suppose X is a geometric random variable with
P (X = x) = (.5)^(x+1), x ∈ {0, 1, 2, . . .}.
(i) Find the generating function GX (t).
(ii) If Gn(t) is the generating function for Zn where Z0 = 1 and
Zn = X1 + · · · + XZn−1 ,
show that
Gn(t) = Gn−1(GX (t)).

### (i) Finding the generating function \( G_X(t) \)

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

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

Given \( P(X = x) = (0.5)^{x+1} \), we substitute this into the formula for the generating function:

\[
G_X(t) = \sum_{x=0}^{\infty} (0.5)^{x+1} t^x
\]

Factor out \( 0.5 \) from the sum:

\[
G_X(t) = 0.5 \sum_{x=0}^{\infty} \left( \frac{t}{2} \right)^x
\]

Recognize that the summation is a geometric series of the form:

\[
\sum_{x=0}^{\infty} r^x = \frac{1}{1 - r} \quad \text{for} \quad |r| < 1
\]

Here, \( r = \frac{t}{2} \), so the generating function becomes:

\[
G_X(t) = 0.5 \left( \frac{1}{1 - \frac{t}{2}} \right)
\]

Thus, the generating function for \( X \) is:

\[
G_X(t) = \frac{0.5}{1 - \frac{t}{2}} = \frac{1}{2 - t}
\]

### (ii) Showing that \( G_n(t) = G_{n-1}(G_X(t)) \)

Given that \( G_n(t) \) is the generating function for \( Z_n \), and \( Z_n = X_1 + X_2 + \dots + X_{Z_{n-1}} \), we aim to show that:

\[
G_n(t) = G_{n-1}(G_X(t))
\]

We proceed by induction on \( n \):

1. **Base Case: \( n = 1 \)**

   By the given definition, \( Z_0 = 1 \), and so \( G_0(t) = t^1 = t \).

2. **Inductive Hypothesis:**
   
   Suppose the result holds for \( n-1 \), i.e., \( G_{n-1}(t) = G_{n-2}(G_X(t)) \).

3. **Inductive Step:**

   We need to show that \( G_n(t) = G_{n-1}(G_X(t)) \).

   By the definition of generating functions, \( G_n(t) = \mathbb{E}[t^{Z_n}] \), and since \( Z_n = X_1 + X_2 + \dots + X_{Z_{n-1}} \), we consider:

   \[
   G_n(t) = \mathbb{E}\left[ t^{Z_n} \right] = \mathbb{E} \left[ t^{X_1 + \dots + X_{Z_{n-1}}} \right]
   \]

   By the **Law of Total Expectation**, we condition on \( Z_{n-1} \):

   \[
   G_n(t) = \mathbb{E} \left[ \mathbb{E}\left[ t^{X_1 + \dots + X_{Z_{n-1}}} \mid Z_{n-1} \right] \right]
   \]

   Using the fact that the \( X_i \) are i.i.d., we get:

   \[
   \mathbb{E} \left[ t^{X_1 + \dots + X_{Z_{n-1}}} \mid Z_{n-1} = k \right] = \left( G_X(t) \right)^k
   \]

   Thus, the generating function becomes:

   \[
   G_n(t) = \mathbb{E} \left[ \left( G_X(t) \right)^{Z_{n-1}} \right]
   \]

   But this is exactly \( G_{n-1}(G_X(t)) \), so we have:

   \[
   G_n(t) = G_{n-1}(G_X(t))
   \]

Thus, by induction, the relation \( G_n(t) = G_{n-1}(G_X(t)) \) holds for all \( n \geq 1 \).

---

Would you like more details on any part of this? Here are some related questions:

1. How do you compute expected values from generating functions?
2. What are the properties of geometric random variables?
3. Can you explain how the total expectation theorem works in this context?
4. What are the conditions for the convergence of a generating function?
5. How does the generating function approach compare to moment-generating functions?

**Tip**: The generating function is a powerful tool for solving recurrence relations and finding distributions of sums of independent random variables.

suppose X is a geometric random variable with P(X = x) = (.5)^(x+1), x ∈ {0, 1, 2, . . .}. (i) Find the generating function GX(t). (ii) If Gn(t) is the generating function for Zn where Z0 = 1 and Zn = X1 + · · · + XZn−1, show that Gn(t) = Gn−1(GX(t)).

Explore the solution for finding the generating function of a geometric random variable with P(X = x) = (0.5)^(x+1) and proving a recurrence relation for generating functions in probability theory. This covers concepts in probability, geometric series, and recurrence relations, suitable for undergraduate students.

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