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 $P(X = 0) = 1 - p$.

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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 $t = 1$.