/*! 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 4 Let \(X\) be a random variable s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Let \(X\) be a random variable such that \(E\left(X^{m}\right)=(m+1) ! 2^{m}, m=1,2,3, \ldots\). Determine the mgf and the distribution of \(X\).

Short Answer

Expert verified
The moment generating function of \(X\) is \(M_X(t) = 2t + 4t^2 + 6t^3 + \dots\) and its distribution is a scaled Exponential distribution with parameter \(\lambda = 2\).

Step by step solution

01

Identify the mgf

From the problem, the \(m\)th moment of the random variable \(X\) is given by \(E(X^m) = (m+1)!2^m. This is the \(m\)th derivative of the mgf evaluated at \(t=0\). So, by setting \(t=0\) in the Taylor expansion of the mgf \(M_X(t)\), we get: \[M_X(t) = \sum_{m=0}^{\infty} \frac{t^m E(X^m)}{m!} = \sum_{m=0}^{\infty} t^m (m + 1)! 2^m\]
02

Simplify the mgf

We simplify the above expression by re-indexing the sum to start from \(m=1\) instead of \(m=0\) (as the zeroth moment \[E(X^0)\] is always 1 for any random variable \(X\)) and rearranging terms:\[M_X(t) = \sum_{m=1}^{\infty} (2t)^m (m + 1) = 2t + 4t^2 + 6t^3 + \dots\]
03

Recognize the distribution

Now we have the form of the mgf. By looking at standard forms of mgf, we can recognize the type of the distribution. The given mgf matches with that of a scaled Exponential distribution with parameter \(\lambda = 2\). So, \(X\) follows a scaled exponential distribution.

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

Compute the measures of skewness and kurtosis of the Poisson distribution with mean \(\mu\).

Let \(X\) and \(Y\) have a bivariate normal distribution with parameters \(\mu_{1}=\) 20, \(\mu_{2}=40, \sigma_{1}^{2}=9, \sigma_{2}^{2}=4\), and \(\rho=0.6\). Find the shortest interval for which \(0.90\) is the conditional probability that \(Y\) is in the interval, given that \(X=22\).

. Let \(X\) have a conditional Burr distribution with fixed parameters \(\beta\) and \(\tau\), given parameter \(\alpha\). (a) If \(\alpha\) has the geometric pdf \(p(1-p)^{\alpha}, \alpha=0,1,2, \ldots\), show that the unconditional distribution of \(X\) is a Burr distribution. (b) If \(\alpha\) has the exponential pdf \(\beta^{-1} e^{-\alpha / \beta}, \alpha>0\), find the unconditional pdf of \(X\).

Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2} .\) Prove that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

. Let \(X\) be \(N(0,1)\). Use the moment-generating-function technique to show that \(Y=X^{2}\) is \(\chi^{2}(1)\) Hint: \(\quad\) Evaluate the integral that represents \(E\left(e^{t X^{2}}\right)\) by writing \(w=x \sqrt{1-2 t}\), \(t<\frac{1}{2}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

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.