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The geometric random variable \(X\) has probability distribution $$ f(x)=(1-p)^{x-1} p, \quad x=1,2, \ldots $$ a. Show that the moment-generating function is $$ M_{X}(t)=\frac{p e^{t}}{1-(1-p) e^{t}} $$ b. Use \(M_{\gamma}(t)\) to find the mean and variance of \(X\).

Short Answer

Expert verified
The geometric MGF is \( M_X(t) = \frac{p e^t}{1-(1-p) e^t} \). The mean is \( \frac{1}{p} \) and the variance is \( \frac{1-p}{p^2} \).

Step by step solution

01

Defining the Moment-Generating Function

The moment-generating function (MGF) of a random variable \(X\) is given by \( M_X(t) = \mathbb{E}[e^{tX}] \). For the geometric distribution, this expectation is computed using the probability mass function \(f(x)=(1-p)^{x-1} p\).
02

Writing the Expectation

First, we write the expectation required for the MGF: \( M_X(t) = \sum_{x=1}^{"symee^{tx}(1-p)^{x-1}p \).
03

Factoring Out Constants

Notice that \(p\) is a constant and can be factored out of the sum. Also, \(e^{tx}\) can be rewritten as \((e^t)^x\), yielding: \( M_X(t) = p e^t \sum_{x=1}^{"](1-p)^{x-1} (e^t)^{x-1}\).
04

Recognizing the Geometric Series

The remaining sum is a geometric series with the first term equal to 1 and common ratio \((1-p)e^t\): \( \sum_{x=0}^{\infty}((1-p)e^t)^x = \frac{1}{1-(1-p)e^t}, \) valid as long as \(|(1-p)e^t| < 1\).
05

Calculating the MGF

Substitute the sum into the expression, giving: \( M_X(t) = \frac{p e^t}{1 - (1-p)e^t} \). This is the moment-generating function of the geometric distribution.
06

Finding the Mean using the MGF

The mean \( \mathbb{E}[X] \) can be found from the MGF by calculating \( M'_X(0) \), the first derivative with respect to \( t \) evaluated at \( t = 0 \).
07

Deriving the First Derivative

Differentiate the MGF: \( M'_X(t) = \frac{d}{dt} \left( \frac{p e^t}{1 - (1-p)e^t} \right) \). Apply the quotient rule or product rule as necessary.
08

Simplifying and Evaluating Mean

Evaluate \( M'_X(t) \) at \( t=0 \) to find \( \mathbb{E}[X] \). The simplification gives \( \mathbb{E}[X] = \frac{1}{p} \).
09

Finding the Variance using the MGF

The variance \( \text{Var}(X) \) can be found using \( \text{Var}(X) = M''_X(0) - (M'_X(0))^2 \).
10

Deriving the Second Derivative

Differentiate the first derivative \( M'_X(t) \) to obtain \( M''_X(t) \), and then evaluate at \( t=0 \). This is a complex derivative that involves product and chain rules.
11

Simplifying and Evaluating Variance

Substitute \( t = 0 \) into \( M''_X(t) \) to find \( M''_X(0) \), and use \( \text{Var}(X) = M''_X(0) - (M'_X(0))^2 \). This yields \( \text{Var}(X) = \frac{1-p}{p^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment-Generating Function (MGF)
The Moment-Generating Function, or MGF, serves as a powerful tool in probability and statistics to summarize the entire probability distribution of a random variable. It is essentially a function that, by taking derivatives, can provide us with moments like the mean and variance.
For a geometric distribution, the MGF is defined as \( M_X(t) = \mathbb{E}[e^{tX}] \). The MGF is driven by the probability mass function, or PMF, which is \( f(x)=(1-p)^{x-1} p \) for a geometric distribution. This shows us how likely each outcome is as we progress from one value of \( x \) to the next.
Here's the real magic: by calculating the expected value of \( e^{tX} \), you can transform the PMF into \( M_X(t) = \frac{p e^t}{1 - (1-p)e^t} \). This elegant expression is valid when \(|(1-p)e^t| < 1\). It's like a sneak peek into the distribution's behavior that tells us a lot without needing the entire distribution in front of us.
Probability Mass Function (PMF)
Understanding the Probability Mass Function (PMF) is crucial when working with discrete random variables, such as the geometric distribution. The PMF tells you the probability that the random variable is exactly equal to a certain value.
In the case of a geometric random variable, which describes the number of Bernoulli trials needed to get a success, the PMF is given by \( f(x) = (1-p)^{x-1} p \) for \( x = 1, 2, \ldots \). This means that the probability decreases exponentially with each increase in \( x \), except for the constant factor \( p \).
Why is this important? Because it neatly encapsulates the likelihood of each scenario in a geometric random experiment. Just like the music notes on a sheet, the PMF dictates the melody of probabilities, allowing us to calculate other important metrics using the MGF.
Mean and Variance
Delving into the Mean and Variance of a geometric distribution gives us a clearer picture of its behavior and spread.
The mean, or average, of our geometric random variable \( X \) can be determined directly from its MGF. By deriving the MGF and evaluating the first derivative at \( t=0 \), we find \( \mathbb{E}[X] = \frac{1}{p} \). This mean value indicates the expected number of trials before the first success.
To find the variance, which tells us how spread out the data points are around the mean, we use the MGF again. Specifically, the variance is calculated as \( \text{Var}(X) = M''_X(0) - (M'_X(0))^2 \). With this setup, we find \( \text{Var}(X) = \frac{1-p}{p^2} \). These values for the mean and variance highlight not only the central tendency of the distribution but also its variability.

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