/*! 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 3 Let \(X_{1}, X_{2}, X_{3}\), and... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 3

Let \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) be four independent random variables, each with pdf \(f(x)=3(1-x)^{2}, 0y)=P\left(X_{i}>y, i=1, \ldots, 4\right)\).

Short Answer

Expert verified
The CDF of \(Y\) is \(P(Y \leq y) = 1 - (1-y)^{12}, 0 < y < 1\) and the PDF of \(Y\) is \(f_{Y}(y) = 12(1-y)^{11}, 0 < y < 1\).

Step by step solution

01

Define the CDF of \(Y\)

The Cumulative Distribution Function (CDF) of \(Y\), \(P(Y \leq y)\) is equal to \(1 - P(Y > y)\), which by the hint is also equal to \(1 - P(X_{i}>y, i=1, \dots , 4)\). Since the variables \(X_{i}\) are independent, the joint probability is the product of individual probabilities. Hence, \(P(Y \leq y) = 1 - P(X_{1} > y)P(X_{2} > y)P(X_{3} > y)P(X_{4} > y)\).
02

Substituting the given PDF

As \(f(x)=3(1-x)^{2}, 0 < x < 1\), the probability of one variable \(X_{i}\) being greater than \(y\) can be expressed as the integral of the PDF from \(y\) to 1, which is \(\int_{y}^{1} 3(1-x)^{2}\, dx = (1-y)^{3}\). Hence we get \(P(X_{i} > y) = (1-y)^{3}\). Substitute this into the equation for the CDF of \(Y\), we get \(P(Y \leq y) = 1 - (1-y)^{12}\), for \(0 < y < 1\).
03

Determine the PDF of \(Y\)

The Probability Density Function (PDF) of \(Y\) can be found by differentiating the CDF \(P(Y \leq y)\) with respect to \(y\).This gives \(f_{Y}(y) = 12(1-y)^{11}, 0 < y < 1\).

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Most popular questions from this chapter

Let the random variables \(X_{1}\) and \(X_{2}\) have the joint pdf \(f\left(x_{1}, x_{2}\right)=1 / \pi\), for \(\left(x_{1}-1\right)^{2}+\left(x_{2}+2\right)^{2}<1\), zero elsewhere. Find \(f_{1}\left(x_{1}\right)\) and \(f_{2}\left(x_{2}\right) .\) Are \(X_{1}\) and \(X_{2}\) independent?

Let \(A_{1}=\\{(x, y): x \leq 2, y \leq 4\\}, A_{2}=\\{(x, y): x \leq 2, y \leq 1\\}, A_{3}=\) \(\\{(x, y): x \leq 0, y \leq 4\\}\), and \(A_{4}=\\{(x, y): x \leq 0 y \leq 1\\}\) be subsets of the space \(\mathcal{A}\) of two random variables \(X\) and \(Y\), which is the entire two-dimensional plane. If \(P\left(A_{1}\right)=\frac{7}{8}, P\left(A_{2}\right)=\frac{4}{8}, P\left(A_{3}\right)=\frac{3}{8}\), and \(P\left(A_{4}\right)=\frac{2}{8}\), find \(P\left(A_{5}\right)\), where \(A_{5}=\\{(x, y): 0

Let \(f\left(x_{1}, x_{2}\right)=21 x_{1}^{2} x_{2}^{3}, 0

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

Let \(X_{1}, X_{2}\) be two random variables with joint \(\mathrm{pmf} p\left(x_{1}, x_{2}\right)=(1 / 2)^{x_{1}+x_{2}}\), for \(1 \leq x_{i}<\infty, i=1,2\), where \(x_{1}\) and \(x_{2}\) are integers, zero elsewhere. Determine the joint mgf of \(X_{1}, X_{2}\). Show that \(M\left(t_{1}, t_{2}\right)=M\left(t_{1}, 0\right) M\left(0, t_{2}\right)\).
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