Problem 86 - An Introduction to Mathematical Statistics and Its Applications

Question

Let \(X_{1}, X_{2}\), and \(X_{3}\) be three independent negative binomial random variables with pdfs $$ p_{X_{i}}(k)=\left(\begin{array}{c} k-1 \\ 2 \end{array}\right)\left(\frac{4}{5}\right)^{3}\left(\frac{1}{5}\right)^{k-3}, \quad k=3,4,5, \ldots $$ for \(i=1,2,3 .\) Define \(X=X_{1}+X_{2}+X_{3} .\) Find \(P(10 \leq\) \(X \leq 12)\). (Hint: Use the moment-generating functions of \(X_{1}, X_{2}\), and \(X_{3}\) to deduce the pdf of \(X_{.)}\)

Step-by-Step Solution

Answer

The probability that the sum of these three negative binomial random variables is between 10 and 12 is: \(P(10 \leq X \leq 12) = \binom{9}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right) + \binom{10}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right)^2 + \binom{11}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right)^3\)

Step 1: Find the Moment-Generating Function of \(X_i\)

The moment-generating function (MGF) of a single random variable \(X_i\) is given by \( M_{X_i}(t) = E[e^{tX_i}]\). Using the provided PDF for \(X_i\), we can compute the MGF to be \(M_{X_i}(t) = \left(\frac{4}{5 - e^t}\right)^3\).

Step 2: Find the Moment-Generating Function of X

Since \(X_1\), \(X_2\), and \(X_3\) are independent, the moment-generating function of their sum \(X = X_1+ X_2+ X_3\) is the product of their individual MGFs. Therefore, we have \(M_X(t) = (M_{X_1}(t))^3 = \left(\frac{4}{5 - e^t}\right)^9\).

Step 3: Find the Probability Distribution Function (pdf) using the MGF

The PDF of X can be retrieved from its MGF \(M_X(t)\) by using inverse Laplace Transform, which is beyond the scope of high school level. However, from the form of \(M_X(t)\), we can see that X has a negative binomial distribution with parameters r=9 and p=4/5.

Step 4: Compute the Required Probability

We need to calculate the probability \(P(10 \leq X \leq 12)\). We know that \(X\) has negative binomial distribution with parameters \(r=9\) and \( p = 4/5\). We use the formula for negative binomial distribution: \[ p_X(k) = \binom{k-1}{r-1}p^r(1-p)^{k-r} \] Hence, the desired probability is: \(P(10 \leq X \leq 12) = p_X(10) + p_X(11) + p_X(12) = \binom{9}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right) + \binom{10}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right)^2 + \binom{11}{8} \left(\frac{4}{5}\right)^9 \left(\frac{1}{5}\right)^3\]

Key Concepts

Moment-Generating Functions

Understanding the concept of Moment-Generating Functions (MGFs) can be immensely useful when studying probability and statistics. An MGF for a random variable, say \(X\), is defined as \(M_X(t) = E[e^{tX}]\). Here, \(e^{tX}\) is an exponential function where \(t\) is a real number. It transforms the random variable into an exponential scale.
MGFs are incredibly important because they offer a convenient way to derive several statistical properties of a random variable, such as mean and variance, simply by taking derivatives. What’s more, the MGF of a sum of independent random variables is the product of their individual MGFs. This concept greatly simplifies calculations for combined distributions.
In our exercise, each \(X_i\) has the MGF \(\left(\frac{4}{5 - e^t}\right)^3\). Since \(X_1\), \(X_2\), and \(X_3\) are independent, we multiply these MGFs to find the MGF for \(X = X_1 + X_2 + X_3\), which is \(\left(\frac{4}{5 - e^t}\right)^9\). This shows how the MGF not only helps in finding individual properties but is also essential when dealing with sums of random variables.

Probability Distribution Function

A Probability Distribution Function (PDF) assigns probabilities to the possible outcomes of a random variable. It's a fundamental concept in statistics and probability. For discrete random variables, the PDF gives the probability that a random variable equals a specific value, while for continuous variables, it provides the probability density.
In the context of the exercise, the focus is on a negative binomial distribution. The PDF for a negative binomial random variable describes the number of trials needed to achieve a fixed number of successes in a series of independent and identically distributed Bernoulli trials.
For our problem, each \(X_i\) has a common PDF. When these variables are summed, the resulting random variable \(X\) still follows a negative binomial distribution, but with different parameters (\(r=9\) and \(p=4/5\)). The PDF allows us to calculate probabilities for specific outcomes. Using this, we computed \(P(10 \leq X \leq 12)\) with the derived PDF equation:

Random Variables

Random Variables are at the core of probability and statistics. Put simply, a random variable is a mathematical representation of possible outcomes of a random phenomenon. They provide a framework for dealing with uncertainty.
Random variables are categorized into two main types:

• Discrete Random Variables: These can take on a finite or countably infinite set of values, like the result of a dice roll.
• Continuous Random Variables: These can assume any real number within a certain range, like a person's weight.

Understanding these helps in determining the kind of probability distribution function to use.
In our given exercise, \(X_1\), \(X_2\), and \(X_3\) are discrete random variables following a negative binomial distribution. We summed these variables to get \(X\), which also turns out to be a discrete random variable. Recognizing the nature of random variables is crucial in selecting appropriate statistical methods and applying probability rules.