Problem 76 - An Introduction to Mathematical Statistics and Its Applications

Question

Sometimes the geometric random variable is defined to be the number of trials, \(X\), preceding the first success. Write down the corresponding pdf and derive the moment-generating function for \(X\) two ways(1) by evaluating \(E\left(e^{2 X}\right)\) directly and (2) by using Theorem \(3.12 .3\).

Step-by-Step Solution

Answer

The pdf of geometric random variable \(X\) defined as the number of trials preceding first success is \(p_X(x) = (1-p)^{x-1}p. Its moment-generating function (MGF) derived by evaluating \(E(e^{2X})\) directly and by using Theorem 3.12.3 is \(\frac{p}{1-(1-p)e^2}\) for \(0 < e^2 < \frac{1}{1-p}\).

Step 1: Derive the PDF

The probability density function (pdf) of a geometric random variable defined as the number of trials preceding the first success can be given as \(p_X(x) = (1-p)^{x-1}p\), where \(p\) is the probability of success in each trial, and \(x\) the number of trials taken until the first success.

Step 2: Find MGF by evaluating \(E(e^{2X})\) directly

The moment-generating function (MGF) for a random variable \(X\) is given by \(M_X(t) = E(e^{tX})\), which in this case will be \(E(e^{2X})\). We can compute this by substituting \(2X\) into \(p_X(x)\) in the integral of the moment-generating function formula: \[\int e^{2x}(1-p)^{x-1}p dx\]. Evaluating this with calculus, the MGF comes out to be \(\frac{p}{1-(1-p)e^2}\) assuming that \(0 < e^2 < \frac{1}{1-p}\).

Step 3: Find MGF using Theorem 3.12.3

Theorem 3.12.3 states that the MGF of a geometrically distributed random variable \(X\), with parameter \(p\), is given by \(M_X(t) = \frac{p}{1-(1-p)e^t}\) which, when substituting \(t=2\), also gives MGF as \(\frac{p}{1-(1-p)e^2}\) assuming that \(0 < e^2 < \frac{1}{1-p}\).

Key Concepts

Probability Density Function

A geometric random variable is a fascinating type of random variable, particularly because it models situations where we're counting the number of trials until the first success occurs in repeated independent trials. The probability density function (pdf) plays a key role here, as it allows us to calculate the likelihood of achieving our first success on the 1st, 2nd, 3rd trial, and so on.

For this type of random variable, we define the pdf as:

• \(p_X(x) = (1-p)^{x-1}p\)

Here, \(p\) is the probability of success in each individual trial while \(x\) is the number of trials before the first success happens. This formula essentially expresses the probability that a certain number of trials are needed until we witness that very first successful outcome.

• The variable \(x\) must be a positive integer since it represents trial counts.
• The factor \((1-p)^{x-1}\) denotes that all the \(x-1\) previous trials have failed.
• The last component, \(p\), accounts for the success on the \(x^{th}\) trial.

It is important to remember that this distribution is memoryless, meaning the probability of its future outcome is not affected by past events although the trials preceding have not seen success.

Moment-Generating Function

The moment-generating function, often abbreviated as MGF, provides a way to summarize all the moments of a random variable. It is an extremely useful function in probability and statistics, offering significant insight into the properties of probability distributions.

For a geometric random variable \(X\), its MGF can be determined by using the formula:

• \(M_X(t) = E(e^{tX})\)

This expresses the expected value of \(e^{tX}\), where \(t\) is a real number. In our specific case based on the exercise, we wanted to find this function with \(t = 2\).

• To evaluate this directly, we use calculus to work through the integral,
  \[\int e^{2x}(1-p)^{x-1}p \, dx\]
• After simplification and assuming a convergence condition holds, we end up with:
  \(\frac{p}{1-(1-p)e^2}\)

This MGF is a clever tool since it bundles all moments of the random variable together into one expression. By differentiating this function, one can obtain moments, which capture important aspects such as mean and variance of \(X\). Keep in mind that MGFs, like the distributions they're based on, have their own conditions of convergence.

Theorem 3.12.3

Theorem 3.12.3 is a powerful tool that simplifies the process of finding the moment-generating function for certain types of random variables, including geometrically distributed ones. According to this theorem, the MGF for a geometric random variable \(X\) with success probability \(p\) is

• \(M_X(t) = \frac{p}{1-(1-p)e^t}\)

This formula closely aligns with the direct evaluation found earlier, only it requires considerably less calculation. When applying this to our exercise, we substitute \(t = 2\) into the theorem's formula to find the MGF directly:

• Substituting results in \(\frac{p}{1-(1-p)e^2}\)
• This only applies if \(0 < e^2 < \frac{1}{1-p}\), ensuring convergence of the MGF.

This theorem underscores a central theme in mathematical statistics—being able to represent complex distributions with simpler, well-defined functions for broader analysis. It highlights the importance of token simplifications that provide accuracy with efficiency.