, of which three should be 0 . b. Determine the marginal probability mass functions by summing numerically. How could these be obtained directly? [Hint: What are the univariate distributions of \(X\) and \(Y ?]\) c. Determine the conditional probability mass function of \(Y\) given \(X=x\) for \(x=0,1,2\). Compare with the \(h(y ; 2-x, 6,10)\) distribution. Intuitively, why should this work? d. Are \(X\) and \(Y\) independent? Explain. e. Determine \(E(Y \mid X=x), x=0,1,2\). Do this numerically and then compare with the use of the formula for the hypergeometric mean, using the hypergeometric distribution given in part (c). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? f. Determine \(V(Y \mid X=x), x=0,1,2\). Do this numerically and then compare with the use of the formula for the hypergeometric variance, using the hypergeometric distribution given in part (c).

The number of customers waiting for gift-wrap service at a department store is an rv \(X\) with possible values \(0,1,2,3,4\) and corresponding probabilities \(.1, .2, .3, .25, .15 .\) A randomly selected customer will have 1,2 , or 3 packages for wrapping with probabilities . 6, .3, and .1, respectively. Let \(Y=\) the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). a. Determine \(P(X=3, Y=3)\), that is, \(p(3,3)\). b. Determine \(p(4,11)\).

A circular sampling region with radius \(X\) is chosen by a biologist, where \(X\) has an exponential distribution with mean value \(10 \mathrm{ft}\). Plants of a certain type occur in this region according to a (spatial) Poisson process with "rate" \(.5\) plant per square foot. Let \(Y\) denote the number of plants in the region. a. Find \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\) b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

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