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.

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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.