We are given the probability mass function (PMF) of a random variable $X$ as:

$p(x) = \frac{1}{2^x}, \quad x = 1, 2, 3, \dots$

1. Moment Generating Function (MGF):

The moment generating function $M_X(t)$ of $X$ is defined as:

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

Substitute $p(x) = \frac{1}{2^x}$:

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

This is a geometric series with first term $a = \frac{e^t}{2}$ and ratio $r = \frac{e^t}{2}$, valid for $|r| < 1$, i.e., $e^t < 2$ or $t < \ln 2$. The sum of the series is:

$M_X(t) = \frac{\frac{e^t}{2}}{1 - \frac{e^t}{2}} = \frac{e^t}{2 - e^t}, \quad \text{for } t < \ln 2.$

2. Expected Value (Mean):

The expected value $\mathbb{E}[X]$ is given by:

$\mathbb{E}[X] = \sum_{x=1}^\infty x \cdot p(x) = \sum_{x=1}^\infty x \cdot \frac{1}{2^x}.$

We can find this using the formula for the expected value of a weighted geometric series. Let:

$S = \sum_{x=1}^\infty x \cdot \frac{1}{2^x}.$

Differentiate the geometric series $\sum_{x=1}^\infty r^x = \frac{r}{1 - r},$ with respect to $r$:

$\frac{d}{dr} \left( \frac{r}{1-r} \right) = \frac{1}{(1-r)^2}.$

For $r = \frac{1}{2}$, we have:

$S = \sum_{x=1}^\infty x \cdot \frac{1}{2^x} = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2.$

Thus, $\mathbb{E}[X] = 2$.

3. Variance:

The variance $\text{Var}(X)$ is given by:

[ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2. ]

To compute $\mathbb{E}[X^2]$, note that:

$\mathbb{E}[X^2] = \sum_{x=1}^\infty x^2 \cdot \frac{1}{2^x}.$

This can also be computed using differentiation. If $S_k = \sum_{x=1}^\infty x^k \cdot r^x$, then $S_k$ is derived from the derivatives of the geometric series. We will compute $S_2$ directly as:

$S_2 = \frac{r(1 + r)}{(1 - r)^3}.$

For $r = \frac{1}{2}$, we have:

$\mathbb{E}[X^2] = \frac{\frac{1}{2}(1 + \frac{1}{2})}{(1 - \frac{1}{2})^3} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{\frac{1}{8}} = 6.$

Thus, the variance is:

[ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = 6 - 2^2 = 6 - 4 = 2. ]

Final Results:

1. Moment Generating Function (MGF): $M_X(t) = \frac{e^t}{2 - e^t}, \, t < \ln 2.$
2. Expected Value: $\mathbb{E}[X] = 2.$
3. Variance: $\text{Var}(X) = 2.$

---

Would you like me to explain any step in more detail or discuss related concepts? 😊

Here are 5 related questions to deepen your understanding:

1. How is the convergence condition of the MGF derived?
2. Can we verify $\mathbb{E}[X]$ using an alternative method?
3. What happens to the MGF when $t \to -\infty$?
4. How would the results change if $p(x) = \frac{1}{3^x}$ instead?
5. How can the variance formula $\text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$ be derived?

Tip: Always check the domain of validity for the MGF to ensure proper usage in solving problems!