/*! 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 549 Suppose the random vector \((\ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the random vector \((\mathrm{X}, \mathrm{Y})\) is distributed with probability density, \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}+\mathrm{y}\) \(0<\mathrm{x}<1\) and \(=0 \quad 0<\mathrm{y}<1\) otherwise. Find \(E[X Y], E[X+Y]\) and \(E(X)\).

Short Answer

Expert verified
The short answer is: E[XY] = \(\frac{7}{12}\) E[X+Y] = \(\frac{5}{3}\) E[X] = \(\frac{5}{6}\)

Step by step solution

01

Compute the Marginal Probability Density Functions of X and Y

To find the marginal probability density functions of X and Y, we will integrate f(x,y) with respect to the other variable. For the marginal probability density function of X, we integrate f(x,y) with respect to y: \(f_X(x) = \int_{0}^{1} (x + y) dy\) For the marginal probability density function of Y, we integrate f(x,y) with respect to x: \(f_Y(y) = \int_{0}^{1} (x + y) dx\)
02

Evaluate the Integrals to Find the Marginal Density Functions of X and Y

Now let's compute both the integrals: \(f_X(x) = \int_{0}^{1} (x + y) dy = \left[ xy + \frac{1}{2}y^2 \right]_0^1 = x + \frac{1}{2}\) \(f_Y(y) = \int_{0}^{1} (x + y) dx = \left[ \frac{1}{2}x^2 + xy \right]_0^1 = \frac{1}{2} + y\)
03

Compute E[XY], E[X+Y], and E[X] Using the Marginal Density Functions

Now we will use the marginal density functions to compute the expected values: \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) dx dy\) \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) dx dy\) \(E[X] = \int_{0}^{1} x f_X(x) dx\)
04

Evaluate the Integrals to Find the Expected Values

Finally, we compute the integrals to find the expected values: \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) dx dy = \frac{7}{12}\) \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) dx dy = \frac{5}{3}\) \(E[X] = \int_{0}^{1} x (x + \frac{1}{2}) dx = \frac{5}{6}\) Hence, the expected values are: E[XY] = \(\frac{7}{12}\) E[X+Y] = \(\frac{5}{3}\) E[X] = \(\frac{5}{6}\)

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Key Concepts

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

Random Variables
Random variables are fundamental elements in probability theory, used to model numerical outcomes of random processes. They are often denoted by capital letters like \(X\) or \(Y\).
In this article, \(X\) and \(Y\) are random variables related through a joint probability density function (pdf) \(f(x, y)\). This function describes the likelihood of \(X\) and \(Y\) taking particular values within a certain range.
For our specific problem, \(f(x, y) = x + y\) within the range of \(0 < x, y < 1\), providing a structured way to determine how probable certain outcomes are.
It's crucial to understand that while random variables can take on a range of values, the pdf gives a snapshot of their behavior over a defined interval. It's a key tool for calculating probabilities and expected outcomes.
Marginal Probability Density Function
Marginal probability density functions (pdfs) allow us to focus on individual random variables from a joint distribution by integrating over the unrelated variable.
To find the marginal pdf of \(X\), denoted \(f_X(x)\), we integrate the joint pdf \(f(x, y)\) with respect to \(y\) over its entire range. Similarly, for the marginal pdf of \(Y\) (\(f_Y(y)\)), we integrate with respect to \(x\).
Here's how it's done:
  • \(f_X(x) = \int_{0}^{1} (x + y) \, dy\)
  • \(f_Y(y) = \int_{0}^{1} (x + y) \, dx\)
This process strips down the joint distribution to focus on a single variable, summarizing its behavior independently of the other. Marginal pdfs are essential for deriving other statistical measures like expected values.
Expected Value
The expected value, or mean, gives the average outcome of a random variable if an experiment is repeated many times.
For a continuous random variable, it's found using the integral of the variable multiplied by its pdf.
For our problem, expected values are found as follows:
  • \(E[XY]\) uses the joint pdf, evaluating \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) \, dx \, dy\)
  • \(E[X+Y]\) evaluates the sum of the variables: \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) \, dx \, dy\)
  • \(E[X]\) uses the marginal pdf of \(X\): \(E[X] = \int_{0}^{1} x \, f_X(x) \, dx\)
The expected value helps us find the central tendency of our distribution, providing insights into what values the random variables are likely to take.
Integral Calculus
Integral calculus is a mathematical technique used to calculate areas under curves, which is critical in probability for finding probabilities and expected values.
When working with probability density functions, integration helps determine:
  • The total probability over a given interval
  • Marginal pdfs by integrating over the irrelevant variables
  • Expected values by integrating the product of the variable and its pdf
For example, finding \(f_X(x)\) and the expected value involves integrating the relevant functions:
\( \int_{0}^{1} (x+y) \, dy \) or \( \int_{0}^{1} x \, f_X(x) \, dx \).
Mastering integration allows us to transform complex relationships in probability theory into calculated probabilities and expected outcomes, making predictions about random processes.

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Most popular questions from this chapter

Assume you have two populations \(\mathrm{N}\left(\mu_{1}, \sigma^{2}\right)\) and \(\mathrm{N}\left(\mu_{2}, \sigma^{2}\right)\). The distributions have the same, but unknown, variance \(\sigma^{2}\). Derive a method for determining a confidence interval for $$ \mu_{1}-\mu_{2} $$

In testing a hypothesis concerned with the value of a population mean, first the level of significance to be used in the test is specified and then the regions of acceptance and rejection for evaluating the obtained sample mean are determined. If the 1 percent level of significance is used, indicate the percentages of sample means in each of the areas of the normal curve, assuming that the population hypothesis is correct, and the test is two-tailed.

Consider a distribution \(\mathrm{N}\left(\mu, \sigma^{2}\right)\) where \(\mu\) is known but \(\sigma^{2}\) is not. Devise a method of producing a confidence interval for \(\sigma^{2}\)

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Consider the joint distribution of \(\mathrm{X}\) and \(\mathrm{Y}\) given in the form of a table below. The cell (i,j) corresponds to the joint probability that \(\mathrm{X}=\mathrm{i}, \mathrm{Y}=\mathrm{j}\), for \(\mathrm{i}=1,2,3, \mathrm{j}=1,2,3\) $$ \begin{array}{|c|c|c|c|} \hline \mathrm{Y}^{\mathrm{X}} & 1 & 2 & 3 \\ \hline 1 & 0 & 1 / 6 & 1 / 6 \\ \hline 2 & 1 / 6 & 0 & 1 / 6 \\ \hline 3 & 1 / 6 & 1 / 6 & 0 \\ \hline \end{array} $$ Check that this is a proper probability distribution. What is the marginal distribution of \(\mathrm{X} ?\) What is the marginal distribution of \(\mathrm{Y}\) ?

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