PROBABILITY, R, STOCHASTIC PROCESSES, BRANCHING PROCESSES

Question

Do you have a friend who seems to answer every single message of yours by giving a very extensive answer? Are you, on the other hand, a person that replies to a message cryptically, but in several seperated messages? Have you had the feeling to avoid answering those long messages, worrying that the communication gets out of hand? In this project we study an asymmetric communication between two persons, called user 1 and user 2 henceforth. They are best friends, so the aim is not to end a friendship, but rather maintain it, while not completely changing one’s habits. One guy will still be giving cryptic answers, the other one is not planning to shorten his answers. These users communicate of course about a range of topics. We select one topic and ask ourselves, will the communication on this very topic eventually end, or not? The behaviour of the two users is described in probabilistic terms as follows: ˆ User 1 regularly checks his email box (or social media app), that is, at every date t = 0, 1, 2, . . . . He selects messages coming from user k with probability λ ∈ (0, 1). However, his replies are manifold, namely equals K + 1, where K is Poisson distributed with parameter z − 1. (So that, on average, he has z replies to each accepted message of user 1.). Use, for example z = 5 and a range of λ, see below. ˆ User 2 rarely answers, namely each day with probability β. However, all messages are answered eventually, and always by sending a single reply (so that an unanswered message will be saved and perhaps answered on the next date.) The question, whether the communication almost surely (that means for almost all scenarios) ends in a (possibly long, but) finite time, can be answered by the theory of branching processes. The branching process that starts with any initial message amount (say i1 messages on user 1’s account, i2 messages on user 2’s account at t = 0, will go extinct in finite time, if and only if λz ≤ 1. Since all messages are treated similarly, the evolution of a single message from user 1, and a single message from user 2 suffices to understand the entire process.

1. To implement the model, start for instance with user 1. The vector µ 1 (0) = (1, 0) describes the population (number of all messages on this topic) at t = 0. This means, there is one incoming message from user 2. At t = 1 user 1 decides to answer or not. If she answers the messages, she draws from a Poisson distributed r.v. (see above), and that gives the number of messages sent to user 2. At t = 2, user 2 acts. [etc.].

2. You can implement this by writing a function that inputs user tag i and outputs the random descendants of a single message from user 2, as a two-vector. Then go ahead and write a function that can deal with a general population input (i1, i2) and outputs a random draw from the population one step ahead.

3. Once you have the population dynamics, you can then choose λ = 1/10, 1/5, 1 and z = 5 and for each of these choices, simulate 100 paths to see, whether extinction really occurs in finite time, or not.

Answer 1

Hi Seamus,

I understand that you're presenting a scenario involving asymmetric communication between two users. The model involves probabilities, branching processes, and the question of whether the communication will almost surely end in a finite time. To address this, we need to implement the model with specified parameters and simulate paths to observe the outcomes.

I'll start by implementing the model for user 1. The vector μ1​(0)=(1,0) represents the initial population, indicating one incoming message from user 2 at t=0. At each step, user 1 decides to answer or not, drawing from a Poisson distribution to determine the number of messages sent to user 2. User 2 then acts, and this process continues.

Let's begin with a function that generates random descendants for a single message from user 2 as a two-vector.

Here's a simple implementation in Python for the first part of the model, where user 1 generates random descendants for a single message from user 2:

import numpy as np

def generate_descendants_user2():

# Generate random descendants for a single message from user 2

poisson_param = 5 # Poisson parameter z - 1

num_descendants = np.random.poisson(poisson_param) + 1

return np.array([0, num_descendants])

# Example usage

descendants = generate_descendants_user2()

print("Descendants for a single message from user 2:", descendants)

This function simulates the number of messages sent by user 1 in response to a single message from user 2, following a Poisson distribution.

Next, we'll move on to the second part, where we'll write a function to handle a general population input and output a random draw from the population one step ahead.

Continuing with the implementation, let's create a function that deals with a general population input (i1,i2) and outputs a random draw from the population one step ahead:

def population_dynamics_step(population):

# Input: Population vector (i1, i2)

# Output: Random draw from the population one step ahead

# User 1 generates random descendants for each message from user 2

descendants_user2 = np.array([generate_descendants_user2() for _ in range(population[1])])

# Calculate the new population

new_population = np.array([

population[0] + sum(descendants_user2[:, 0]), # i1 updated

sum(descendants_user2[:, 1]) # i2 updated

])

return new_population

# Example usage

initial_population = np.array([1, 0]) # Initial population (i1, i2)

new_population = population_dynamics_step(initial_population)

print("New population:", new_population)

This function takes the current population as input and calculates the new population one step ahead based on the communication dynamics.

Now, we can move on to simulating paths for different choices of λ and z, as well as checking whether extinction occurs in finite time by simulating 100 paths for each scenario.