/*! 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
The constants \(c_{1}\) and \(c_{2}\) are 2 and 5 respectively. The joint pdf of \(X_{1}\) and \(X_{2}\) is 10x_{1}x_{2}^{2}. The conditional probability \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2} | X_{2}=\frac{5}{8}\right)\) is 0.512. The marginal probability \(P\left(\frac{1}{4}<X_{1}<\frac{1}{2}\right)\) is obtained by calculating the marginal pdf of \(X_{1}\).

Step by step solution

01

- Determine the constants \(c_{1}\) and \(c_{2}\)

These constants are the normalization factors of the pdfs. To find them, integrate each function over their respective range and set the answer equal to 1, and solve for \(c_{1}\) and \(c_{2}\). This is because the integral of a pdf over its range must be 1 by definition. This process gives: \nFor \(c_{1}\): \(\int_{0}^{x_{2}} c_{1} x_{1} / x_{2}^{2} dx_{1} = 1\)\nWhich simplifies to \(c_{1} = 2\)\nFor \(c_{2}\): \(\int_{0}^{1} c_{2} x_{2}^{4} dx_{2} = 1\)\nWhich simplifies to \(c_{2} = 5\)
02

- Determine the joint pdf of \(X_{1}\) and \(X_{2}\)

The joint pdf \(f_{1,2}\(x_{1}, x_{2}\)\) can be obtained using the formula of conditional and marginal pdfs as follows: \(f_{1,2}(x_{1}, x_{2}) = f_{1 | 2}(x_{1} | x_{2})f_{2}(x_{2})\), substituting \(c_{1}\) and \(c_{2}\) derived values gives, \(f_{1,2}(x_{1}, x_{2}) = (2x_{1}/x_{2}^{2}).(5x_{2}^{4}) = 10x_{1}x_{2}^{2}\) for \(0<x_{1}<x_{2}<1\), zero elsewhere.
03

- Calculate the probability \(P\left(\frac{1}{4}

The conditional probability is found by integrating the conditional pdf from \(X_{1} = \frac{1}{4}\) to \(X_{1} = \frac{1}{2}\) for \(X_{2} = \frac{5}{8}\):\n\(\int_{\frac{1}{4}}^{\frac{1}{2}} 2x_{1} / (\frac{5}{8})^{2} dx_{1} = 0.512\)
04

- Calculate the probability \(P\left(\frac{1}{4}

The marginal probability is calculated by first obtaining the marginal pdf of \(X_{1}\) by integrating the joint pdf \(f_{1,2}(x_{1}, x_{2})\) over all possible values of \(X_{2}\) and then use this marginal pdf to compute \(\int_{1/4}^{1/2} f_{1}(x_{1}) dx_{1}\)

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Most popular questions from this chapter

In each case compute the correlation coefficient of \(X\) and \(Y\). Let \(X\) and \(Y\) have the joint pmf described as follows: \begin{tabular}{c|cccccc} \((x, y)\) & \((1,1)\) & \((1,2)\) & \((1,3)\) & \((2,1)\) & \((2,2)\) & \((2,3)\) \\ \hline\(p(x, y)\) & \(\frac{2}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{1}{15}\) & \(\frac{1}{15}\) & \(\frac{4}{15}\) \end{tabular} and \(p(x, y)\) is equal to zero elsewhere. (a) Find the means \(\mu_{1}\) and \(\mu_{2}\), the variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), and the correlation coefficient \(\rho\). (b) Compute \(E(Y \mid X=1), E(Y \mid X=2)\), and the line \(\mu_{2}+\rho\left(\sigma_{2} / \sigma_{1}\right)\left(x-\mu_{1}\right) .\) Do the points \([k, E(Y \mid X=k)], k=1,2\), lie on this line?

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=3 x, 0

Let \(X_{1}, X_{2}\) be two random variables with the joint \(\operatorname{pmf} p\left(x_{1}, x_{2}\right)=\left(x_{1}+\right.\) \(\left.x_{2}\right) / 12\), for \(x_{1}=1,2, \quad x_{2}=1,2\), zero elsewhere. Compute \(E\left(X_{1}\right), E\left(X_{1}^{2}\right), E\left(X_{2}\right)\) \(E\left(X_{2}^{2}\right)\), and \(E\left(X_{1} X_{2}\right) .\) Is \(E\left(X_{1} X_{2}\right)=E\left(X_{1}\right) E\left(X_{2}\right) ?\) Find \(E\left(2 X_{1}-6 X_{2}^{2}+7 X_{1} X_{2}\right)\)

Let \(X_{1}, X_{2}, X_{3}\) be iid with common pdf \(f(x)=e^{-x}, x>0,0\) elsewhere. Find the joint pdf of \(Y_{1}=X_{1}, Y_{2}=X_{1}+X_{2}\), and \(Y_{3}=X_{1}+X_{2}+X_{3}\).

2.6.2. Let \(f\left(x_{1}, x_{2}, x_{3}\right)=\exp \left[-\left(x_{1}+x_{2}+x_{3}\right)\right], 0
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