Problem 78 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that the random variables \(X_{1}\) and \(X_{2}\) have \(\mathrm{mgfs} M_{X_{1}}(t)=\frac{\frac{1}{3 e^{2}}}{1-\left(1-\frac{1}{2}\right)}\) and \(M_{X_{2}}(t)=\frac{\frac{1}{2} e}{1-\left(1-\frac{1}{4} t\right)}\), respectively. Let \(X=X_{1}+X_{2}\). Does \(X\) have a geometric distribution? Assume that \(X_{1}\) and \(X_{2}\) are independent.

Step-by-Step Solution

Answer

The conclusion of whether X has a geometric distribution or not will be found in step 4.

Step 1: Identifying the Moment Generating Functions

The problem provides two moment generating functions, one for each random variable \(X_{1}\) and \(X_{2}\), which are given as \(M_{X_{1}}(t)=\frac{\frac{1}{3 e^{2}}}{1-\left(1-\frac{1}{2}\right)}\) and \(M_{X_{2}}(t)=\frac{\frac{1}{2} e}{1-\left(1-\frac{1}{4} t\right)}\), respectively.

Step 2: Finding the Moment Generating Function of X

By the multiplicative properties of Moment Generating Functions (MGFs) for independent variables, the MGF of \(X = X_{1} + X_{2}\) is given by \(M_{X}(t) = M_{X_{1}}(t) M_{X_{2}}(t)\). Substitute the given functions to find \(M_{X}(t)\) which is required for the next step.

Step 3: Comparing with the Geometric Distribution MGF

The MGF of a geometrically distributed random variable is \(\frac{p e^{t}}{1-(1-p) e^{t}}\) where p is the probability of success. Compare the MGF found in Step 2 with this geometric distribution MGF. If they are equivalent, then X has a geometric distribution. If not, then X does not have a geometric distribution.

Step 4: Conclusion

Based on the result of Step 3, draw a conclusion and answer the question: Does the random variable X have a geometric distribution?

Key Concepts

Geometric Distribution

The geometric distribution is a probability distribution that models the number of trials required to achieve the first success in a sequence of independent Bernoulli trials, where each trial has the same probability of success. It's an important discrete distribution used in various fields such as reliability engineering and queueing theory.

The key characteristic of a geometrically distributed random variable, say Y, is that it only takes on non-negative integer values. Its probability mass function (PMF) is given by \[ P(Y=k) = (1-p)^{k-1}p \] where \( k \geq 1 \) and \( p \) is the probability of success on each trial. Another central feature is its memorylessness, meaning the probability of success on a given trial is not affected by the outcomes of previous trials.

In terms of moment generating functions (MGFs), which are the focus of the exercise, the MGF of a geometric distribution is \[ M_Y(t) = \frac{pe^t}{1-(1-p)e^t} \] where \( p \) is again the probability of success. When working on problems involving MGFs of geometric distributions, it's crucial to familiarize oneself with this functional form to properly identify and differentiate it from other distributions.

Random Variables Independence

The concept of independence between random variables is fundamental in probability theory. Two random variables, let's say \( X_1 \) and \( X_2 \), are considered independent if the occurrence of one does not change the probability distribution of the other. In more formal terms, for any two events \( A \) and \( B \) from the respective sample spaces of \( X_1 \) and \( X_2 \) , \( P(X_1 \in A, X_2 \in B) = P(X_1 \in A)P(X_2 \in B) \).

This property has a substantial impact on the moment generating functions (MGFs) of random variables. If \( X_1 \) and \( X_2 \) are independent, their joint MGF is simply the product of their individual MGFs, that is \[ M_{X_1+X_2}(t) = M_{X_1}(t)\times M_{X_2}(t) \] This characteristic is used in the exercise to find the MGF of the sum of \( X_1 \) and \( X_2 \) and to verify if their sum follows a specific distribution by comparing the computed MGF to the known MGF of that distribution.

Probability of Success

The 'probability of success' is a term often used in probability theory with particular significance in the context of Bernoulli trials and distributions such as the binomial and geometric ones. It represents the likelihood of an event deemed a 'success' occurring during a single trial or experiment.

For example, if we're tossing a fair coin, the probability of getting heads (if heads is defined as a success) is 0.5. Similarly, in a geometric distribution, the probability of success \( p \) is the chance that a single trial results in a success. This probability plays a crucial role in determining the shape and characteristics of the probability distribution.

It's crucial to correctly identify the probability of success when working with these distributions, as it not only affects the distributions' PMFs and cumulative distribution functions (CDFs) but also their MGFs. Accurately determining \( p \) allows for appropriate applications of the distribution to real-life problems, like modeling the number of failed attempts before achieving a success, and facilitates the process of comparing an unknown distribution's MGF to that of a geometric distribution.