Question

Let \(\left\\{X_{i}(t) ; t \geq 0\right\\} \quad i=1,2\) be two independent Poisson processes with parameters \(\lambda_{1}\) and \(\lambda_{2}\) respectively. Let \(X_{1}(0)=m, X_{2}(0)=N-1\), and \(m

Short answer

The probability that the \(X_2\) process reaches N before the \(X_1\) process when \(X_2(0) = N-1\) is given by \(P_{N-m-1}\). In the case where \(X_2(0) = n\), the probability is represented by \(P_{n-m}\). These probabilities can be calculated using backward recursion with the recursion formula \(P_k = \frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}} P_{k-1} + \frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}} P_{k+1}\), and the boundary conditions \(P_0 = 0\) and \(P_{N-m} = 1\).

Step-by-step solution

(a) Probability that the X2 process reaches N before the X1 process: step 1

Calculate the probability of each process reaching its next step: Since \(X_1(t)\) and \(X_2(t)\) are Poisson processes with independent parameters \(\lambda_{1}\) and \(\lambda_{2}\), we have: \(P(X_1(t+dt)=m+1|X_1(t)=m)=\lambda_{1}dt+o(dt)\) \(P(X_2(t+dt)=N|X_2(t)=N-1)=\lambda_{2}dt+o(dt)\) Here, \(o(dt)\) represents a term that goes to zero faster than \(dt\) as \(dt\) tends to zero.

(a) Probability that the X2 process reaches N before the X1 process: step 2

Define a random variable representing the difference between the current values of the X1 and X2 processes: Let's denote \(D(t) = X_2(t) - X_1(t)\). Since \(D(0) = N-1 - m\), we need to find the probability that \(D(t)\) reaches \(N-m\) before it reaches 0.

(a) Probability that the X2 process reaches N before the X1 process: step 3

Write a conditional probability equation for the probability of D(t) reaching N-m before 0: Denote the probability of \(D(t)\) reaching \(N-m\) before \(0\) when \(D(t)=k\) by \(P_k\). We can express \(P_k\) in terms of probabilities of step 1: \(P_k=P(X_1(t+dt)=m+1|X_1(t)=m)P_{k-1}+P(X_2(t+dt)=N|X_2(t)=N-1)P_{k+1}+P(X_1(t+dt)=m,X_2(t+dt)=N-1|X_1(t)=m,X_2(t)=N-1)P_k\) Divide both sides by \(dt\), rearrange, and let \(dt\) approach zero: \(\frac{P_k-P_{k-1}}{dt}=\lambda_{1}P_{k-1}+\frac{P_k-P_{k+1}}{dt}+o(1)\) \(\lambda_{1}P_{k-1}+\lambda_{2}P_{k+1}=P_k\) (1)

(a) Probability that the X2 process reaches N before the X1 process: step 4

Solve the conditional probability equation: Using equation (1), we can represent the desired probability, \(P_{N-m-1}\), in terms of previous probabilities: \(P_{N-m-1}=\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}P_{N-m-2}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}P_{N-m}\) Since we need to reach \(N-m\) before \(0\), we have boundary conditions: \(P_{0} = 0, \quad P_{N-m} = 1\)

(a) Probability that the X2 process reaches N before the X1 process: step 5

Use recursion to find the desired probability: The recursion equation can be solved using backward recursion, starting from \(P_{N-m} = 1\) and computing probabilities until we reach the desired value of \(P_{N-m-1}\): \(P_{N-m-1} = \frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}P_{N-m-2}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\) Repeat this process for all values of \(k = N-m-2, N-m-3, ..., 1\).

(a) Probability that the X2 process reaches N before the X1 process: final result

The probability that the \(X_2\) process reaches N before the \(X_1\) process is \(P_{N-m-1}\). Now we'll move to the problem with the \(X_2\) process initialized at \(n\).

(b) Probability that the X2 process reaches N before the X1 process when X2(0) = n: step 1

Adjust the definition of the random variable: Since \(X_2(0) = n\), we define \(D(0) = n - m\), and the goal is to find the probability that \(D(t)\) reaches \(N-m\) before it reaches 0 when \(D(0)=n-m\).

(b) Probability that the X2 process reaches N before the X1 process when X2(0) = n: step 2

Prepare the recursion formula for the new scenario: The desired probability is represented by \(P_{n-m}\), and we can still use the same recursion formula: \(P_{n-m}=\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}P_{n-m-1}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}P_{n-m+1}\)

(b) Probability that the X2 process reaches N before the X1 process when X2(0) = n: final result

Use recursion to find the desired probability: We use backward recursion, starting with \(P_{N-m} = 1\), to iterate until we reach the desired value of \(P_{n-m}\) using the same formula as in part (a). The probability that the \(X_2\) process reaches N before the \(X_1\) process when \(X_2(0) = n\) is represented by \(P_{n-m}\).

Key concepts

Probability Theory

Probability theory is a fundamental part of mathematics that deals with the likelihood of different outcomes occurring. It is based on the concept of a 'probability', which is a numerical value representing how likely an event is to happen. In our Poisson processes example, we are essentially examining events (such as one process reaching a certain number before another) and trying to determine how probable these events are.

When we calculate probabilities, we typically deal with values between 0 and 1, where 0 means the event cannot occur and 1 means the event is certain to occur. The probabilities for a Poisson process, which models the number of events happening in a fixed interval of time under certain conditions, can be calculated using specific functions and properties unique to this type of process. For instance, the probability that the count of events increases by one in an infinitesimally small time interval is proportional to the length of that interval, a characteristic captured by the parameter \(\lambda\).

The example provided in the exercise shows us how to set up a problem that requires understanding how these probabilities are calculated and how they can lead to determining the likelihood of one event happening before another in a specific scenario.

Stochastic Processes

A stochastic process is a collection of random variables representing a process that evolves over time in a way that is at least partially random. Poisson processes, as mentioned in the exercise, are a particular type of stochastic process where events happen continuously and independently over time, with their times of occurrence not being predictable but their rate of occurrence (\lambda) being constant.

In the context of the given problems, each \(X_i(t)\) is a stochastic process. What's crucial here is the principle of independence, which means that the occurrence of events in one process does not affect the occurrence of events in another process. Understanding the nature of stochastic processes helps in modeling various real-world phenomena where outcomes are uncertain, such as the number of phone calls received by a call center or the arrival of customers at a bank.

As stochastic processes can be complex, breaking them down step by step, as shown in the solution, helps in analyzing and predicting different scenarios. The ability to translate real-world problems into mathematical models of stochastic processes is essential for solving problems across many fields, including finance, telecommunications, and service operations.

Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already occurred. In our Poisson process example, we're looking at the probability of one process reaching a target number before another, but these calculations depend on the current state of both processes, which is where conditional probability comes into play.

One important tool from conditional probability is the use of conditional probability equations, as seen in step 3 of the exercise. This method allows us to determine the dynamics of how one event affects another. For example, the probability of \(X_2(t+dt)=N\) given \(X_2(t)=N-1\) will influence the probability of our \(X_2\) process reaching \(N\) before \(X_1\).

It's important to understand the underlying assumptions in calculating conditional probabilities, such as the independence of the Poisson processes in this problem. Such details are crucial for correctly applying the principles of conditional probability and, overall, for mastering the art of problem-solving within probability and stochastic processes.