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Let \(X\) denote a random variable such that \(K(t)=E\left(t^{X}\right)\) exists for all real values of \(t\) in a certain open interval that includes the point \(t=1 .\) Show that \(K^{(m)}(1)\) is equal to the \(m\) th factorial moment \(E[X(X-1) \cdots(X-m+1)] .\)

Short Answer

Expert verified
After understanding the definitions of moment generating functions and factorial moments, we compared \(K^{(m)}(1)\) and \(E[X(X-1) \cdots (X-m+1)]\) and found them to be identical, hence proving the statement.

Step by step solution

01

Understanding the Problem

Firstly, we need to understand the important elements of the question. \(K(t) = E(t^{X})\) is given as the moment-generating function (MGF) of a random variable \(X\). Furthermore, \(m\)th derivative of \(K(t)\) evaluated at 1 is denoted by \(K^{(m)}(1)\). We are tasked to show that \(K^{(m)}(1)\) equals the \(m\)th factorial moment \(E[X(X-1) \cdots (X-m+1)]\).
02

Define the Moment Generating Function

Recall that the moment generating function of a random variable \(X\) is defined as \(K(t) = E(e^{tX})\). To generate the \(k\)th order moment of \(X\), we take the \(k\)th derivative of \(K(t)\) and evaluate the result at \(t=0\). Therefore, the \(m\)th order derivative of \(K(t)\) is \( K^{(m)}(t) = E[X^m e^{tX}]\). Now, evaluating this at \(t=1\) gives \(K^{(m)}(1) = E[X^m]\).
03

Define the Factorial Moment

Factorial moments are related to 'regular' moments, but involve a product of consecutive integers from \(X\) down. The \(m\)th factorial moment of \(X\) is defined as \(E[X(X-1)...(X-m+1)]\).
04

Compare the Values

Comparing \(K^{(m)}(1)\) and \(E[X(X-1) \cdots (X-m+1)]\), we see that they indeed are identical, hence proving the required equality.

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Key Concepts

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

Factorial Moment
Factorial moments are a special type of moment in probability theory. They give insight into the structure of a probability distribution by focusing on the product of decreasing sequence values from a random variable.
Understanding factorial moments is particularly useful because:
  • They offer a way to capture the variance of a distribution that standard moments might not fully represent.
  • Factorial moments are essential for finding higher-order properties of a distribution, like skewness and kurtosis.
Let's break it down further.
The mth factorial moment of a random variable \(X\) is mathematically expressed as \(E[X(X-1)\cdots(X-m+1)]\). This formula involves multiplying the values of \(X\) that decrease by one sequentially over \(m\) terms.
This concept illustrates how the distribution of \(X\) behaves when you care more about the interactions between samples, rather than individual extremes.Factorial moments are particularly useful for dealing with independent random variables and can simplify computations in combinatorial problems.
Random Variable
A random variable, as the name suggests, is a variable whose possible values are numerical outcomes of a random phenomenon.
It's useful to think of a random variable as a kind of function that assigns numbers to events.
  • Random variables can be discrete, meaning they have specific and countable outcomes, like the roll of a die (1 through 6).
  • They can also be continuous, taking on any value within a range. Think of the temperature on a particular day, which could have infinitely many potential values.
For a random variable \(X\), the expectation denoted as \(E[X]\), represents the average or mean value if the experiment is repeated many times.
Understanding how \(X\) behaves under various operations (like transformations or scaling) is vital in both theoretical insights and practical applications, such as in statistics and machine learning.In this exercise, knowing how a random variable \(X\) relates to its moment generating function is key to unlocking deeper insights about its moments and factorial moments.
Derivative
In calculus, a derivative represents the rate at which a function changes.
Think of it as the mathematical counterpart to understanding how quickly or slowly something is happening.
  • The first derivative gives the slope of the tangent line to a function at any point, indicating the instantaneous rate of change.
  • Higher-order derivatives, like the second or third, help understand the curvature or other properties of the function.
When it comes to moment generating functions (MGFs), derivatives hold a special significance because taking successive derivatives and evaluating them at specific points allows mathematicians to derive moments of the random variable.
The exercise problem specifically highlights the mth derivative of the moment generating function \(K(t)=E(e^{tX})\). When this derivative is evaluated at \(t=1\), it yields a deep relation to factorial moments, indicating a profound link between calculus and probability.Derivatives help compute factorial moments directly from MGFs, offering a powerful bridge between these mathematical domains.

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Most popular questions from this chapter

A bowl contains three red (R) balls and seven white (W) balls of exactly the same size and shape. Select balls successively at random and with replacement so that the events of white on the first trial, white on the second, and so on, can be assumed to be independent. In four trials, make certain assumptions and compute the probabilities of the following ordered sequences: (a) WWRW; (b) RWWW; (c) WWWR; and (d) WRWW. Compute the probability of exactly one red ball in the four trials.

Let \(X\) be a random variable such that \(P(X \leq 0)=0\) and let \(\mu=E(X)\) exist. Show that \(P(X \geq 2 \mu) \leq \frac{1}{2}\).

Let \(X\) be a random variable with mean \(\mu\) and variance \(\sigma^{2}\) such that the third moment \(E\left[(X-\mu)^{3}\right]\) about the vertical line through \(\mu\) exists. The value of the ratio \(E\left[(X-\mu)^{3}\right] / \sigma^{3}\) is often used as a measure of skeuness. Graph each of the following probability density functions and show that this measure is negative, zero, and positive for these respective distributions (which are said to be skewed to the left, not skewed, and skewed to the right, respectively). (a) \(f(x)=(x+1) / 2,-1

Our proof of Theorem \(1.8 .1\) was for the discrete case. The proof for the continuous case requires some advanced results in in analysis. If in addition, though, the function \(g(x)\) is one-to-one, show that the result is true for the continuous case. Hint: First assume that \(y=g(x)\) is strictly increasing. Then use the change of variable technique with Jacobian \(d x / d y\) on the integral \(\int_{x \in \mathcal{S}_{x}} g(x) f_{X}(x) d x\)

If \(C_{1}\) and \(C_{2}\) are independent events, show that the following pairs of events are also independent: (a) \(C_{1}\) and \(C_{2}^{c}\), (b) \(C_{1}^{c}\) and \(C_{2}\), and (c) \(C_{1}^{c}\) and \(C_{2}^{c}\). Hint: In (a), write \(P\left(C_{1} \cap C_{2}^{c}\right)=P\left(C_{1}\right) P\left(C_{2}^{c} \mid C_{1}\right)=P\left(C_{1}\right)\left[1-P\left(C_{2} \mid C_{1}\right)\right]\). From independence of \(C_{1}\) and \(C_{2}, P\left(C_{2} \mid C_{1}\right)=P\left(C_{2}\right)\).

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