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?

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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.