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Chapter 1: Problem 96

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

Short Answer

Expert verified
\R^{(\infty)}(0)\ is equal to the m-th moment of the distribution about the point \(b\)

Step by step solution

01

Understand the Problem

We are given that X is a random variable such that \(R(t)=E\left(e^{t(x-b)}\right)\) exists for \(-h<t<h\). We are tasked with showing that \(R^{(\infty)}(0)\) is equal to the m-th moment of the distribution about the point \(b\), where \(m\) is a positive integer. We need to apply the mathematical logic used in calculating moments of a distribution.
02

Identify the m-th moment

The m-th moment of the distribution about the point \(b\) is given by \(E\left[(X-b)^m\right]\).
03

Work with the function \(R(t)\)

We notice that \(R(t)\) can be expressed as a Taylor series about \(t=0\), i.e, \(R(t) = R(0) + R'(0)t + \frac{R''(0)t^2}{2!} + \cdots = \sum_{n=0}^{\infty} \frac{R^{(n)}(0)t^n}{n!} \) for \(-h<t<h\).
04

Calculate \(R^{(\infty)}(0)\)

Taking the m-th derivative of \(R(t)\) yields \(R^{(m)}(t) = E\left[ (x-b)^m e^{t(x-b)} \right]\). Setting \(t=0\) gives us \(R^{(m)}(0) = E\left[ (x-b)^m \right]\). Hence, \(R^{(\infty)}(0)\) is equal to the m-th moment of the distribution about point \(b\).
05

Result

Hence, we have shown as required that \(R^{(\infty)}(0)\) is equal to the m-th moment of the distribution about the point \(b\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
In probability theory, a random variable is a fundamental concept that represents a variable whose value is subject to randomness. Think of it as a bridge connecting abstract probability events to tangible outcomes. Random variables can take on different types of values鈥攄iscrete or continuous.
  • Discrete random variables: These have specific, countable outcomes. For instance, the number of heads when flipping a coin three times.
  • Continuous random variables: These can take on any value within a given range. For example, the exact height of students in a class.
In the context of the given problem, we consider a random variable \(X\) and explore the concept of moments, which help summarize the distribution's shape.
Exponential Function in Statistics
The exponential function, denoted as \(e^x\), is a critical component in probability and statistics due to its unique mathematical properties. It is frequently used when dealing with continuous growth or decay processes.
  • Natural exponential: The base \(e\) (approximately 2.718) is an essential constant that appears naturally in many mathematical contexts.
  • Moment generating functions: A function of the form \(E[e^{t(X-b)}]\) is pivotal in finding statistical moments, where \(E\) denotes the expected value.
In our exercise, we utilize the exponential function in the form of \(R(t) = E[e^{t(x-b)}]\) to derive moments from the distribution of the random variable \(X\), using the variable \(t\) to adjust the 'center' of these calculations around a point \(b\).
Taylor Series in Probability
The Taylor series is a mathematical tool that allows us to approximate functions as power series. In probability and statistics, the Taylor series is often used to approximate moment generating functions.
  • Basic form: A Taylor series expands a function \(f(t)\) about a point, typically \(t=0\), as \(f(t) = f(0) + f'(0)t + \frac{f''(0)t^2}{2!} + \ldots\)
  • Connecting to moments: In our problem, this concept helps us express \(R(t)\) as a series, allowing us to extract moments by evaluating derivatives at \(t=0\).
For example, finding \(R^{(m)}(t)\) by differentiating the Taylor series of \(R(t)\), then setting \(t=0\), provides the \(m\)-th moment, \(E[(X-b)^m]\), essentially accessing information about the distribution's characteristics.

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Most popular questions from this chapter

A positive integer from one to six is to be chosen by casting a die. Thus the elements \(c\) of the sample space \(\mathscr{E}\) are \(1,2,3,4,5,6 .\) Let \(C_{1}=\) \(\\{c ; c=1,2,3,4\\}, C_{2}=\\{c ; c=3,4,5,6\\}\). If the probability set function \(P\) assigns a probability of \(\frac{1}{6}\) to each of the elements of \(\mathscr{6}\), compute \(P\left(C_{1}\right), P\left(C_{2}\right)\), \(P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right)\)

Show that the moment-generating function of the random variable \(X\) having the p.d.f. \(f(x)=\frac{1}{3},-1

Let a random variable \(X\) of the continuous type have a p.d.f. \(f(x)\) whose graph is symmetric with respect to \(x=c .\) If the mean value of \(X\) exists, show that \(E(X)=c .\) Hint. Show that \(E(X-c)\) equals zero by writing \(E(X-c)\) as the sum of two integrals: one from \(-\infty\) to \(c\) and the other from \(c\) to \(\infty .\) In the first, let \(y=c-x ;\) and, in the second, \(z=x-c\). Finally, use the symmetry condition \(f(c-y)=f(c+y)\) in the first.

A coin is to be tossed as many times as is necessary to turn up one head. Thus the elements \(c\) of the sample space \(\mathscr{C}\) are \(\mathrm{H}, \mathrm{TH}, \mathrm{TTH}, \mathrm{TTTH}\), and so forth. Let the probability set function \(P\) assign to these elements the respective probabilities \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\), and so forth. Show that \(P(\%)=1\). Let \(C_{1}=\\{c ; c\) is \(H, T H, T T H, T T T H\), or \(T T T T H\\} .\) Compute \(P\left(C_{1}\right) .\) Let \(C_{2}=\) \(\\{c ; c\) is \(T T T T H\) or \(T T T T T H\\} .\) Compute \(P\left(C_{2}\right), P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right\rangle .\)

Let \(f\left(x_{1}, x_{2}, x_{3}\right)=\exp \left[-\left(x_{1}+x_{2}+x_{3}\right)\right], 0
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