Question

Let \(\left\\{X_{1}(t), t \geq 0\right\\}\) and \(\left\\{X_{2}(t), t \geq 0\right\\}\) be independent Poisson processes with common intensity, \(\lambda\). Suppose that \(X_{1}(0)=9\) and \(X_{2}(0)=5 .\) What is the probability that the \(X_{1}\)-process reaches level 10 before the \(X_{2}\)-process does?

Short answer

The probability is \(\frac{1}{6}\).

Step-by-step solution

Understand Poisson Process

A Poisson process is a mathematical model for a series of events, where events occur randomly over a continuous time interval. Here, both \(X_1(t)\) and \(X_2(t)\) are Poisson processes with a common intensity \(\lambda\). This means that for each process, the number of events follows a Poisson distribution.

Define Goal

We are tasked with finding the probability that the \(X_1\)-process reaches level 10 before the \(X_2\)-process reaches level 10. Currently, \(X_1(0) = 9\) and \(X_2(0) = 5\). Thus, \(X_1(t)\) needs 1 additional event to achieve level 10, while \(X_2(t)\) needs 5 additional events.

Analyze Individual Event Processes

For \(X_1(t)\), reaching level 10 implies it experiences 1 event before \(X_2(t)\) experiences 5 events. Each event in \(X_1(t)\) occurs independently with rate \(\lambda\), and events in \(X_2(t)\) also occur independently with the same rate \(\lambda\).

Use Race Towards 10

Since both processes are independent and the number of events follow an exponential distribution with parameter \(\lambda\), this becomes a "race" to see which process reaches its required event level first. \(X_1(t)\) reaching 10 before \(X_2(t)\) can be thought of as \(T_1 < T_5\), where \(T_1\) is the time \(X_1(t)\) takes to increase by 1 event and \(T_5\) is the time \(X_2(t)\) takes to increase by 5 events.

Calculate Probability

The probability that \(T_1 < T_5\) can be evaluated using the exponential distribution properties. Given \(T_1\) follows \(\text{Exponential} \left( \lambda \right)\) and \(T_5\) follows \(\text{Exponential} \left( 5\lambda \right)\), \[ P(T_1 < T_5) = \frac{\lambda}{\lambda + 5\lambda} = \frac{1}{6}\]

Finalize Solution

The probability that the \(X_1\)-process reaches level 10 before the \(X_2\)-process does is \(\frac{1}{6}\). This is based on the concept of exponential race, where the process with a fewer required increment has an advantage.

Key concepts

Independence of Random Variables

Independence of random variables is a fundamental concept in probability theory. Two random variables are independent if the occurrence of an event in one variable does not affect the probability of an event in the other variable. This concept is crucial when dealing with Poisson processes because it simplifies the understanding of the event interactions.

In the context of the problem, we have two independent Poisson processes, \(X_1(t)\) and \(X_2(t)\), with a common intensity, \(\lambda\). What does this mean in practice? It means that the events in one process occur without any influence from the other process. Whenever you see that processes are independent, you can think of them as two separate, non-interacting paths.

• For \(X_1(t)\): Events occurring have no effect on \(X_2(t)\).
• For \(X_2(t)\): Events occurring have no effect on \(X_1(t)\).

This independence allows us to consider the probability that each process reaches a specific level without needing to account for the interactions between the processes.

Exponential Distribution

The exponential distribution is essential when dealing with the time between events in a Poisson process. The exponential distribution characterizes the time until the next event occurs, and it has a remarkable property of being 'memoryless'. This means the probability of an event occurring in the future is independent of the past.

When addressing the problem, job is made easier because each \(T_1\) and \(T_5\), the times taken for \(X_1(t)\) to reach level 10 and \(X_2(t)\) to reach level 5 events respectively, follow the exponential distribution:

• For \(T_1\): Follows \(\text{Exponential}(\lambda)\).
• For \(T_5\): Follows \(\text{Exponential}(5\lambda)\).

Why different rates? This reflects the different number of events needed for each process. \(X_1\) simply requires one more event, hence a rate of \(\lambda\), while \(X_2\) requires five, with a rate of \(5\lambda\). Remember, the smaller the \(\lambda\) value, the longer a process is likely to take, given the same conditions.

Event Rate

The concept of event rate, represented by \(\lambda\), is crucial in understanding the dynamics of Poisson processes and, by extension, the exponential distribution. The event rate \(\lambda\) indicates the average number of events occurring in a unit time period.

In the Poisson process, the event rate is a measure of intensity or frequency of events. Both \(X_1(t)\) and \(X_2(t)\) operate with the same event rate, \(\lambda\).

• Higher \(\lambda\): Means more frequent events.
• Lower \(\lambda\): Means less frequent events.

In this exercise, we have to comprehend how the event rates influence the time taken for each process to reach their required levels. In practical terms, we are dealing with exponential random variables,\(T_1\) and \(T_5\), with rates of \(\lambda\) and \(5\lambda\) respectively. The rate being higher for \(X_2(t)\) indicates the requirement for more events within the same time period compared to \(X_1(t)\), setting the stage for the probability comparison that was calculated.