Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(X\) be a random variable such that \(R(t)=E\left(e^{t(X-b)}\right)\) exists for \(t\) such that \(-h

Short Answer

Expert verified
The mth derivative of the Moment Generating Function (MGF) of X-b at t=0 is equal to the mth moment of the distribution about the point b, which can be shown by taking the derivatives of the MGF and understanding the properties of the MGF and moments.

Step by step solution

01

Understand Moment Generating Function

We know that the Moment Generating Function (MGF) for a random variable \(X\) is represented by \(M(t) = E(e^{tX})\). For random variable \(X\) shifted by an amount \(b\), the MGF \(R(t) = E(e^{t(X-b)})\).
02

Derive the Moment Generating Function

We are given R(t), the MGF of (X-b). In order to find the mth moment, we need to take the mth derivative of R(t) with respect to \(t\)
03

Apply Principles of Derivative

The \(m\)th derivative of a MGF evaluated at \(t=0\) gives the \(m\)th moment about zero. Mathematically, that can be represented as \(R^{(m)}(0)=E[(X-b)^{m}]\).
04

Define Moment about the Point

Here, E[(X-b)^m] represents the \(m\)th moment of the distribution about the point \(b\). This justifies our statement given in the problem.

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 Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free