Question

Let \(X\) have the pmf \(p(x)=1 / k, x=1,2,3, \ldots, k\), zero elsewhere. Show that the mgf is $$M(t)=\left\\{\begin{array}{ll}\frac{e^{t}\left(1-e^{k t}\right)}{k\left(1-e^{t}\right)} & t \neq 0 \\ 1 & t=0\end{array}\right.$$

Short answer

The moment generating function \(M(t)\) for the given probability mass function \(p(x) = 1/k\), where \(x = 1,2,3,...,k\) can be directly computed and simplified resulting in the function \(M(t) = \frac{e^{t}(1-e^{kt})}{k(1-e^{t})}\) for \(t \neq 0\), and \(1\) for \(t = 0\).

Step-by-step solution

Define the Problem

The moment-generating function (MGF) of a random variable \(X\) is defined as \(M(t) = E[e^{tX}]\). In this case, the random variable \(X\) has been defined with a probability mass function (pmf) \(p(x) = 1/k\) for \(x=1,2,3,...,k\), and 0 elsewhere. The task is to prove that the MGF of \(X\) has the form, \(M(t)=\frac{e^{t}\left(1-e^{kt}\right)}{k\left(1-e^{t}\right)}\) for \(t \neq 0\) and \(1\) for \(t=0\).

Compute the MGF based on the pmf and Simplify

Substitute values from the pmf into the formula for \(M(t)\). \(M(t) = E[e^{tX}]\) = \(\Sigma_{x=1}^{k} e^{tx} * p(x)\), where \(p(x) = 1/k\) for \(x=1,2,3,...,k\). This equals \(M(t)\) = \(\frac{1}{k}\Sigma_{x=1}^{k} e^{tx}\), which is a geometric series. The sum of the first \(k\) terms in a geometric series \(ar^{n-1}\) is \(\frac{a(r^{k}-1)}{r-1}\). Using \(a=1\), \(r=e^{t}\) and \(n=1\) to \(k\), we obtain the sum as \(\frac{e^{t}\left(e^{kt}-1\right)}{e^{t}-1}\). Finally, cancelling out the \(e^{t}\) terms gives \(M(t) = \frac{e^{t}\left(1-e^{kt}\right)}{k\left(1-e^{t}\right)}\) which is what we were to prove.

Handle special case for \(t=0\)

Now, make sure that \(M(t)\) is defined for \(t=0\). Direct substitution of \(t=0\) into the MGF gives a form \(0/0\), which is undefined. However, using L'Hopital's rule or recognizing the limit form from calculus \(\lim_{t \to 0}\frac{sin(t)}{t} = 1\), we can verify that \(M(t)\) simplifies to \(1\), when \(t=0\).

Key concepts

Probability Mass Function (PMF)

When working with random variables, it's crucial to understand how their probabilities are distributed. The probability mass function (PMF) is the tool we use for discrete random variables. It assigns probabilities to each possible value that the variable can take.

For a discrete random variable \(X\), the PMF, denoted as \(p(x)\), defines the probability that \(X\) takes the specific value of \(x\):

• The PMF ensures that the sum of all probabilities equals 1.
• For example, in the original exercise, \(p(x) = 1/k\) for \(x = 1, 2, 3, \ldots, k\), meaning each outcome is equally likely.

This formula essentially means every chosen value of \(x\) from 1 to \(k\) has the same chance of occurring, making it a uniform distribution. Understanding this concept is vital for calculating the moment-generating function (MGF) in the given problem.

Random Variable

A random variable is a numerical description of the outcomes of a random phenomenon. Think of it as a variable that can take different values, determined by the randomness of an experiment or event.

There are two types of random variables: discrete and continuous. Our focus here is on discrete random variables, which take on a finite or countably infinite set of values. In the exercise, the random variable \(X\) can take on any of the values \(1, 2, 3, \ldots, k\).

• Each of these values corresponds to an event whose probability is given by the PMF.
• Understanding how a random variable behaves helps us compute its MGF, which is derived from these probabilities.

The discrete random variable's behavior is the basis for creating various probability models and predicting possible outcomes, allowing for deeper analysis in the field of probability and statistics.

Geometric Series

A geometric series is a series of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. It's instrumental in solving the exercise at hand.

In the context of the problem, we encounter a geometric series when finding the sum \(\Sigma_{x=1}^{k} e^{tx}\). The formula for the sum of the first \(k\) terms of a geometric series is given by: \[ S_k = a \frac{r^{k}-1}{r-1} \] where \(a\) is the first term and \(r\) is the common ratio. In our case:

• \(a = e^t\) and \(r = e^t\),
• which leads to \(S_k = \frac{e^{t}(e^{kt}-1)}{e^{t}-1}\).

Understanding how geometric series work enables us to simplify expressions and solve for the moment-generating function (MGF), a key aspect of this problem.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool in calculus when dealing with certain types of indeterminate forms. If you encounter a limit of the form \(0/0\) or \(\infty/\infty\), L'Hôpital's rule can be applied. It allows you to find these limits by differentiating the numerator and the denominator separately.

In the context of the exercise, when \(t = 0\), direct substitution in the MGF formula leads us to a \(0/0\) form. Here's where L'Hôpital's Rule becomes invaluable:

• The rule states that \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\), given the limits exist.
• This helps us determine that \(M(t) = 1\) when \(t = 0\).

This concludes our proof and confirms the MGF as defined, ensuring that it's correctly evaluated even at the special case \(t = 0\).