/*! 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 2 Let \(f_{1 \mid 2}\left(x_{1} \m... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 2

Let \(f_{1 \mid 2}\left(x_{1} \mid x_{2}\right)=c_{1} x_{1} / x_{2}^{2}, 0

Short Answer

Expert verified
\(c_{1} = 2x_{2}\), \(c_{2} = 5\), the joint pdf of \(X_{1}\) and \(X_{2}\) is \(2x_{1}x_{2}^{3}\), \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2} \mid X_{2}=\frac{5}{8}\right) = 0.4375\), and \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2}\right) = 0.1302\)

Step by step solution

01

Determine the Constants \(c_{1}\) and \(c_{2}\)

For any pdf, the integral over all possible values must equal 1. Thus, we set up the following integrals to find \(c_{1}\) and \(c_{2}\): \[\int_{0}^{1} c_{2} x_{2}^4 dx_{2} = 1\] Solving, we find that \(c_{2} = 5\). For \(c_{1}\), we have \[\int_{0}^{x_{2}} c_{1} \frac{x_{1}}{x_{2}^{2}} dx_{1} = 1\] which gives \(c_{1} = 2 x_{2}\)
02

Compute the Joint pdf of \(X_{1}\) and \(X_{2}\)

The joint pdf of \(X_{1}\) and \(X_{2}\) can be obtained by multiplying the marginal pdf of \(X_{2}\) (which we have already found) with the conditional pdf of \(X_{1}\) (given \(X_{2} = x_{2}\)), so \(f_{1,2}(x_{1}, x_{2}) = f_{1 \mid 2}(x_{1} \mid x_{2}) f_{2}(x_{2})=2 x_{1} x_{2}^{3}\)
03

Calculate \(P\left(\frac{1}{4}

We want to calculate this specific probability. Plug \(x_{2} = \frac{5}{8}\) into the pdf \(f_{1 \mid 2}(x_{1} \mid x_{2})\), and then integrate from \(\frac{1}{4}\) to \(\frac{1}{2}\), so \[P\left(\frac{1}{4}<X_{1}<\frac{1}{2} \mid X_{2}=\frac{5}{8}\right)=\int_{1/4}^{1/2} f_{1 \mid 2}(x_{1} \mid x_{2}) dx_{1}= 0.4375\]
04

Calculate \(P\left(\frac{1}{4}

This probability can be calculated by integrating the joint pdf \(f_{1,2}(x_{1},x_{2})\) over the appropriate region. Here the region is a strip from \(\frac{1}{4}\) to \(\frac{1}{2}\) in the \(x_{1}\) direction, bounded by \(x_{2}=x_{1}\) and \(x_{2}=1\) in the \(x_{2}\) direction. So we obtain \[P\left(\frac{1}{4}<X_{1}<\frac{1}{2}\right)=\int_{1/4}^{1/2} \int_{x_{1}}^{1} f_{1,2}(x_{1},x_{2}) dx_{2} dx_{1} = 0.1302\]

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Most popular questions from this chapter

Let \(X\) and \(Y\) have the pdf \(f(x, y)=1,0

Let \(X_{1}, X_{2}\), and \(X_{3}\) be three random variables with means, variances, and correlation coefficients, denoted by \(\mu_{1}, \mu_{2}, \mu_{3} ; \sigma_{1}^{2}, \sigma_{2}^{2}, \sigma_{3}^{2}\); and \(\rho_{12}, \rho_{13}, \rho_{23}\), respectively. For constants \(b_{2}\) and \(b_{3}\), suppose \(E\left(X_{1}-\mu_{1} \mid x_{2}, x_{3}\right)=b_{2}\left(x_{2}-\mu_{2}\right)+b_{3}\left(x_{3}-\mu_{3}\right)\) Determine \(b_{2}\) and \(b_{3}\) in terms of the variances and the correlation coefficients.

A fair die is cast at random three independent times. Let the random variable \(X_{i}\) be equal to the number of spots that appear on the \(i\) th trial, \(i=1,2,3\). Let the random variable \(Y\) be equal to \(\max \left(X_{i}\right) .\) Find the cdf and the pmf of \(Y\). Hint: \(P(Y \leq y)=P\left(X_{i} \leq y, i=1,2,3\right)\).

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Let \(X_{1}\) and \(X_{2}\) have the joint pdf \(f\left(x_{1}, x_{2}\right)=15 x_{1}^{2} x_{2}, 0
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