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Chapter 1: Problem 12

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.

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Most popular questions from this chapter

Let a random variable \(X\) of the continuous type have a pdf \(f(x)\) 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.

Two distinct integers are chosen at random and without replacement from the first six positive integers. Compute the expected value of the absolute value of the difference of these two numbers.

A mode of a distribution of one random variable \(X\) is a value of \(x\) that maximizes the pdf or pmf. For \(X\) of the continuous type, \(f(x)\) must be continuous. If there is only one such \(x\), it is called the mode of the distribution. Find the mode of each of the following distributions: (a) \(p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(f(x)=12 x^{2}(1-x), 0

A coin is tossed two independent times, each resulting in a tail (T) or a head (H). The sample space consists of four ordered pairs: TT, TH, HT, HH. Making certain assumptions, compute the probability of each of these ordered pairs. What is the probability of at least one head?

If \(X\) is a random variable such that \(E(X)=3\) and \(E\left(X^{2}\right)=13\), use Chebyshev's inequality to determine a lower bound for the probability \(P(-2<\) \(X<8)\)
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