a problem related to finding the moment-generating function (MGF), mean, and variance of a discrete random variable �*X* with a given probability mass function (PMF).

Given:

The probability mass function (PMF) �(�)*f*(*x*) is provided as:

f(x)=1/2*(2/3)^x , x = 1,2,3.....

To solve this problem, we'll follow these steps:

1. **Find the Moment-Generating Function (MGF) of \( X \)**.
2. **Determine the Mean (\( \mu \))** of \( X \) using the MGF.
3. **Determine the Variance (\( \sigma^2 \))** of \( X \) using the MGF.

### 1. Finding the Moment-Generating Function (MGF) \( M_X(t) \)

The MGF of a discrete random variable \( X \) is defined as:

\[
M_X(t) = \mathbb{E}[e^{tX}] = \sum_{x=1}^{\infty} e^{tx} f(x)
\]

Given the PMF \( f(x) = \frac{1}{2} \left(\frac{2}{3}\right)^x \), the MGF becomes:

\[
M_X(t) = \sum_{x=1}^{\infty} e^{tx} \cdot \frac{1}{2} \left(\frac{2}{3}\right)^x
\]

Simplifying the expression inside the sum:

\[
M_X(t) = \frac{1}{2} \sum_{x=1}^{\infty} \left(e^{t} \cdot \frac{2}{3}\right)^x = \frac{1}{2} \sum_{x=1}^{\infty} \left(\frac{2e^t}{3}\right)^x
\]

This is a geometric series with the common ratio \( r = \frac{2e^t}{3} \). The sum of an infinite geometric series \( \sum_{x=1}^{\infty} r^x \) is \( \frac{r}{1-r} \) for \( |r| < 1 \).

So,

\[
M_X(t) = \frac{1}{2} \cdot \frac{\frac{2e^t}{3}}{1 - \frac{2e^t}{3}} = \frac{e^t}{3 - 2e^t}
\]

### 2. Finding the Mean (\( \mu \)) using MGF

The mean of \( X \), \( \mathbb{E}[X] \), can be found by taking the first derivative of the MGF with respect to \( t \) and evaluating at \( t = 0 \):

\[
\mu = M'_X(0)
\]

Let's find the derivative \( M'_X(t) \):

\[
M'_X(t) = \frac{d}{dt} \left(\frac{e^t}{3 - 2e^t}\right)
\]

Using the quotient rule:

\[
M'_X(t) = \frac{(3 - 2e^t) \cdot e^t - e^t \cdot (-2e^t)}{(3 - 2e^t)^2} = \frac{3e^t}{(3 - 2e^t)^2}
\]

Evaluating at \( t = 0 \):

\[
\mu = M'_X(0) = \frac{3e^0}{(3 - 2e^0)^2} = \frac{3}{9} = \frac{1}{3}
\]

### 3. Finding the Variance (\( \sigma^2 \)) using MGF

The variance of \( X \), \( \text{Var}(X) \), can be found using:

\[
\text{Var}(X) = M''_X(0) - [M'_X(0)]^2
\]

First, we find the second derivative \( M''_X(t) \):

\[
M''_X(t) = \frac{d}{dt} \left(\frac{3e^t}{(3 - 2e^t)^2}\right)
\]

This derivative is a bit complex, but by applying the quotient rule again:

\[
M''_X(t) = \frac{(3 - 2e^t)^2 \cdot 3e^t - 3e^t \cdot 2(3 - 2e^t)(-2e^t)}{(3 - 2e^t)^4}
\]

Evaluating \( M''_X(t) \) at \( t = 0 \):

\[
M''_X(0) = \frac{3(3)}{9^2} + \frac{12}{27} = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}
\]

Thus, the variance is:

\[
\text{Var}(X) = M''_X(0) - [M'_X(0)]^2 = \frac{5}{9} - \left(\frac{1}{3}\right)^2 = \frac{5}{9} - \frac{1}{9} = \frac{4}{9}
\]

### Summary of Results:
- **MGF**: \( M_X(t) = \frac{e^t}{3 - 2e^t} \)
- **Mean** (\( \mu \)): \( \frac{1}{3} \)
- **Variance** (\( \sigma^2 \)): \( \frac{4}{9} \)

Would you like a detailed explanation of any part, or do you have any questions?

---

Here are five related questions:
1. How would you find the MGF for a different PMF?
2. Can the MGF be used to find higher moments like skewness and kurtosis?
3. What happens to the MGF if \( X \) is a continuous random variable?
4. How is the variance related to the second moment of a distribution?
5. Can you derive the MGF for a binomial distribution?

**Tip:** The moment-generating function is a powerful tool because it can uniquely determine the distribution of a random variable.

Learn how to calculate the moment-generating function (MGF), mean, and variance of a discrete random variable X with a given probability mass function (PMF). This detailed solution covers the use of the MGF formula, properties of geometric series, and specific calculations for mean and variance. Suitable for advanced undergraduate students and researchers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation