Math Problem Statement

1. This question is about a branching process and involves deriving generating functions, finding probabilities of extinction, and solving related equations. 2. It also includes a random walk problem focusing on generating functions and hitting times.

Solution

To assist with the mathematical questions in the image, I'll solve them step-by-step:


1. Branching Process

(i) Generating Function GX(t)G_X(t)

The generating function for XX, given X{0,1,2,3}X \in \{0, 1, 2, 3\} with the specified probabilities, is:

GX(t)=E[tX]=k=03P(X=k)tkG_X(t) = \mathbb{E}[t^X] = \sum_{k=0}^3 P(X=k)t^k

Substituting the given probabilities:

GX(t)=18t0+12t1+14t2+18t3G_X(t) = \frac{1}{8}t^0 + \frac{1}{2}t^1 + \frac{1}{4}t^2 + \frac{1}{8}t^3

GX(t)=18+12t+14t2+18t3G_X(t) = \frac{1}{8} + \frac{1}{2}t + \frac{1}{4}t^2 + \frac{1}{8}t^3


(ii) Probability of Extinction Using Total Probability Law

Let η\eta be the probability of extinction. Using the law of total probability:

η=j=03P(Z1=j)ηj\eta = \sum_{j=0}^3 P(Z_1 = j) \eta^j

The generating function GX(η)G_X(\eta) gives the simplification:

η=GX(η)=18+12η+14η2+18η3\eta = G_X(\eta) = \frac{1}{8} + \frac{1}{2}\eta + \frac{1}{4}\eta^2 + \frac{1}{8}\eta^3

The equation η=GX(η)\eta = G_X(\eta) must be solved to find η\eta, which represents the probability of extinction.


(iii) Probability of Extinction

This corresponds to finding the smallest non-negative root of the equation η=GX(η)\eta = G_X(\eta). Solve numerically or algebraically:

η=18+12η+14η2+18η3\eta = \frac{1}{8} + \frac{1}{2}\eta + \frac{1}{4}\eta^2 + \frac{1}{8}\eta^3

Rearranging:

18η3+14η212η+18=0\frac{1}{8}\eta^3 + \frac{1}{4}\eta^2 - \frac{1}{2}\eta + \frac{1}{8} = 0

This cubic equation gives η=1\eta = 1 as a trivial root, which corresponds to the extinction probability if there are no further individuals. Other roots may also need verification depending on the process context.


2. Simple Random Walk

(i) Using Total Probability for P(T0=n)P(T_0 = n)

Let T0T_0 be the first hitting time of zero for SnS_n. Using total probability and the fact that X1=±1X_1 = \pm 1 with equal probabilities:

P(T0=n)=12P(T0=n1)+12P(T1=n1)P(T_0 = n) = \frac{1}{2}P(T_0 = n-1) + \frac{1}{2}P(T_1 = n-1)

Given symmetry, P(T0=n)=P(T1=n1)P(T_0 = n) = P(T_1 = n-1), so:

P(T0=n)=P(T1=n1)P(T_0 = n) = P(T_1 = n-1)


(ii) Showing G0(t)=tG1(t)G_0(t) = tG_1(t)

The generating function for T0T_0 satisfies the recurrence:

G0(t)=n=1P(T0=n)tnG_0(t) = \sum_{n=1}^\infty P(T_0 = n)t^n

From part (i), P(T0=n)=P(T1=n1)P(T_0 = n) = P(T_1 = n-1), so:

G0(t)=tG1(t)G_0(t) = tG_1(t)


(iii) Generating Function for TaT_a (for a>1a > 1)

Define TaT_a as the first hitting time of aa. The recurrence relation for TaT_a depends on the state and follows similar logic. The generating function can be derived as a function of aa and analyzed for specific values of aa.


Would you like detailed numerical solutions or expansions for these steps? Here are five related questions you might want to explore:

  1. How can we numerically solve the cubic equation for extinction probability?
  2. Can we generalize GX(t)G_X(t) for other branching processes?
  3. What are the conditions under which extinction is certain?
  4. How do we interpret G0(t)=tG1(t)G_0(t) = tG_1(t) in the context of first hitting times?
  5. How does the random walk's behavior change as aa increases?

Tip: Generating functions are powerful tools for solving recurrence relations and analyzing probabilities systematically.

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Math Problem Analysis

Mathematical Concepts

Branching Processes
Probability Theory
Generating Functions
Random Walks
Law of Total Probability

Formulas

Generating Function G_X(t) = E[t^X] = Σ P(X=k)t^k
Extinction Probability: η = G_X(η)
Random Walk Recurrence: P(T_0=n) = P(T_1=n-1)
Generating Function G_0(t) = tG_1(t)

Theorems

Law of Total Probability
Generating Function Analysis for Random Walks

Suitable Grade Level

Undergraduate Level