/*! 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 77 In Exercise 5.9 we determined th... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise 5.9 we determined that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find $$\text { a. } E\left(Y_{1}\right) \text { and } E\left(Y_{2}\right)$$ $$\text { b. } V\left(Y_{1}\right) \text { and } V\left(Y_{2}\right)$$ $$\text { c. } E\left(Y_{1}-3 Y_{2}\right)$$

Short Answer

Expert verified
a. \(E(Y_1) = \frac{1}{4}\), \(E(Y_2) = \frac{1}{2}\); b. \(V(Y_1) = \frac{49}{240}\), \(V(Y_2) = \frac{1}{20}\); c. \(E(Y_1 - 3Y_2) = -\frac{5}{4}\).

Step by step solution

01

Verify Joint PDF Validity

First, we will confirm that the given function is a valid joint probability density function (PDF) by integrating it over the entire space. The joint PDF \( f(y_1, y_2) = 6(1-y_2) \) is defined for \( 0 \leq y_1 \leq y_2 \leq 1 \). The integration is done as follows:\[ \int_0^1 \int_0^{y_2} 6(1-y_2) \; dy_1 \, dy_2 \]Integrate with respect to \( y_1 \) first:\[ \int_0^1 [6(1-y_2) y_1]_{y_1=0}^{y_1=y_2} \, dy_2 = \int_0^1 6y_2(1-y_2) \, dy_2 \]Simplify and integrate with respect to \( y_2 \):\[ 6 \int_0^1 (y_2 - y_2^2) \, dy_2 = 6 \left[ \frac{y_2^2}{2} - \frac{y_2^3}{3} \right]_{0}^{1} = 6 \left( \frac{1}{2} - \frac{1}{3} \right) = 6 \times \frac{1}{6} = 1 \]The integral equals 1, confirming a valid joint PDF.
02

Find Marginal PDFs

To find the marginal PDFs, we marginalize the joint PDF over the other variable.For \( Y_1 \):\[ f_{Y_1}(y_1) = \int_{y_1}^1 6(1-y_2) \, dy_2 \]Integrate:\[ = 6 \left[ (y_2 - \frac{y_2^2}{2}) \right]_{y_1}^{1} = 6 \left( 1 - \frac{1}{2} - (y_1 - \frac{y_1^2}{2}) \right) = 3y_1^2 - 6y_1 + 3 \]Now for \( Y_2 \):\[ f_{Y_2}(y_2) = \int_0^{y_2} 6(1-y_2) \, dy_1 \]Integrate:\[ = 6(1-y_2) \int_0^{y_2} \, dy_1 = 6y_2(1-y_2) \]
03

Calculate Expected Value E(Y1)

Find the expected value \( E(Y_1) \) using its marginal PDF:\[ E(Y_1) = \int_0^1 y_1 f_{Y_1}(y_1) \, dy_1 = \int_0^1 y_1(3y_1^2 - 6y_1 + 3) \, dy_1 \]Simplify and integrate:\[ = \int_0^1 (3y_1^3 - 6y_1^2 + 3y_1) \, dy_1 = \left[ \frac{3y_1^4}{4} - 2y_1^3 + \frac{3y_1^2}{2} \right]_{0}^{1} \]Calculate:\[ = \left( \frac{3}{4} - 2 + \frac{3}{2} \right) = \frac{1}{4} \]
04

Calculate Expected Value E(Y2)

Find the expected value \( E(Y_2) \) using its marginal PDF:\[ E(Y_2) = \int_0^1 y_2 f_{Y_2}(y_2) \, dy_2 = \int_0^1 y_2(6y_2(1-y_2)) \, dy_2 \]Expand and integrate:\[ = \int_0^1 (6y_2^2 - 6y_2^3) \, dy_2 = \left[ 2y_2^3 - \frac{3y_2^4}{2} \right]_{0}^{1} \]Calculate:\[ = \left( 2 - \frac{3}{2} \right) = \frac{1}{2} \]
05

Calculate Variance V(Y1)

Calculate the variance of \( Y_1 \):First, find \( E(Y_1^2) \):\[ E(Y_1^2) = \int_0^1 y_1^2 (3y_1^2 - 6y_1 + 3) \, dy_1 = \int_0^1 (3y_1^4 - 6y_1^3 + 3y_1^2) \, dy_1 \]Integrate:\[ = \left[ \frac{3y_1^5}{5} - \frac{3y_1^4}{2} + y_1^3 \right]_{0}^{1} = \frac{3}{5} - \frac{3}{2} + 1 \]Calculate:\[ = \frac{8}{30} = \frac{4}{15} \]Now find variance:\[ V(Y_1) = E(Y_1^2) - (E(Y_1))^2 = \frac{4}{15} - \left(\frac{1}{4}\right)^2 = \frac{4}{15} - \frac{1}{16} = \frac{64}{240} - \frac{15}{240} \]Calculate final result:\[ = \frac{49}{240} \]
06

Calculate Variance V(Y2)

Calculate the variance of \( Y_2 \):First, compute \( E(Y_2^2) \):\[ E(Y_2^2) = \int_0^1 y_2^2 (6y_2(1-y_2)) \, dy_2 = \int_0^1 (6y_2^3 - 6y_2^4) \, dy_2 \]Integrate:\[ = \left[ \frac{3y_2^4}{2} - \frac{6y_2^5}{5} \right]_{0}^{1} = \left( \frac{3}{2} - \frac{6}{5} \right) \]Calculate:\[ = \frac{15}{10} - \frac{12}{10} = \frac{3}{10} \]Now find variance:\[ V(Y_2) = E(Y_2^2) - (E(Y_2))^2 = \frac{3}{10} - \left(\frac{1}{2}\right)^2 = \frac{3}{10} - \frac{1}{4} = \frac{6}{20} - \frac{5}{20} \]Calculate final result:\[ = \frac{1}{20} \]
07

Calculate E(Y1 - 3Y2)

Use the linearity of expectation to find \( E(Y_1 - 3Y_2) \):\[ E(Y_1 - 3Y_2) = E(Y_1) - 3E(Y_2) \]Substitute the expected values:\[ = \frac{1}{4} - 3 \times \frac{1}{2} \]Calculate the result:\[ = \frac{1}{4} - \frac{3}{2} = \frac{1 - 6}{4} = -\frac{5}{4} \]

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Key Concepts

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

Expected Value
The expected value, often called the mean, is a fundamental concept in probability and statistics. It provides a measure of the central tendency of a random variable. In simpler terms, the expected value is the average outcome you'd expect if you were to repeat an experiment infinitely many times.

To compute the expected value for continuous random variables like in this exercise, we use integration. For a random variable \( Y_1 \), its expected value \( E(Y_1) \) is calculated by integrating the product of the variable and its probability density function over its range:\[E(Y_1) = \int_{-fty}^{fty} y_1 f(y_1) \, dy_1\]where \( f(y_1) \) is the marginal probability density function of \( Y_1 \).

In our solution, we first find the marginal probability density functions before calculating \( E(Y_1) \) and \( E(Y_2) \). This process helps us to accurately understand the behavior of each variable independently.
Variance
Variance is a measure of how much a random variable's values deviate from their expected value. It quantifies the spread of a distribution. A larger variance indicates that values are more spread out; a smaller variance means they are closer to the mean. Variance is crucial when assessing the reliability and predictability of random outcomes.

Mathematically, variance of a random variable \( Y_1 \) is defined as:\[V(Y_1) = E(Y_1^2) - (E(Y_1))^2\]The term \( E(Y_1^2) \) is the expected value of the square of the random variable, and \( (E(Y_1))^2 \) is the square of the expected value of the random variable.

In the solved exercise, we determined \( V(Y_1) \) and \( V(Y_2) \) by first finding \( E(Y_1^2) \) and \( E(Y_2^2) \) through integration and then applying the variance formula. The calculations reveal how scattered or concentrated the values can be around the mean.
Marginal Probability Density Function
The marginal probability density function (PDF) describes the probability distribution of a subset of a collection of random variables, disregarding the presence of other variables. In simpler terms, it provides the probability density of a single variable within a multi-variable distribution.

To extract the marginal PDF of a variable, you integrate the joint PDF over the range of the other variable(s). For instance, if you have two variables, \( Y_1 \) and \( Y_2 \), to find the marginal PDF of \( Y_1 \), you integrate the joint PDF \( f(y_1, y_2) \) with respect to \( y_2 \):\[f_{Y_1}(y_1) = \int_{y_1}^{1} f(y_1, y_2) \, dy_2\]This results in a function dependent only on \( y_1 \).

In this exercise, we calculated both \( f_{Y_1}(y_1) \) and \( f_{Y_2}(y_2) \) to help isolate the behaviors of \( Y_1 \) and \( Y_2 \) independently.
Linearity of Expectation
Linearity of expectation is a very useful property in probability theory, particularly because it holds regardless of whether the random variables are independent. This property states that the expected value of a sum of random variables is equal to the sum of their expected values. Formally, for any two random variables \( X \) and \( Y \):\[E(X + Y) = E(X) + E(Y)\]This can be extended to:\[E(aX + bY) = aE(X) + bE(Y)\]where \( a \) and \( b \) are constants.

For the given exercise, linearity of expectation was used to calculate \( E(Y_1 - 3Y_2) \). Instead of needing complex calculations for this combination, we simply took the expected value of \( Y_1 \) and subtracted three times the expected value of \( Y_2 \). This significantly simplifies the problem-solving process and demonstrates the power of this concept.

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Most popular questions from this chapter

Let \(Y_{1}\) and \(Y_{2}\) have a bivariate normal distribution. a. Show that the marginal distribution of \(Y_{1}\) is normal with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\) b. What is the marginal distribution of \(Y_{2} ?\)

A box contains four balls, numbered 1 through 4 . One ball is selected at random from this box. Let \(X_{1}=1\) if ball1or ball2is drawn, \(X_{2}=1\) if ball1or ball3is drawn, \(X_{3}=1\) if ball1or ball4is drawn. The \(X_{i}\) values are zero otherwise. Show that any two of the random variables \(X_{1}, X_{2},\) and \(X_{3}\) are independent but that the three together are not.

An electronic system has one each of two different types of components in joint operation. Let \(Y_{1}\) and \(Y_{2}\) denote the random lengths of life of the components of type 1 and type II, respectively. The joint density function is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} (1 / 8) y_{1} e^{-\left(y_{1}+y_{1}\right) / 2}, & y_{1}>0, y_{2}>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ (Measurements are in hundreds of hours.) Find \(P\left(Y_{1}>1, Y_{2}>1\right)\).

Let \(Y_{1}\) and \(Y_{2}\) be jointly distributed random variables with finite variances. a. Show that \(\left[E\left(Y_{1} Y_{2}\right)\right]^{2} \leq E\left(Y_{1}^{2}\right) E\left(Y_{2}^{2}\right) .\) [Hint: Observe that \(E\left[\left(t Y_{1}-Y_{2}\right)^{2}\right] \geq 0\) for any real number t or, equivalently, $$t^{2} E\left(Y_{1}^{2}\right)-2 t E\left(Y_{1} Y_{2}\right)+E\left(Y_{2}^{2}\right) \geq 0$$ This is a quadratic expression of the form \(A t^{2}+B t+C\); and because it is nonnegative, we must have \(B^{2}-4 A C \leq 0 .\) The preceding inequality follows directly.] b. Let \(\rho\) denote the correlation coefficient of \(Y_{1}\) and \(Y_{2} .\) Using the inequality in part (a), show that \(\rho^{2} \leq 1\)

A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three possible exits. Exit 1 presents a reward of food, but exits 2 and 3 do not. (If the rat eventually selects exit 1 almost every time, learning may have taken place.) Let \(Y_{i}\) denote the number of times exit \(i\) is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run. a. Find the probability that \(n=6\) runs result in \(Y_{1}=3, Y_{2}=1,\) and \(Y_{3}=2\). b. For general \(n\), find \(E\left(Y_{1}\right)\) and \(V\left(Y_{1}\right)\). c. Find \(\operatorname{Cov}\left(Y_{2}, Y_{3}\right)\) for general \(n\). d. To check for the rat's preference between exits 2 and \(3,\) we may look at \(Y_{2}-Y_{3} .\) Find \(E\left(Y_{2}-Y_{3}\right)\) and \(V\left(Y_{2}-Y_{3}\right)\) for general \(n\).

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