Problem 91 - Applied Statistics and Probability for Engineers

Question

A random variable \(X\) has the discrete uniform distribution $$ f(x)=\frac{1}{m}, \quad x=1,2, \ldots, m $$ (a) Show that the moment-generating function is \(M_{X}(t)=\frac{e^{t}\left(1-e^{t m}\right)}{m\left(1-e^{t}\right)}\) (b) Use \(M_{X}(t)\) to find the mean and variance of \(X\).

Step-by-Step Solution

Answer

The mgf is \( \frac{e^{t}(1-e^{tm})}{m(1-e^{t})} \), mean is \( \frac{m+1}{2} \), and variance is \( \frac{m^2-1}{12} \)."

Step 1: Define the Moment Generating Function

To find the moment-generating function (mgf) of a discrete uniform distribution, which is defined for a random variable \(X\) taking values \(1, 2, \ldots, m\), we start with the formula for the mgf: \(M_X(t) = E[e^{tX}] = \sum_{x=1}^m e^{tx} f(x)\). Since \(f(x) = \frac{1}{m}\) (the probability of each outcome), the mgf becomes \(M_X(t) = \frac{1}{m} \sum_{x=1}^m e^{tx}\).

Step 2: Calculate the Sum

The sum \(\sum_{x=1}^m e^{tx}\) is a geometric series with common ratio \(e^t\) and \(m\) terms. The sum of a geometric series can be expressed as \(\frac{e^{t(m+1)} - e^t}{e^t - 1}\). Thus, substituting it in, \(M_X(t) = \frac{1}{m} \cdot \frac{e^{t(m+1)}-e^t}{e^t-1}\). Simplifying further, we get \(M_X(t) = \frac{e^t(1-e^{tm})}{m(1-e^t)}\).

Step 3: Find the Mean Using MGF

The mean of \(X\), \(E(X)\), can be found by taking the first derivative of \(M_X(t)\) with respect to \(t\) and evaluating at \(t = 0\). Thus, compute \(M'_X(t)\) and find \(M'_X(0)\). Calculating the derivative and setting \(t = 0\), we find \(E(X) = \frac{m+1}{2}\).

Step 4: Find the Variance Using MGF

Variance \(Var(X)\) is found by evaluating \(Var(X) = M''_X(0) - [M'_X(0)]^2\). We must compute the second derivative \(M''_X(t)\) and evaluate it at \(t = 0\). After calculation, the variance is \(Var(X) = \frac{m^2-1}{12}\).

Key Concepts

Discrete Uniform Distribution

A discrete uniform distribution is a type of probability distribution where each outcome in a finite set is equally likely. This means if you have a random variable \(X\), which can take on any of \(m\) distinct values \((1, 2, \ldots, m)\), each of these values has an equal probability of occurring.
For example, when you roll a fair die, the outcome is a discrete uniform distribution. Whether you roll a 1, 2, 3, 4, 5, or 6, each result has the same probability, which is \(\frac{1}{6}\). Now, if we expand this idea to any integer-valued variable from 1 to \(m\), the probability \(f(x)\) for each is \(\frac{1}{m}\).

Discrete uniform distributions are straightforward and ideal for examples involving fairness and equal likelihood among outcomes. Whenever you need to understand scenarios where outcomes are equally possible, such as picking a card from a deck (if you ignore suits), a discrete uniform distribution applies.

Mean and Variance Calculation

To compute the mean \(E(X)\) and variance \(Var(X)\) of a discrete uniform distribution, you can utilize the moment-generating function (MGF). This powerful tool helps you find characteristic properties of a distribution.
For a discrete uniform distribution, the mean is intuitive; it's simply the average of the first and last value in the distribution. Therefore, for our example, it is computed as \(E(X) = \frac{m+1}{2}\). This makes sense if you think of it as the middle value in an evenly spaced sequence.
The variance, on the other hand, measures how spread out the values are from the mean. Using MGF, variance is found via \(Var(X) = M''_X(0) - [M'_X(0)]^2\). For a random variable \(X\) with a discrete uniform distribution from 1 to \(m\), the variance works out to be \(\frac{m^2-1}{12}\). This tells us the average of the squared differences from the mean, giving insight into the distribution's "spread".

Geometric Series

A geometric series plays an important role in finding the moment-generating function of a discrete uniform distribution. Simply put, a geometric series is a series of terms each multiplied by a fixed number, the common ratio. When you're dealing with sums like \(\sum_{x=1}^m e^{tx}\) in the MGF, understanding how to sum a series efficiently can reduce a lot of effort.
The sum of a geometric series \(a + ar + ar^2 + \ldots + ar^{n-1}\) is given by the formula \(\frac{ar^n - a}{r - 1}\). In the context of the MGF, each term is \(e^{tx}\), leading to the common ratio \(r = e^t\). By applying the geometric series sum formula, you can solve for the series \(\sum_{x=1}^m e^{tx}\), which simplifies into useful expressions for further derivations.
Remember, this concept is especially useful in probability and calculus, allowing you to handle infinite and finite series efficiently, especially in contexts like those encountered with MGFs and other statistical tools.