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Give a method for simulating a hypergeometric random variable.

Short Answer

Expert verified
To simulate a hypergeometric random variable, follow these steps: 1. Define the parameters: population size (N), number of successes in the population (K), and number of draws without replacement (n). 2. Create a population of N objects, with K successes and N-K failures. 3. Randomly draw n objects from the population without replacement. 4. Count the number of successes in the drawn sample. This count represents the realization of the hypergeometric random variable. Use a programming language like Python and utilize built-in random sampling functions to perform the simulation.

Step by step solution

01

Understand the Hypergeometric Distribution Parameters

A hypergeometric distribution has three parameters: 1. N: The size of the population. 2. K: The number of successes in the population. 3. n: The number of draws (i.e., the sample size).
02

Define the Process for Simulating a Hypergeometric Random Variable

To simulate a hypergeometric random variable, we will follow these steps: 1. Create a population of N objects, with K successes and N-K failures. 2. Randomly draw n objects from the population without replacement. 3. Count the number of successes in the drawn sample. 4. The count of successes represents the realization of the hypergeometric random variable.
03

Use a Programming Language for the Simulation

Choose a programming language to perform the simulation, such as Python. Most programming languages have built-in functions for random sampling that can be utilized for simulating the hypergeometric distribution.
04

Example Simulation of a Hypergeometric Random Variable

In this example, we use Python to simulate a hypergeometric random variable. Consider the following scenario: Population size (N): 50 Number of successes in the population (K): 20 Number of draws without replacement (n): 10 The code to simulate this hypergeometric random variable is as follows: ```python import random # Define the parameters N = 50 K = 20 n = 10 # Create the population population = ['success'] * K + ['failure'] * (N - K) # Draw n samples without replacement from the population sample = random.sample(population, n) # Count the number of successes in the sample hypergeometric_variable = sample.count('success') print("The realization of the hypergeometric random variable is:", hypergeometric_variable) ``` When executing this code, you will get a realization of a hypergeometric random variable based on the defined parameters. Running the code multiple times will give different realizations, illustrating the variability in the outcomes of the hypergeometric distribution.

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Most popular questions from this chapter

Suppose we are able to simulate independent random variables \(X\) and \(Y .\) If we simulate \(2 k\) independent random variables \(X_{1}, \ldots, X_{k}\) and \(Y_{1}, \ldots, Y_{k}\), where the \(X_{i}\) have the same distribution as does \(X\), and the \(Y_{j}\) have the same distribution as does \(Y\), how would you use them to estimate \(P(X

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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 \(\lambda(t) \leqslant \lambda\), then we end up with the approach given in method 1 of Section \(11.5\).

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