Recall that a geometric rv has pmf p(x)=p(1-p)^x-1 for x=1,2,3, …In Example 2.12, it was shown t

Recall that a geometric rv has pmf p(x)=p(1-p)^x-1 for...

Recall that a geometric rv has pmf $p(x)=p(1-p)^{x-1}$ for $x=1,2,3, \ldots$ In Example $2.12,$ it was shown that the cdf of this distribution is given by $F(x)=1-(1-p)^{x}$ for positive integers $x .$ (a) Write a program that implements the inverse cdf method to simulate a geometric distribution. Your program should have as inputs the numerical value of $p$ and the desired sample size. (b) Use your program to simulate $10,000$ values from a geometric rv $X$ with $p=.85 .$ From these values, estimate each of the following: $P(X \leq 2), E(X), \operatorname{SD}(X) .$ How do these compare to the corresponding exact values?

Recall that a geometric rv has pmf $p(x)=p(1-p)^{x-1}$ for $x=1,2,3, \ldots$ In Example $2.12,$ it was shown that the cdf of this distribution is given by $F(x)=1-(1-p)^{x}$ for positive integers $x .$
(a) Write a program that implements the inverse cdf method to simulate a geometric distribution. Your program should have as inputs the numerical value of $p$ and the desired sample size.
(b) Use your program to simulate $10,000$ values from a geometric rv $X$ with $p=.85 .$ From these values, estimate each of the following: $P(X \leq 2), E(X), \operatorname{SD}(X) .$ How do these compare to the corresponding exact values?

Key Concepts

Geometric Distribution

The geometric distribution is a discrete probability distribution that models the number of independent Bernoulli trials needed to achieve the first success. It is characterized by a constant probability of success on each trial and has a probability mass function (pmf) of the form p(x) = p(1-p)^(x-1) for x = 1, 2, 3, …, making it particularly useful in scenarios involving waiting times and run lengths.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a probability distribution gives the probability that a random variable is less than or equal to a given value. For the geometric distribution, the CDF can be derived from the pmf and is typically represented as F(x) = 1 - (1-p)^x for positive integers, summarizing the probability of achieving at least one success within x trials.

Inverse CDF Method

The inverse CDF method, also known as inverse transform sampling, is a simulation technique used to generate random numbers from a specified probability distribution. This method relies on the property that if U is a uniform random variable on [0, 1], then X = F^(-1)(U) has the distribution with CDF F. It is particularly useful when the inverse of the CDF can be computed or approximated.

Monte Carlo Simulation

Monte Carlo simulation involves using repeated random sampling to approximate the statistical properties of a distribution. In the context of this problem, simulation is used to generate a sample of values from the geometric distribution, which can then be used to estimate probabilities, expected values, and standard deviations. Such simulation techniques allow for the comparison of sample estimates with exact theoretical values.

Transcript

00:01 In this exercise, we want to simulate the geometric random variable using the inverse cdf method.

00:09 The cdf of the geometric random variable is given by this equation, and using the inverse cdf method, we want to find x such that this inequality holds.

00:31 Another way of stating this is to find x as the biggest integer that is less than or equal to the inverse cdf of u.

00:59 So the process for drawing a random variable from the geometric random variable is first you draw a uniform random variable and then two you find x to satisfy this statement.

01:30 So actually this should be greater than.

01:38 So now if we set f at x equal to you, we have 1 minus 1 minus p to the exponent x equals u, 1 minus p to the exponent x equals u minus 1.

02:11 1 minus p to the exponent x equals 1 minus u.

02:15 And now if we take the base 1 minus p logarithm of both sides, we have, and this is our inverse cdf.

02:44 So now i have made some code in r to achieve the simulation.

02:49 So the code looks like this.

02:57 So we start out by setting the number of simulations to some number, and we define the probability of success.

03:06 And then we run n uniform random variables.

03:11 So u is a vector that contains n uniform random variables.

03:16 And then for each of these random variables, we find the inverse cdf, so we find the value that we want to assign to x.

03:26 So x gives us a vector that is n values.

03:31 Each of these values are a random draw from the geometric random variable with probability success p.

03:38 And there should be one more bracket in here.

03:47 That's part a.

03:50 For part b we were asked to simulate the program using 10 ,000 values and probability of success, 0 .85 to determine three things.

04:00 The probability that x is at most of 2, the expected value for x and the standard deviation for x.

04:07 So i already have the probability of 0 .85 in here.

04:12 And so that would be 10 ,000 simulations...

Please give Ace some feedback