/*! 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

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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 \(\frac{e^{t+1}}{2-e}\), the mean is \(\frac{2}{(2-e)^2}\), and the variance is \(\frac{2}{(2-e)^3}\)

Step by step solution

01

Find the Moment Generating Function (mgf)

Let's denote the moment generating function \(M_x(t)\) of \(X\). Here, \(M_x(t) = E\{e^{tX}\} = \sum_x e^{tx} p(x)\). Let's substitute \(p(x)\) into the formula, we get \(M_x(t) =\sum_{x=1}^\infty e^t (\frac{1}{2})^x = e^{t} \sum_{x=1}^\infty (\frac{e}{2})^x\). Now, let's calculate the sum of the geometric series \(\sum_{x=1}^\infty (a)^x = \frac{a}{1-a}\) where \(a = \frac{e}{2}\), thus \(M_x(t) = \frac{e^t\cdot e/2}{1-e/2} = \frac{e^{t+1}}{2-e}\)
02

Find the mean

The mean or expected value of a random variable \(X\) with mgf \(M(t)\) is given by \(E(X) = M'(0)\), where \(M'(t)\) is the derivative of the mgf. Let's derive the mgf \(M_x(t) = \frac{e^{t+1}}{2-e}\) and evaluate at \(t = 0\), we get \(E(X) = \frac{e}{(2-e)^2}\) evaluated at \(t = 0\) gives us \(E(X) = \frac{2}{(2-e)^2}\)
03

Find the variance

The variance of a random variable \(X\) with mgf \(M(t)\) is given by \(Var(X) = M''(0) - [M'(0)]^2\), where \(M''(t)\) is the second derivative of the mgf. Let's derive \(M'(t)\) to find \(M''(t) = \frac{2e^{t+1}}{(2-e)^3}\) and evaluate at \(t = 0\), we get \(M''(0) = \frac{4}{(2-e)^3}\). Thus, \(Var(X) = \frac{4}{(2-e)^3} - \left(\frac{2}{(2-e)^2}\right)^2 = \frac{2}{(2-e)^3}\)

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