/*! 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 Give a method for simulating a n... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give a method for simulating a negative binomial random variable.

Short Answer

Expert verified
To simulate a negative binomial random variable, follow these steps: 1. Define the parameters: probability of success \(p\) (0 < p ≤ 1) and target number of successes \(r\). Initialize a counter for the number of failures \(F\) and successes \(S\) to 0. 2. Generate a random number \(U\) uniformly distributed between 0 and 1, using a built-in function (e.g., `random.random()` in Python). 3. Compare \(U\) to \(p\). If \(U ≤ p\), increment successes (S = S + 1). If \(U > p\), increment failures (F = F + 1). 4. Check if the target number of successes \(r\) has been reached. If yes, the negative binomial random variable is equal to the number of failures \(F\). 5. If the target number of successes is not reached, go back to Step 2 and repeat until the target number of successes is achieved.

Step by step solution

01

1. Define parameters and necessary variables

First, we need to define the two parameters of the negative binomial distribution: the probability of success \(p\) (0 < p ≤ 1) and the target number of successes \(r\). Additionally, initialize a counter for the number of failures \(F\) and the number of successes \(S\) to 0.
02

2. Generate a random number

Now we will generate a random number \(U\) uniformly distributed between 0 and 1. Many programming languages have built-in functions that can be used for this purpose (e.g., Python uses `random.random()`).
03

3. Compare the generated random number to the probability of success

Next, we will compare the randomly generated number \(U\) to the probability of success \(p\). If \(U\) is less than or equal to \(p\), it means we have achieved a success, so we will increment the number of successes \(S\) by 1 (S = S + 1). If \(U > p\), it means we have encountered a failure, so we will increment the number of failures \(F\) by 1 (F = F + 1).
04

4. Check if the target number of successes has been reached

After updating the number of successes or failures in the previous step, we will check if the target number of successes \(r\) has been reached. If yes, we will stop the simulation, and the negative binomial random variable will be equal to the number of failures, \(F\).
05

5. If the target number is not reached, go back to Step 2

If the target number of successes has not been reached, we will go back to Step 2 and continue the simulation by generating a new random number, comparing it to the probability of success, and updating the number of successes and failures accordingly. This process will be repeated until the target number of successes is achieved.

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 91Ó°ÊÓ!

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

If \(f\) is the density function of a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\), show that the tilted density \(f_{t}\) is the density of a normal random variable with mean \(\mu+\sigma^{2} t\) and variance \(\sigma^{2}\).

Suppose it is relatively easy to simulate from \(F_{i}\) for each \(i=1, \ldots, n\). How can we simulate from (a) \(F(x)=\Pi_{i=1}^{n} F_{i}(x) ?\) (b) \(F(x)=1-\Pi_{i=1}^{n}\left(1-F_{i}(x)\right)\) ? (c) Give two methods for simulating from the distribution \(F(x)=x^{n}\), \(0

Verify that if we use the hazard rate approach to simulate the event times of a nonhomogeneous Poisson process whose intensity function \(\lambda(t)\) is such that \(\lambda(t) \leq \lambda\), then we end up with the approach given in method 1 of Section \(11.5 .\)

Let \(X_{1,+\cdots}, X_{n}\) be independent random variables with \(E\left[X_{i}\right]=\theta\), \(\operatorname{Var}\left(X_{i}\right)=\sigma_{i}^{2}, i=1, \ldots, n\), and consider estimates of \(\theta\) of the form \(\sum_{i=1}^{n} \lambda_{i} X_{i}\) where \(\sum_{i=1}^{n} \lambda_{i}=1 .\) Show that \(\operatorname{Var}\left(\sum_{i=1}^{n} \lambda_{i} X_{i}\right)\) is minimized when \(\lambda_{i}=\left(1 / \sigma_{i}^{2}\right) /\left(\sum_{j=1}^{n} 1 / \sigma_{j}^{2}\right), i=1, \ldots, n\) Possible Hint: If you cannot ?o this for general \(n\), try it first when \(n=2\) The following two problems are concerned with the estimation of \(\int_{0}^{1} g(x) d x=E[g(U)]\) where \(U\) is uniform \((0,1) .\)

For a nonhomogeneous Poisson process with intensity function \(\lambda(t)\), \(t \geq 0\), where \(\int_{0}^{\infty} \lambda(t) d t=\infty\), let \(X_{1}, X_{2}, \ldots\) denote the sequence of times at which events occur. (a) Show that \(\int_{0}^{x_{1}} \lambda(t) d t\) is exponential with rate 1 . (b) Show that \(\int_{X_{i}-1}^{X_{j}} \lambda(t) d t, i \geq 1\), are independent exponentials with rate 1, where \(X_{0}=0\). In words, independent of the past, the additional amount of hazard that must be experienced until an event occurs is exponential with rate \(1 .\)

See all solutions

Recommended explanations on Math Textbooks

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.