/*! 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 26 In Exercise 5.8 , we derived the... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise 5.8 , we derived the fact that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find a. the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. \(P\left(Y_{1} \leq 1 / 2 | Y_{2} \geq 3 / 4\right)\) c. the conditional density function of \(Y_{1}\) given \(Y_{2}=y_{2}\) d. the conditional density function of \(Y_{2}\) given \(Y_{1}=y_{1}\) e. \(P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)\)

Short Answer

Expert verified
a. Marginals: \(f_{Y_1}(y_1)=2y_1\), \(f_{Y_2}(y_2)=2y_2\) b. \(\frac{1}{28}\) c. \(f_{Y_1|Y_2}(y_1|y_2)=2y_1\) d. \(f_{Y_2|Y_1}(y_2|y_1)=2y_2\) e. \(\frac{9}{16}\)

Step by step solution

01

Find Marginal Density for Y1

To find the marginal density function of \(Y_1\), denoted as \(f_{Y_1}(y_1)\), integrate the joint density function \(f(y_1, y_2) = 4y_1y_2\) with respect to \(y_2\) over its range [0,1]. \[ f_{Y_1}(y_1) = \int_{0}^{1} 4y_1y_2\, dy_2 = 4y_1\cdot \frac{1}{2} = 2y_1, \quad 0 \leq y_1 \leq 1 \]
02

Find Marginal Density for Y2

To find the marginal density function of \(Y_2\), denoted as \(f_{Y_2}(y_2)\), integrate the joint density function \(f(y_1, y_2) = 4y_1y_2\) with respect to \(y_1\) over its range [0,1]. \[ f_{Y_2}(y_2) = \int_{0}^{1} 4y_1y_2\, dy_1 = 4y_2\cdot \frac{1}{2} = 2y_2, \quad 0 \leq y_2 \leq 1 \]
03

Calculate Conditional Probability for Part (b)

To find \(P(Y_1 \leq \frac{1}{2} | Y_2 \geq \frac{3}{4})\), we need the conditional distribution. Since \(Y_2 \geq \frac{3}{4}\), consider the range \(\frac{3}{4} \leq y_2 \leq 1\). Calculate: \[ P(Y_1 \leq \frac{1}{2} | Y_2 \geq \frac{3}{4}) = \frac{\int_{3/4}^{1} \int_{0}^{1/2} 4y_1 y_2 \, dy_1 \, dy_2}{\int_{3/4}^{1} \int_{0}^{1} 4y_1 y_2 \, dy_1 \, dy_2} \] Evaluate both integrals: \[ \text{Numerator: } \int_{3/4}^{1} \left( \frac{4y_2}{8} \right) \, dy_2 = \frac{1}{2} \int_{3/4}^{1} y_2 \, dy_2 = \frac{1}{2} \left[ \frac{y_2^2}{2} \right]_{3/4}^{1} = \frac{1}{128} \] \[ \text{Denominator: } \int_{3/4}^{1} \left( 2y_2 \right) \, dy_2 = \left[ \frac{y_2^2}{2} \right]_{3/4}^{1} = \frac{7}{32} \] Therefore, \[ P(Y_1 \leq \frac{1}{2} | Y_2 \geq \frac{3}{4}) = \frac{\frac{1}{128}}{\frac{7}{32}} = \frac{1}{28} \]
04

Find Conditional Density Function of Y1 given Y2

To find the conditional density function \(f_{Y_1|Y_2}(y_1|y_2)\), divide the joint density by the marginal density of \(Y_2\): \[ f_{Y_1|Y_2}(y_1|y_2) = \frac{f(y_1, y_2)}{f_{Y_2}(y_2)} = \frac{4y_1y_2}{2y_2} = 2y_1 \] for \(0 \leq y_1 \leq 1\) and \(0 \leq y_2 \leq 1\).
05

Find Conditional Density Function of Y2 given Y1

To find the conditional density function \(f_{Y_2|Y_1}(y_2|y_1)\), divide the joint density by the marginal density of \(Y_1\): \[ f_{Y_2|Y_1}(y_2|y_1) = \frac{f(y_1, y_2)}{f_{Y_1}(y_1)} = \frac{4y_1y_2}{2y_1} = 2y_2 \] for \(0 \leq y_1 \leq 1\) and \(0 \leq y_2 \leq 1\).
06

Calculate Conditional Probability for Part (e)

For \(P(Y_1 \leq \frac{3}{4} | Y_2 = \frac{1}{2})\), use the conditional density we just found. Evaluate: \[ P(Y_1 \leq \frac{3}{4} | Y_2 = \frac{1}{2}) = \int_{0}^{3/4} f_{Y_1|Y_2}(y_1|1/2) \, dy_1 = \int_{0}^{3/4} 2y_1 \, dy_1 \] Solve this integral: \[ = \left[y_1^2 \right]_0^{3/4} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

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Key Concepts

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

Marginal Density Functions
Understanding marginal density functions is crucial when working with joint probability distributions. When you have a joint probability density function, such as \( f(y_1, y_2) = 4y_1y_2 \), it describes the likelihood of pairs \((y_1, y_2)\) occurring together within defined ranges. However, sometimes we are only interested in the probability of one variable regardless of the other. This is where marginal density functions come in handy.

To find the marginal density function of a single variable, say \( Y_1 \), we integrate the joint density function over the entire range of the other variable. This basically "sums" the probabilities across all values of \( Y_2 \), giving us \( f_{Y_1}(y_1) \).

  • For \( Y_1 \): \( f_{Y_1}(y_1) = \int_{0}^{1} 4y_1y_2 \, dy_2 = 2y_1 \)
  • For \( Y_2 \): \( f_{Y_2}(y_2) = \int_{0}^{1} 4y_1y_2 \, dy_1 = 2y_2 \)
These functions tell us how the chances of \( Y_1 \) and \( Y_2 \) being particular values behave separately.
Conditional Probability
Conditional probability is about finding the probability of an event occurring given that another event has already occurred. In terms of joint distributions, it can tell us the likelihood of one variable meeting a condition when we know something about the other variable.

Here's an example of how you compute that: If you want to determine \( P(Y_1 \leq 1/2 | Y_2 \geq 3/4) \), you start by defining the condition that \( Y_2 \) must be in the range \([3/4, 1]\).

  • First, evaluate the "numerator": \( \int_{3/4}^{1} \int_{0}^{1/2} 4y_1 y_2 \, dy_1 \, dy_2 \)
  • Next, compute the "denominator": \( \int_{3/4}^{1} \int_{0}^{1} 4y_1 y_2 \, dy_1 \, dy_2 \)
  • Divide these results to get the conditional probability.
Thus, you know the chances of \( Y_1 \) staying below \( 1/2 \) when \( Y_2 \) is at least \( 3/4 \), which helps understand how probabilities change in the presence of certain known conditions.
Conditional Density Function
Conditional density functions are key to understanding how one variable behaves given a specific value of another variable. They are derived from joint probability density functions by dividing the joint density by the marginal density of the given condition.

For the conditional density function of \( Y_1 \) given \( Y_2 = y_2 \), you would find:

  • \( f_{Y_1|Y_2}(y_1|y_2) = \frac{f(y_1, y_2)}{f_{Y_2}(y_2)} = \frac{4y_1y_2}{2y_2} = 2y_1 \)


  • Similarly, for \( Y_2 \) given \( Y_1 \), you will compute:

  • \( f_{Y_2|Y_1}(y_2|y_1) = \frac{f(y_1, y_2)}{f_{Y_1}(y_1)} = \frac{4y_1y_2}{2y_1} = 2y_2 \)


  • This shows how to simplify complex joint behaviors by focusing on one variable under a given circumstance of another, making it easier to predict outcomes in conditional contexts.

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    Most popular questions from this chapter

    We considered two individuals who each tossed a coin until the first head appears. Let \(Y_{1}\) and \(Y_{2}\) denote the number of times that persons \(A\) and \(B\) toss the coin, respectively. If heads occurs with probability \(p\) and tails occurs with probability \(q=1-p,\) it is reasonable to conclude that \(Y_{1}\) and \(Y_{2}\) are independent and that each has a geometric distribution with parameter p. Consider \(Y_{1}-Y_{2}\), the difference in the number of tosses required by the two individuals. a. Find \(E\left(Y_{1}\right), E\left(Y_{2}\right),\) and \(E\left(Y_{1}-Y_{2}\right)\) b. Find \(E\left(Y_{1}^{2}\right), E\left(Y_{2}^{2}\right),\) and \(E\left(Y_{1} Y_{2}\right)\) (recall that \(Y_{1}\) and \(Y_{2}\) are independent). c. Find \(E\left(Y_{1}-Y_{2}\right)^{2}\) and \(V\left(Y_{1}-Y_{2}\right)\) d. Give an interval that will contain \(Y_{1}-Y_{2}\) with probability at least \(8 / 9\)

    Suppose that \(Y_{1}\) and \(Y_{2}\) have correlation coefficient \(\rho=.2 .\) What is is the value of the correlation coefficient between a. \(1+2 Y_{1}\) and \(3+4 Y_{2} ?\) b. \(1+2 Y_{1}\) and \(3-4 Y_{2} ?\) c. \(1-2 Y_{1}\) and \(3-4 Y_{2} ?\)

    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 \(Y_{2} ?\)

    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 $$\text { a. } E\left(Y_{1}\right) \text { and } E\left(Y_{2}\right)$$ $$\text { b. } V\left(Y_{1}\right) \text { and } V\left(Y_{2}\right)$$ $$\text { c. } E\left(Y_{1}-3 Y_{2}\right)$$

    In Exercise \(5.16, Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\). Find the expected value of this measure of productivity.

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