/*! 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 7 Let \(F\) be the distribution fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 7

Let \(F\) be the distribution function $$F(x)=x^{n} \quad 0

Short Answer

Expert verified
(a) For simulating a random variable with distribution F using the inversion method, first find the inverse function F鈦宦(y) = y^(1/n) for 0 < y < 1. Generate a uniformly distributed random number Y in the interval [0, 1] and compute X = F鈦宦(Y) = Y^(1/n). (b) Since U鈧, U鈧, ..., U鈧 are independent random numbers, P{max(U鈧, U鈧, ..., U鈧) 鈮 x} = x * x * ... * x = x鈦. (c) To simulate a random variable X with distribution F, generate n independent uniformly distributed random numbers U鈧, U鈧, ..., U鈧 in the interval [0, 1], and calculate X = max{U鈧, U鈧, ..., U鈧檥. The random variable X has the desired distribution F(x) = x鈦, 0 < x < 1.

Step by step solution

01

(a) Inversion method for generating a random variable with distribution F

To generate a random variable that follows a given distribution function F(x), we can use the inversion method. First, we find the inverse function of F(x), denoted as F鈦宦(y). Then, we generate a uniformly distributed random number Y in the interval [0, 1] and evaluate F鈦宦(Y) to get our random variable X with the desired distribution. 1. Compute the inverse function of F(x): F(x) = y = x鈦, where 0 < x < 1 and 0 < y < 1. Now, we solve for x: x = y^(1/n) So, the inverse function is F鈦宦(y) = y^(1/n) for 0 < y < 1. 2. Generate a uniformly distributed random number Y in the interval [0, 1]. 3. Compute X = F鈦宦(Y) = Y^(1/n) X will have the desired distribution F.
02

(b) Probability of maximum of independent random numbers

We want to show that the probability P{max(U鈧, U鈧, ..., U鈧) 鈮 x} = x鈦. Since U鈧, U鈧, ..., U鈧 are independent random numbers, we can rewrite the probability as: P{max(U鈧, U鈧, ..., U鈧) 鈮 x} = P{U鈧 鈮 x, U鈧 鈮 x, ..., U鈧 鈮 x} = P{U鈧 鈮 x} * P{U鈧 鈮 x} * ... * P{U鈧 鈮 x} Since U鈧, U鈧, ..., U鈧 are uniformly distributed between 0 and 1: P{Ui 鈮 x} = x for all i = 1, 2, ..., n. Now, the probability becomes: P{max(U鈧, U鈧, ..., U鈧) 鈮 x} = x * x * ... * x = x鈦
03

(c) Second method for simulating a random variable with distribution F

According to the result in part (b), if we generate n independent uniformly distributed random numbers U鈧, U鈧, ..., U鈧 in the interval [0, 1], and take the maximum of them, the resulting maximum value follows the distribution F(x) = x鈦, 0 < x < 1. To simulate a random variable X with distribution F, do the following: 1. Generate n independent uniformly distributed random numbers U鈧, U鈧, ..., U鈧 in the interval [0, 1]. 2. Calculate X = max{U鈧, U鈧, ..., U鈧檥. 3. The random variable X has the desired distribution F(x) = x鈦, 0 < x < 1.

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!

Key Concepts

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

Random Variable Simulation
Simulating random variables is key in many statistical applications. To create a random variable matching a specific distribution, we rely on different techniques. One common method is the inversion method, helpful when dealing with continuous distributions, like in our exercise.
The process starts with a known distribution function, such as \( F(x) = x^n \). We then compute its inverse \( F^{-1}(y) \). The inversion method involves these steps:
  • First, determine the inverse of the function \( F(x) \). For our function, this becomes \( x = y^{1/n} \).
  • Next, generate a random number \( Y \) uniformly distributed on the interval [0,1].
  • Finally, calculate \( X = F^{-1}(Y) = Y^{1/n} \). This \( X \) is the random variable that follows the desired distribution.
Through this process, a uniformly distributed random number is transformed into a random variable with the specified distribution.
Independent Random Numbers
Independent random numbers play an essential role in probability theory and simulations. Independent numbers mean the outcome of one random event does not influence another.
For instance, with independent random numbers \( U_1, U_2, \ldots, U_n \), certain probabilities can be simplified. Given these numbers are independent and uniformly distributed on [0,1], we find a useful probability:
  • The probability \( P(\max(U_1, U_2, \ldots, U_n) \leq x) \) simplifies to \( x^n \).
  • This equation emerges because \( P(U_i \leq x) \) equals \( x \) for all \( i \), leading to \( P(U_1 \leq x) \cdot P(U_2 \leq x) \cdot \ldots \cdot P(U_n \leq x) = (x)^n \).
This principle is crucial when simulating random variables based on multiple independent uniform distributions.
Uniform Distribution
The uniform distribution is a foundational concept in probability and statistics, central to simulations and analyses across various fields. A random number generated from a uniform distribution means every number within the specified range is equally likely.
In the interval [0,1], generating independent uniform random numbers assists in a broad array of simulations.
  • Each number \( U \) in [0,1] has a probability of \( U_i \leq x \), which can be written as \( x \).
  • This uniformity allows us to create complex distributions by transforming these basic uniform random variables.
By understanding how uniform distribution functions, we can efficiently simulate complex systems, manipulate probability distributions, and predict outcomes by harnessing these basic principles of uniform probability.

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

The following algorithm will generate a random permutation of the elements \(1,2, \ldots, n .\) It is somewhat faster than the one presented in Example \(1 \mathrm{a}\) but is such that no position is fixed until the algorithm ends. In this algorithm, \(P(i)\) can be interpreted as the element in position \(i\) Step 1. Set \(k=1\) Step \(2 .\) Set \(P(1)=1\) Step \(3 .\) If \(k=n,\) stop. Otherwise, let \(k=k+1\) Step 4. Generate a random number \(U\) and let $$\begin{aligned}P(k) &=P([k U]+1) \\\P([k U]+1) &=k\end{aligned}$$ Go to step 3 (a) Explain in words what the algorithm is doing. (b) Show that at iteration \(k\) - that is, when the value of \(P(k)\) is initially \(\operatorname{set}-P(1), P(2), \ldots\) \(P(k)\) is a random permutation of \(1,2, \ldots, k\) Hint: Use induction and argue that $$\begin{array}{l}P_{k}\left\\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\right\\} \\ \quad=P_{k-1}\left\\{i_{1}, i_{2}, \ldots, i_{j-1}, i, i_{j}, \ldots, i_{k-2}\right\\} \frac{1}{k} \\ \quad=\frac{1}{k !} \text { by the induction hypothesis } \end{array}$$

Let \((X, Y)\) be uniformly distributed in the circle of radius 1 centered at the origin. Its joint density is thus $$f(x, y)=\frac{1}{\pi} \quad 0 \leq x^{2}+y^{2} \leq 1$$Let \(R=\left(X^{2}+Y^{2}\right)^{1 / 2}\) and \(\theta=\tan ^{-1}(Y / X)\) denote the polar coordinates of \((X, Y) .\) Show that \(R\) and \(\theta\) are independent, with \(R^{2}\) being uniform on (0,1) and \(\theta\) being uniform on \((0,2 \pi)\)

Let \(X\) be a random variable on (0,1) whose density is \(f(x) .\) Show that we can estimate \(\int_{0}^{1} g(x) d x\) by simulating \(X\) and then taking \(g(X) / f(X)\) as our estimate. This method, called importance sampling, tries to choose \(f\) similar in shape to \(g,\) so that \(g(X) / f(X)\) has a small variance.

Use the inverse transformation method to present an approach for generating a random variable from the Weibull distribution $$F(t)=1-e^{-a t^{\beta}} \quad t \geq 0$$

Give a technique for simulating a random variable having the probability density function $$f(x)=\left\\{\begin{array}{ll}\frac{1}{2}(x-2) & 2 \leq x \leq 3 \\ \frac{1}{2}\left(2-\frac{x}{3}\right) & 3
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

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.