/*! 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} Q1E Let 聽\({\bf{f}}\left( {{{\bf{x}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Q1E (page 787)

Let \({\bf{f}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{ = cg}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) be a joint p.d.f for \(\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{,}}\)each \({x_{2\,}}\)let\({{\bf{h}}_{{\bf{2}}\,}}\left( {{{\bf{x}}_{{\bf{1}}\,}}} \right){\bf{ = g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) that is what we get by considering \({\bf{g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) as a function of \({{\bf{x}}_{{\bf{1}}\,}}\)for fixed \({{\bf{x}}_{2\,}}\)show that there is a multiplicative factor \({{\bf{c}}_{{\bf{2}}\,}}\)that does not depend on such that is the conditional p.d.f of \({{\bf{x}}_{{\bf{1}}\,}}\) given \(\left( {{{\bf{x}}_{{\bf{2}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\)

Short Answer

Expert verified

That is what we get by considering.

\({g_1}\left( {{x_1}\mid {x_2}} \right)\frac{{\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}} = \frac{{cg\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}}.\)

\({x_2}\,\,\,\)is fixed, and for each \({x_2}\,\,\,\)

\({h_2}\left( {{x_1}} \right)\, = g\left( {{x_1},{x_2}} \right),\)

\({c_2} = \frac{c}{{{f_2}\left( {{x_2}} \right)\,}}.\)

Step by step solution

01

Definition of the conditional probability density function

The random variable's probability distribution is updated by taking into account some information that gives rise to a conditional probability distribution. A conditional probability density function can characterize this distribution.

The conditional probability density function of \({X_2}\,\), given that \({X_2} = {x_2}\,\,\)

\({g_1}\left( {{x_1}\mid {x_2}} \right)\frac{{\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}} = \frac{{cg\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}}.\)

\({x_2}\,\,\,\)is fixed, and for each \({x_2}\,\,\,\)

\({h_2}\left( {{x_1}} \right)\, = g\left( {{x_1},{x_2}} \right),\)

The conditional probability density function of \({X_2}\,\),given that \({X_2} = {x_2}\,\,\,\,\)

02

Conditional probability

A conditional probability density function can characterize this distribution.

Then the conditional probability density function of \({X_2}\,\)is

given that \({X_2} = {x_2}\,\,\,\,\)

\({g_1}\left( {{x_1}\mid {x_2}} \right)\,\, = \frac{{cg\left( {{x_1},{x_2}} \right)}}{{{f_2}\left( {{x_2}} \right)\,\,\,\,}}{h_2}\left( {{x_1}} \right)\, = {c_2}\)

\({c_2} = \frac{c}{{{f_2}\left( {{x_2}} \right)\,}}.\)

hence,

\({c_2} = \frac{c}{{{f_2}\left( {{x_2}} \right)\,}}.\)

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 91影视!

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

In Example 12.6.7, let \(\left( {{X^*},{Y^*}} \right)\) be a random draw from the sample distribution\({F_n}\) . Prove that the correlation \({X^*} and {Y^*}\) is R in Eq. (12.6.2).

Use the beef hot dog data in Exercise 7 of Sec. 8.5. Form 10,000 nonparametric bootstrap samples to solve this exercise.

a. Approximate a 90 percent percentile bootstrap confidence interval for the median calorie count in beef hot dogs.

b. Approximate a 90 percent percentile-t bootstrap confidence interval for the median calorie count in beef hot dogs.

c. Compare these intervals to the 90 percent interval formed using the assumption that the data came from a normal distribution.

In Example 12.5.6, we modeled the parameters \({\tau _1}, \ldots {\tau _\pi }\) as i.i.d. having the gamma distribution with parameters \({\alpha _0}\) , and \({\beta _0}.\) We could have added a level to the hierarchical model that would allow the \({\tau _\iota }\) 's to come from a distribution with an unknown parameter. For example, suppose that we model the \({\tau _\iota }\) 's as conditionally independent, having the gamma distribution with parameters \({\alpha _0}\) and \(\beta \) given \(\beta \). Let \(\beta \) be independent of \(\psi \) and \({\mu _1}, \ldots ,{\mu _p}\) with \(\beta \) having the prior distributions as specified in Example 12.5.6.

a. Write the product of the likelihood and the prior as a function of the parameters \({\mu _1}, \ldots ,{\mu _p},{\tau _1}, \ldots ,{\tau _\pi },\psi ,\) \(\beta \).

b. Find the conditional distributions of each parameter given all of the others. Hint: For all the parameters besides \(\beta \), the distributions should be almost identical to those given in Example 12.5.6. It wherever \({\beta _0}.\) appears, of course, something will have to change.

c. Use a prior distribution in which and \({\psi _0} = 170.\) Fit the model to the hot dog calorie data from Example 11.6.2. Compute the posterior means of the four \({\mu _i}'s\) and

\(1/{\tau _i}^\prime s.\)

Let \({x_1},...,{x_n}\) be the observed values of a random sample \(X = \left( {{x_1},...,{x_n}} \right)\) . Let \({F_n}\)be the sample c.d.f. Let \(\,{j_1},...\,,{j_n}\) be a random sample with replacement from the numbers \(\left\{ {1,.....,n} \right\}\) Define\({x_i}^ * = x{j_i}\) for \(i = 1,..,n.\)ashow that \({x^ * } = \left( {{x_1}^ * ,...,{x_n}^ * } \right)\) is an i.i.d. sample from the distribution\({F_n}\)

Use the blood pressure data in Table 9.2 that was described in Exercise 10 of Sec. 9.6. Suppose now that we are not confident that the variances are the same for the two treatment groups. Perform a parametric bootstrap analysis of the sort done in Example 12.6.10. Use v=10,000 bootstrap simulations.

a. Estimate the probability of type I error for a two-sample t-test whose nominal level is \({\alpha _0} = 0.1.\)

b. Correct the level of the two-sample t-test by computing the appropriate quantile of the bootstrap distribution of \(\left| {{U^{(i)}}} \right|.\)

c. Compute the standard simulation error for the quantile in part (b).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation
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.