/*! 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 Let \(p(x)=\left(\frac{1}{2}\rig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 2

Let \(p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere, be the pmf of the random variable \(X\). Find the mgf, the mean, and the variance of \(X\).

Short Answer

Expert verified
The moment generating function (mgf) of \(X\) is \(M(t) = \frac{e^t}{2-e^t}\), the mean of \(X\) is \(0.5\), and the variance of \(X\) is \(0.75\).

Step by step solution

01

Moment Generating Function Calculation

The moment generating function \(M(t)\) is defined by the expected value of \(e^{tX}\). For a discrete random variable, \(M(t)\) can be calculated as:\n\n\(M(t) = \sum_{x=1}^{\infty}e^{tx}p(x)\)\n\nSubstituting \(p(x) = \left(\frac{1}{2}\right)^{x}\) into the equitation gives us:\n\n\(M(t) = \sum_{x=1}^{\infty}e^{tx}\left(\frac{1}{2}\right)^{x} = \sum_{x=1}^{\infty}\left(\frac{e^t}{2}\right)^{x}.\)\n\nThis is a geometric series with common ratio \(r=\frac{e^t}{2}\), so it converges when \(|r| < 1\), i.e., when \(-\ln(2)<t<\ln(2)\). The sum of the series is then given by:\n\n\(M(t) = \frac{\frac{e^t}{2}}{1-\frac{e^t}{2}} = \frac{e^t}{2-e^t}\).
02

Calculation of the Mean

The mean of a random variable is the first moment about the origin, \(E[X]\), and can be obtained by differentiating the moment generating function with respect to \(t\) and evaluating at \(t=0\), because \(M'(t)|_{t=0} = E[X]\). So, we differentiate \(M(t) = \frac{e^t}{2-e^t}\) to get:\n\n\(M'(t) = \frac{2e^{2t}}{(2-e^t)^2}\).\n\nEvaluating at \(t=0\), we get:\n\n\(M'(0) = E[X] = \frac{2}{4} = \frac{1}{2}\).
03

Calculation of the Variance

The variance of a random variable is the second central moment, \(E[X^2] - (E[X])^2\), and can be obtained by differentiating the moment generating function twice with respect to \(t\) and evaluating at \(t=0\), because \(M''(t)|_{t=0} = E[X^2]\). Differentiating \(M'(t) = \frac{2e^{2t}}{(2-e^t)^2}\) we get:\n\n\(M''(t) = \frac{8e^{3t}}{(2-e^t)^3}\).\n\nEvaluating at \(t=0\), we get \(M''(0) = E[X^2] = 2\). Therefore, the variance is given by:\n\nVar(X) = E[X^2] - (E[X])^2 = 2 - \left(\frac{1}{2}\right)^2 = 2 - \frac{1}{4} = \frac{3}{4}\).\n

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!

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

Let \(f(x)=\frac{1}{3},-1

Let \(X\) equal the number of heads in four independent flips of a coin. Using certain assumptions, determine the pmf of \(X\) and compute the probability that \(X\) is equal to an odd number.

Divide a line segment into two parts by selecting a point at random. Find the probability that the larger segment is at least three times the shorter. Assume a uniform distribution.

Let \(X\) have the exponential pdf, \(f(x)=\beta^{-1} \exp \\{-x / \beta\\}, 0

Let a bowl contain 10 chips of the same size and shape. One and only one of these chips is red. Continue to draw chips from the bowl, one at a time and at random and without replacement, until the red chip is drawn. (a) Find the pmf of \(X\), the number of trials needed to draw the red chip. (b) Compute \(P(X \leq 4)\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

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.