Let's break this problem into smaller sections for clarity. Here's an explanation for the questions given in the image:

Question 1: Branching Process

(i) Obtain the generating function for $X$: $G_X(t) = E(t^X)$.

The probabilities for $X$ are:

• $P(X = 0) = \frac{1}{8}$,
• $P(X = 1) = \frac{1}{2}$,
• $P(X = 2) = \frac{1}{4}$,
• $P(X = 3) = \frac{1}{8}$.

The generating function is: $G_X(t) = E(t^X) = \sum_{x=0}^{3} P(X = x) \cdot t^x.$ Substitute the values: $G_X(t) = \frac{1}{8}t^0 + \frac{1}{2}t^1 + \frac{1}{4}t^2 + \frac{1}{8}t^3.$ Simplify: $G_X(t) = \frac{1}{8} + \frac{1}{2}t + \frac{1}{4}t^2 + \frac{1}{8}t^3.$

(ii) Use the law of total probability to show the probability of extinction $\eta$.

Let $\eta$ be the probability of extinction. Using the total probability law: $\eta = \sum_{j=0}^{\infty} P(Z_1 = j) \cdot P(\text{extinction} | Z_1 = j).$ For $Z_1 = j$, the offspring of $Z_0$, the extinction probability depends on $G_X(\eta)$, the generating function evaluated at $\eta$. Therefore: $\eta = G_X(\eta).$ Substituting $G_X(t)$ from part (i) into this equation gives the extinction equation.

(iii) Find the probability of extinction.

Solve $\eta = G_X(\eta)$: $\eta = \frac{1}{8} + \frac{1}{2}\eta + \frac{1}{4}\eta^2 + \frac{1}{8}\eta^3.$ This is a cubic equation. The extinction probability $\eta$ is the smallest non-negative root of this equation.

Question 2: Simple Random Walk

(i) Use the law of total probability to show $P(T_0 = n) = P(T_1 = n-1)$.

The random walk satisfies $S_0 = 0$ and moves up or down with equal probability. For $T_0 = n$ (the first return to 0 at step $n$): $P(T_0 = n) = P(X_1 = 1) \cdot P(T_1 = n-1) + P(X_1 = -1) \cdot P(T_1 = n-1).$ Since $P(X_1 = 1) = P(X_1 = -1) = \frac{1}{2}$, we get: $P(T_0 = n) = P(T_1 = n-1).$

(ii) Show $G_0(t) = t G_1(t)$.

Let $G_0(t)$ and $G_1(t)$ be the generating functions for $T_0$ and $T_1$, respectively. From part (i), $P(T_0 = n) = P(T_1 = n-1)$. The generating function is: $G_0(t) = \sum_{n=1}^\infty P(T_0 = n) t^n.$ Substitute $P(T_0 = n) = P(T_1 = n-1)$: $G_0(t) = \sum_{n=1}^\infty P(T_1 = n-1) t^n.$ Reindex $n \to n-1$ to match $T_1$: $G_0(t) = t \sum_{n=0}^\infty P(T_1 = n) t^n.$ Thus: $G_0(t) = t G_1(t).$

(iii) Generating function for $T_a$ where $a > 1$.

For $T_a$, the minimum time $n > 1$ such that $S_n = a$, the generating function can be expressed recursively using symmetry and first-step analysis. Use similar steps to previous parts to derive this.

Would you like further clarification, or should I expand on solving the cubic equation for extinction probability or deriving $T_a$'s generating function?

Related Questions:

1. How do generating functions help in analyzing stochastic processes?
2. What are the key properties of branching processes in extinction problems?
3. Can the extinction probability $\eta$ be determined without solving the cubic equation exactly?
4. How does symmetry simplify calculations in simple random walks?
5. What are real-world applications of branching processes and random walks?

Tip:

When solving equations with generating functions, substituting values like $t = 1$ or $t = \eta$ often simplifies the problem significantly.