Let us solve the problem step-by-step:

(i) Finding the Generating Function of $N$:

Given $N \sim \text{Poisson}(\lambda)$, the probability mass function of $N$ is: $P(N = n) = \frac{\lambda^n}{n!} e^{-\lambda}.$

The generating function $G_N(t) = E(t^N)$ is: $G_N(t) = \sum_{n=0}^\infty P(N = n) t^n = \sum_{n=0}^\infty \frac{\lambda^n}{n!} e^{-\lambda} t^n.$

Factor $e^{-\lambda}$ outside the summation: $G_N(t) = e^{-\lambda} \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!}.$

The series $\sum_{n=0}^\infty \frac{x^n}{n!}$ is the Taylor expansion of $e^x$. Substituting $x = \lambda t$, we get: $G_N(t) = e^{-\lambda} e^{\lambda t} = e^{\lambda (t-1)}.$

(ii) Generating Function of $S = X_1 + X_2 + \cdots + X_N$:

Each $X_i$ is independent and identical, with generating function: $G_X(t) = \frac{1}{2} + \frac{1}{2} t.$

Using the property of the Poisson distribution and generating functions: $G_S(t) = G_N(G_X(t)).$

Substituting $G_N(t) = e^{\lambda (t-1)}$ and $G_X(t) = \frac{1}{2} + \frac{1}{2}t$: $G_S(t) = G_N\left( G_X(t) \right) = e^{\lambda (G_X(t) - 1)} = e^{\lambda \left( \frac{1}{2} + \frac{1}{2}t - 1 \right)}.$

Simplify: $G_S(t) = e^{\lambda \left( -\frac{1}{2} + \frac{1}{2}t \right)} = e^{\frac{\lambda}{2} (t - 1)}.$

This proves the result: $G_S(t) = e^{\frac{\lambda}{2}(1 - t)}.$

(iii) Distribution of $S$:

From the generating function $G_S(t) = e^{\frac{\lambda}{2} (t - 1)}$, we recognize this as the generating function of a Poisson random variable with mean $\frac{\lambda}{2}$. Thus: $S \sim \text{Poisson}\left( \frac{\lambda}{2} \right).$

Let me know if you'd like further clarifications or a more detailed explanation.

5 Questions to Expand Understanding:

1. What are the properties of generating functions for independent random variables?
2. How does the generating function of $S$ relate to the generating function of $N$?
3. Why is the Poisson distribution well-suited for problems involving sums of random variables?
4. Can we extend this approach to non-Poisson distributions? How would it differ?
5. What is the intuition behind the result that $S \sim \text{Poisson}(\frac{\lambda}{2})$?

Tip:

For generating functions, always remember that substitutions like $G_N(G_X(t))$ allow you to elegantly handle nested random variables, especially in problems involving sums and independent random variables.