	A random variable X has the probability function p(x)= 1/2^x  ; x =1,2,3...
	Find moment generating function 
	Find expected value
	Find variance


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!

A random variable X has the probability function p(x)= 1/2^x  ; x =1,2,3...
Find moment generating function
Find expected value
Find variance

Explore the solution for finding the moment generating function, expected value, and variance of a random variable X with the probability mass function p(x) = 1/2^x for x = 1, 2, 3, ... Detailed steps for calculating each result, including use of geometric series and differentiation techniques. Suitable for college or university students studying probability theory.

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