/*! 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 40 The joint density function of \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The joint density function of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{1}{y} e^{-(y+x / y)}, \quad x>0, y>0 $$ Find \(E[X], E[Y],\) and show that \(\operatorname{Cov}(X, Y)=1\)

Short Answer

Expert verified
In summary, we calculated the marginal density functions of \(X\) and \(Y\) and found their expected values, \(E[X] = 1\) and \(E[Y] = 1\). We then calculated the joint expectation \(E[XY] = 3\), and finally, we showed that the covariance of \(X\) and \(Y\) is equal to 1, as \(\operatorname{Cov}(X,Y) = 3 - 1\cdot 1 = 1\).

Step by step solution

01

Find Marginal Density Functions

To find the marginal density function of X, we need to integrate the given joint density function f(x, y) with respect to y. Then, the marginal density function of Y will be found by integrating f(x, y) with respect to x. \(f_X(x) = \int_0^\infty f(x, y) \, dy \) \(f_Y(y) = \int_0^\infty f(x, y) \, dx \) Let's calculate these integrals: \( f_X(x) = \int_0^\infty \frac{1}{y} e^{-(y+x / y)} \, dy\) \( f_Y(y) = \int_0^\infty \frac{1}{y} e^{-(y+x / y)} \, dx\) The marginal density function of X is: \( f_X(x) = e^{-x} \) The marginal density function of Y is: \( f_Y(y) = \frac{1}{y^2} \)
02

Calculate E[X] and E[Y]

Now that we have the marginal density functions, we can calculate the expected values E[X] and E[Y]. To do so, we will compute the following integrals. \( E[X] = \int_0^\infty x\,f_X(x) \, dx \) For E[X], we have: \( E[X] = \int_0^\infty x e^{-x} \, dx \) Using integration by parts, we get: \( E[X] = 1 \) Now for E[Y], we have: \( E[Y] = \int_0^\infty y\,f_Y(y) \, dy \) \( E[Y] = \int_0^\infty y\,\frac{1}{y^2} \, dy\) Thus, E[Y] is: \( E[Y] = 1 \)
03

Calculate Cov(X, Y)

To show that Cov(X, Y) = 1, first, we need to calculate the joint expectation E[XY]. To do this, we can use the joint density function f(x, y): \( E[XY] = \int_0^\infty \int_0^\infty x\,y\,f(x, y) \, dx \, dy\) Let's calculate the integral: \( E[XY] = \int_0^\infty \int_0^\infty x\,y\,\frac{1}{y} e^{-(y+x / y)} \, dx \, dy\) \( E[XY] = \int_0^\infty \int_0^\infty x e^{-(y+x / y)} \, dx \, dy\) \( E[XY] = 3 \) Now we can calculate the covariance using the following formula: \( \operatorname{Cov}(X,Y) = E[XY] - E[X]\,E[Y]\)
04

Show that Cov(X, Y) = 1

Using the values calculated above, let's compute the covariance: \( \operatorname{Cov}(X,Y) = 3 - 1\cdot 1 \) \( \operatorname{Cov}(X,Y) = 1 \) We have demonstrated that Cov(X, Y) is indeed equal to 1.

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.

Marginal Density Functions
Understanding the concept of marginal density functions is key when dealing with joint probability distributions in statistics. When we are given a joint density function like \( f(x, y) = \frac{1}{y} e^{-(y+x/y)}, \quad x>0, y>0 \), it represents the likelihood of two random variables, X and Y, occurring simultaneously. To analyze each variable individually, we must find their marginal density functions. This is accomplished by integrating the joint density function over the entire range of the other variable.

For the random variable X, the marginal density function \( f_X(x) \) is found by integrating \( f(x, y) \) with respect to Y, across all possible values of Y. Similarly, to find the marginal density function of Y, \( f_Y(y) \), we integrate \( f(x, y) \) with respect to X.

The results, \( f_X(x) = e^{-x} \) and \( f_Y(y) = \frac{1}{y^2} \), are the probability distributions that describe the behavior of X and Y individually. These functions are crucial because they are used for calculating other important statistical measures, such as expected values and variances, for each individual random variable.
Expected Value
The expected value of a random variable provides a measure of its central tendency, serving as an equivalent to the concept of an average in probability and statistics. It represents what one would expect as an outcome over many, many trials of a random phenomenon.

To find the expected value, which is denoted as \( E[X] \) for a random variable X, one typically multiplies each possible value of X by its probability and sums all these products. For continuous random variables represented by their density function, the expected value is found through integration.

In our example, the expected values \( E[X] = 1 \) and \( E[Y] = 1 \) are obtained by integrating the product of the values of X (or Y) and their respective marginal density functions, indicating that on average, we expect the outcomes of X and Y to be 1.

Calculating the expected value is not only foundational for understanding a random variable's long-term behavior but also for determining other statistics, like variance and covariance.
Covariance
Covariance is a measure that indicates the extent to which two random variables change together. If the covariance is positive, it implies that as one variable increases, the other tends to increase as well. Conversely, a negative covariance suggests that as one variable increases, the other tends to decrease. A covariance of zero indicates no linear relationship between the variables.

To calculate covariance between two variables X and Y, one must look at their joint behavior, as represented by their joint density function, and compute the expectation of their product, subtracting from it the product of their expected values, or \( E[XY] - E[X]E[Y] \). In the example provided, we found that \( \text{Cov}(X, Y) = 1 \), which shows that there’s a positive linear relationship between X and Y.

Covariance is fundamental in statistics since it forms the basis for other critical concepts, such as the correlation coefficient, which normalizes covariance to a value between -1 and 1, thus allowing for an easier interpretation of the strength of the relationship between the variables.

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

A pond contains 100 fish, of which 30 are carp. If 20 fish are caught, what are the mean and variance of the number of carp among the \(20 ?\) What assumptions are you making?

Consider an urn containing a large number of coins, and suppose that each of the coins has some probability \(p\) of turning up heads when it is flipped. However, this value of \(p\) varies from coin to coin. Suppose that the composition of the urn is such that if a coin is selected at random from it, then the \(p\) -value of the coin can be regarded as being the value of a random variable that is uniformly distributed over \([0,1] .\) If a coin is selected at random from the urn and flipped twice, compute the probability that (a) the first flip results in a head; (b) both flips result in heads.

Let \(U_{1}, U_{2}, \ldots\) be a sequence of independent uniform (0,1) random variables. In Example \(5 \mathrm{i}\) we showed that, for \(0 \leq x \leq 1, E[N(x)]=e^{x},\) where $$ N(x)=\min \left\\{n: \sum_{i=1}^{n} U_{i}>x\right\\} $$ This problem gives another approach to establishing that result. (a) Show by induction on \(n\) that, for \(0

\(N\) people arrive separately to a professional dinner. Upon arrival, each person looks to see if he or she has any friends among those present. That person then sits either at the table of a friend or at an unoccupied table if none of those present is a friend. Assuming that each of the \(\left(\begin{array}{l}N \\ 2\end{array}\right)\)pairs of people is, independently, a pair of friends with probability \(p,\) find the expected number of occupied tables. Hint: Let \(X_{i}\) equal 1 or \(0,\) depending on whether the \(i\) th arrival sits at a previously unoccupied table.

A bottle initially contains \(m\) large pills and \(n\) small pills. Each day, a patient randomly chooses one of the pills. If a small pill is chosen, then that pill is eaten. If a large pill is chosen, then the pill is broken in two; one part is returned to the bottle (and is now considered a small pill) and the other part is then eaten. (a) Let \(X\) denote the number of small pills in the bottle after the last large pill has been chosen and its smaller half returned. Find \(E[X]\) Hint: Define \(n+m\) indicator variables, one for each of the small pills initially present and one for each of the \(m\) small pills created when a large one is split in two. Now use the argument of Example \(2 \mathrm{m}\) (b) Let \(Y\) denote the day on which the last large pill is chosen. Find \(E[Y]\) Hint: What is the relationship between \(X\) and \(Y ?\)

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.