Suppose X is a geometric random variable with X ∈ {0, 1, 2, . . .} and with
probability mass function
P(X = x) = (1 − p) px, x = 0, 1, 2, . . . ,
for some 0 < p < 1.
Find
GX (t) = E(tX )
and indicate for what values of t are appropriate.
What is, and explain, the value of GX (1).
What is GX (0)

We are given that \( X \) is a geometric random variable taking values in \( \{0, 1, 2, \dots\} \) with the probability mass function:

\[
P(X = x) = (1 - p) p^x \quad \text{for } x = 0, 1, 2, \dots,
\]
where \( 0 < p < 1 \).

### Part 1: Finding \( G_X(t) = E(t^X) \)

The **probability generating function** \( G_X(t) \) of a discrete random variable \( X \) is defined as:

\[
G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X = x) t^x
\]

Using the given PMF for \( X \), we can write:

\[
G_X(t) = \sum_{x=0}^{\infty} (1 - p) p^x t^x
\]

Factor out \( (1 - p) \):

\[
G_X(t) = (1 - p) \sum_{x=0}^{\infty} (p t)^x
\]

This is a geometric series with first term 1 and common ratio \( p t \). The sum of an infinite geometric series is valid when the absolute value of the common ratio is less than 1, i.e., when \( |p t| < 1 \), and it is given by:

\[
\sum_{x=0}^{\infty} (p t)^x = \frac{1}{1 - p t}, \quad \text{for } |p t| < 1
\]

Thus, we have:

\[
G_X(t) = \frac{1 - p}{1 - p t}, \quad \text{for } |t| < \frac{1}{p}
\]

### Part 2: Finding \( G_X(1) \)

We substitute \( t = 1 \) into the expression for \( G_X(t) \):

\[
G_X(1) = \frac{1 - p}{1 - p \cdot 1} = \frac{1 - p}{1 - p} = 1
\]

This makes sense because \( G_X(1) = E(1^X) = E(1) = 1 \), which reflects the total probability.

### Part 3: Finding \( G_X(0) \)

We substitute \( t = 0 \) into the expression for \( G_X(t) \):

\[
G_X(0) = \frac{1 - p}{1 - p \cdot 0} = \frac{1 - p}{1} = 1 - p
\]

This represents the probability that \( X = 0 \), since \( G_X(0) = E(0^X) \), and \( P(X = 0) = 1 - p \).

### Summary of Results:
1. \( G_X(t) = \frac{1 - p}{1 - p t}, \quad |t| < \frac{1}{p} \)
2. \( G_X(1) = 1 \)
3. \( G_X(0) = 1 - p \)

### Explanation of \( G_X(1) \):
- \( G_X(1) \) represents the sum of all probabilities, which must equal 1 because the total probability over all possible values of a probability distribution is always 1.

### Explanation of \( G_X(0) \):
- \( G_X(0) \) corresponds to the probability that \( X = 0 \), which is \( P(X = 0) = 1 - p \).

---

Would you like more details on any step or have any further questions? Here are some related questions for deeper understanding:

1. How does the geometric series formula apply to probability generating functions?
2. What is the significance of the domain restriction \( |t| < \frac{1}{p} \)?
3. How would you compute higher moments using the generating function?
4. What does the first derivative of \( G_X(t) \) tell you about the distribution?
5. How does this geometric distribution relate to the negative binomial distribution?

**Tip**: The probability generating function is useful for calculating moments (like the mean and variance) of a random variable by taking derivatives at \( t = 1 \).

Suppose X is a geometric random variable with X ∈ {0, 1, 2, . . .} and with probability mass function P(X = x) = (1 − p) p^x, x = 0, 1, 2, . . . , for some 0 < p < 1. Find GX (t) = E(t^X) and indicate for what values of t are appropriate. What is, and explain, the value of GX (1). What is GX (0)?

Explore the probability generating function for a geometric random variable, including calculations for G_X(t), G_X(1), and G_X(0). Learn how the geometric series and probability theory apply to this problem. Suitable for undergraduate students in probability theory.

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