Question

Sven waits for the bus. The waiting time, \(T\), until a bus comes is \(U(0, a)\)-distributed. While he waits he tries to get a ride from cars that pass by according to a Poisson process with intensity \(\lambda\). The probability of a passing car picking him up is \(p\). Determine the probability that Sven is picked up by some car before the bus arrives. Remark. All necessary independence assumptions are permitted.

Short answer

Sven's probability of being picked up before the bus arrives is \(1 - \frac{1 - e^{-\lambda pa}}{\lambda pa}\).

Step-by-step solution

Define the Random Variables

The time Sven waits for the bus is defined by the random variable \(T\), which follows a uniform distribution from 0 to \(a\), i.e., \(T \sim U(0, a)\). The occurrence of passing cars follows a Poisson process with intensity \(\lambda\).

Determine Expected Number of Cars

Since cars pass with an intensity of \(\lambda\), the expected number of cars that pass by time \(t\) is given by the Poisson process as \(\lambda t\).

Calculate Probability of Being Picked Up by a Car

For each car that passes, the probability of it picking up Sven is \(p\). Therefore, the probability that Sven is picked up by any one car is \(1 - (1-p)\). For \(n\) cars, the probability of not being picked by any car is \((1-p)^n\). Consequently, the probability of being picked up after \(n\) trials is \(1 - (1-p)^n\).

Calculate Total Probability of Being Picked Up Before Time t

The total probability that Sven is picked up by some car before time \(t\) can be expressed through the cumulative distribution involving the Poisson distribution: \(1 - e^{-\lambda t p}\). This formula follows from the fact that the events of being picked up are independent Poisson events.

Integrate Over Possible Waiting Times

Since \(T\) is uniformly distributed from 0 to \(a\), the probability Sven is picked up by a car before the bus arrives is the average of the probabilities over all possible waiting times: \[ P(\text{picked up before bus}) = \frac{1}{a} \int_{0}^{a} (1 - e^{-\lambda t p}) \, dt \].

Evaluate the Integral

Now, evaluate the integral: \[ \int_{0}^{a} (1 - e^{-\lambda t p}) \, dt = \int_{0}^{a} 1 \, dt - \int_{0}^{a} e^{-\lambda t p} \, dt = a + \frac{1}{\lambda p} \left( e^{-\lambda pa} - 1 \right) \].

Simplify the Expression

Substitute back into the integral formula: \[ P(\text{picked up before bus}) = \frac{1}{a} \left( a + \frac{1}{\lambda p} \left( e^{-\lambda pa} - 1 \right) \right) = 1 - \frac{1}{a}\frac{1-e^{-\lambda pa}}{\lambda p}\].

Final Probability Formula

Thus, the probability that Sven is picked up by a car before the bus arrives is: \[ 1 - \frac{1 - e^{-\lambda pa}}{\lambda pa} \].

Key concepts

Poisson process

The Poisson process is a fundamental concept in probability theory, often used to model random events occurring over time or space. In Sven’s bus scenario, cars passing by him are considered events. With a Poisson process, these events are assumed to happen independently and at a constant average rate. This rate is denoted by \( \lambda \), known as the intensity.This means if the intensity \( \lambda \) is higher, cars will pass more frequently, increasing the chances of Sven getting a ride before the bus arrives. One of the key properties of a Poisson process is that the number of events happening in a fixed period follows a Poisson distribution.

Uniform distribution

The uniform distribution is one of the simplest probability distributions. It is typically used when each outcome in a range is equally likely. For Sven's wait time for the bus, denoted as \( T \), it follows a uniform distribution from 0 to \( a \), represented as \( T \sim U(0, a) \).Since the distribution is uniform, Sven has an equal chance of waiting any time duration between 0 and \( a \) for his bus. When calculating probabilities for time-based events, such as the chance of getting picked up, we account for all possible waiting times within this uniform range.

Random variables

In probability, a random variable is a variable whose possible values are numerical and determined by the outcomes of a random phenomenon. In the context of Sven waiting for a bus, we define \( T \) as a random variable representing the time Sven waits for the bus.Additionally, the number of cars passing by follows a Poisson process, and the decision each car makes, whether to pick Sven up or not, can also be viewed as a random variable. Understanding random variables is crucial as they help quantify scenarios and their likelihoods in probability calculations.

Integration

Integration is a mathematical process used to find accumulated quantities, such as area under a curve, which in probability theory, often represents the accumulated probability.In Sven’s case, since his waiting time \( T \) is uniformly distributed, we integrate over the interval from 0 to \( a \) to average the probabilities of being picked up over time. This integration determines how likely Sven is to be picked up by a car before the bus arrives, across all possible wait times for the bus.

Expected value

The expected value is a fundamental concept in probability, representing the average outcome of a random process if it were repeated many times. It's a measure of the center of a probability distribution.For the Poisson process of cars passing by Sven with intensity \( \lambda \), the expected number of cars passing by time \( t \) is \( \lambda t \). This concept helps calculate the probability of outcomes, such as one or more cars stopping to give Sven a ride based on the average number of cars passing during his wait time. Understanding expected value allows us to predict patterns and outcomes in random processes effectively.