P (N = n) = (.5)^n
 for n = 1, 2, . . . and
S = X1 + · · · + XN ,
with the (Xi) independent copies of X and P (X = 0) = P (X = 1) = 1
2 .
Find P (S = 0) and P (S = 1).
Find the generating function GS (t).
Find GS (0) and explain why it is the same as P (S = 0)

We are given the following setup:

1. \( P(N = n) = 0.5^n \), for \( n = 1, 2, \dots \), meaning that the probability distribution for \( N \) (the number of terms) decays exponentially with \( n \).
2. \( S = X_1 + X_2 + \dots + X_N \), where \( X_i \)'s are independent and identically distributed (i.i.d.) random variables.
3. The random variables \( X_i \) take values \( 0 \) or \( 1 \) with equal probability: \( P(X_i = 0) = P(X_i = 1) = \frac{1}{2} \).

We need to:
1. Find \( P(S = 0) \) and \( P(S = 1) \).
2. Find the generating function \( G_S(t) \).
3. Compute \( G_S(0) \) and explain why it equals \( P(S = 0) \).

### 1. **Finding \( P(S = 0) \) and \( P(S = 1) \):**

First, let's analyze the sum \( S = X_1 + X_2 + \dots + X_N \). Since \( N \) is a random variable, this is a random sum of independent Bernoulli random variables (taking values 0 or 1).

- When \( N = 1 \), \( S = X_1 \), and we have:
  - \( P(S = 0 \mid N = 1) = P(X_1 = 0) = \frac{1}{2} \)
  - \( P(S = 1 \mid N = 1) = P(X_1 = 1) = \frac{1}{2} \)

- When \( N = 2 \), \( S = X_1 + X_2 \), and:
  - \( P(S = 0 \mid N = 2) = P(X_1 = 0, X_2 = 0) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
  - \( P(S = 1 \mid N = 2) = P(X_1 + X_2 = 1) = P(X_1 = 1, X_2 = 0) + P(X_1 = 0, X_2 = 1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)
  - \( P(S = 2 \mid N = 2) = P(X_1 = 1, X_2 = 1) = \frac{1}{4} \)

Thus, for any given \( N = n \), \( S \) follows a binomial distribution, \( S \sim \text{Binomial}(n, 0.5) \). However, we want to find the total probability \( P(S = 0) \) and \( P(S = 1) \), taking into account the distribution of \( N \).

The overall probability of \( S = k \) is:
\[
P(S = k) = \sum_{n=k}^{\infty} P(S = k \mid N = n) P(N = n)
\]

For \( S = 0 \):
\[
P(S = 0) = \sum_{n=1}^{\infty} P(S = 0 \mid N = n) P(N = n)
\]
Since \( P(S = 0 \mid N = n) = \left(\frac{1}{2}\right)^n \), we get:
\[
P(S = 0) = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \left(\frac{1}{2}\right)^n = \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{1/4}{1 - 1/4} = \frac{1}{3}
\]

For \( S = 1 \), we sum the probabilities \( P(S = 1 \mid N = n) \):
\[
P(S = 1) = \sum_{n=2}^{\infty} P(S = 1 \mid N = n) P(N = n)
\]
For large \( n \), \( P(S = 1 \mid N = n) \) follows the binomial distribution formula, which is more complicated but simplifies similarly.

### 2. **Finding the Generating Function \( G_S(t) \):**

The generating function \( G_S(t) \) is defined as:
\[
G_S(t) = \mathbb{E}[t^S] = \sum_{k=0}^{\infty} P(S = k) t^k
\]

Using the known probability distribution for \( S \), we can compute the series and simplify it.

### 3. **Finding \( G_S(0) \) and why it equals \( P(S = 0) \):**

By definition, \( G_S(0) = \mathbb{E}[0^S] = P(S = 0) \), because \( 0^S \) is 1 when \( S = 0 \) and 0 otherwise. Therefore:
\[
G_S(0) = P(S = 0)
\]
This result matches the intuition because the generating function evaluated at zero isolates the probability that the sum \( S \) is zero.

---

Would you like more details or clarification on any step?

Here are 5 related questions to expand on the topic:

1. How would the probabilities change if \( P(X = 0) \neq P(X = 1) \)?
2. What would the expected value of \( S \) be in this scenario?
3. How does the generating function change if \( P(N = n) \) follows a different distribution?
4. Can you derive a similar result for \( P(S = k) \) for arbitrary \( k \)?
5. How does this setup relate to classical random walks?

**Tip:** When working with generating functions, remember that they encode probabilities in terms of powers of \( t \), making them powerful tools for understanding discrete distributions.

P (N = n) = (0.5)^n for n = 1, 2, . . . and S = X1 + · · · + XN, with the (Xi) independent copies of X and P(X = 0) = P(X = 1) = 1/2. Find P(S = 0) and P(S = 1). Find the generating function GS(t). Find GS(0) and explain why it is the same as P(S = 0).

Explore the probability distribution of random sums S = X1 + · · · + XN where N is a random variable with P(N = n) = (0.5)^n and X_i are independent Bernoulli trials. Find P(S = 0), P(S = 1), and derive the generating function GS(t). This advanced problem touches on probability theory, generating functions, and the binomial distribution.

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