Question

Cars pass a certain street location according to a Poisson process with rate \(\lambda\). A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next \(T\) time units. (a) Find the probability that her waiting time is \(0 .\) (b) Find her expected waiting time. Hint: Condition on the time of the first car.

Short answer

(a) The probability that her waiting time is 0 is: \[P(waiting\: time = 0) = e^{-\lambda T}\] (b) The expected waiting time for the woman to cross the street is: \[E(waiting\: time) = -T e^{-\lambda T} + \frac{1}{\lambda}\left(1 - e^{-\lambda T} \right)\]

Step-by-step solution

Find the probability of no cars in the next T time units

We know that the number of cars passing the location follows a Poisson distribution with a rate of λ. The probability that no cars will pass in the next T time units can be calculated using the Poisson probability mass function: \[P(X = 0) = \frac{e^{-\lambda T}( \lambda T)^0}{0!} = e^{-\lambda T}\] where X is the number of cars passing by.

Part (a): Probability that her waiting time is 0

The woman wants to cross the street when she sees no cars coming by in the next T time units. Her waiting time would be 0 if there are no cars coming in the next T time units at the very moment she arrives. We've deduced that the probability no cars will pass in the next T time units is \(e^{-\lambda T}\), so that's her probability of waiting time being 0. Thus, the probability that her waiting time is 0 is: \[P(waiting\: time = 0) = e^{-\lambda T}\]

Part (b): Finding her expected waiting time

To find her expected waiting time, we need to condition on the time of the first car. Let Y be her waiting time, and let X be the time until the first car passes. We want to find E(Y). We can write the expected waiting time as: \[E(Y) = E(Y|X = x) P(X = x)\] Since we are conditioning on the time of the first car, we need to determine the probability density function for X. Given that cars follow a Poisson process with rate λ, the inter-arrival time of cars follows an exponential distribution with parameter λ, i.e., \[f_X(x)=\lambda e^{-\lambda x}\] for x ≥ 0. Now, let's compute the expected waiting time given X = x: If \(x > T\), then the woman can start crossing the street immediately and her waiting time will be 0, i.e., \[E(Y | X = x) = 0\] If \(0 \leq x \leq T\), the woman will have to wait until the time x has elapsed before starting to cross, i.e., \[E(Y | X = x) = x\] The expected waiting time is then given by the following integral: \[E(Y) = \int_{0}^{T} x \lambda e^{-\lambda x} dx + \int_{T}^{\infty} 0 \lambda e^{-\lambda x} dx\] Since the second integral is multiplied by 0, it does not contribute to the total, and we are left with: \[E(Y) = \lambda \int_{0}^{T} x e^{-\lambda x} dx\] To solve this integral, we can use integration by parts, considering: \[u=x, dv=e^{-\lambda x} dx\Rightarrow du=dx, v=-\frac{1}{\lambda}e^{-\lambda x}\] Integration by parts gives: \[E(Y) = \lambda \left[-\frac{x}{\lambda}e^{-\lambda x} \Big|_0^T + \frac{1}{\lambda}\int_{0}^{T} e^{-\lambda x} dx\right] = -x e^{-\lambda x} \Big|_0^T + \frac{1}{\lambda} \left( -\frac{1}{\lambda} e^{-\lambda x} \Big|_0^T \right)\] Evaluating the limits for the integrals: \[E(Y) = -T e^{-\lambda T} + \frac{1}{\lambda}\left(1 - e^{-\lambda T} \right)\] This gives us the expected waiting time for the woman to cross the street: \[E(waiting\: time) = -T e^{-\lambda T} + \frac{1}{\lambda}\left(1 - e^{-\lambda T} \right)\]

Key concepts

Poisson Distribution

The Poisson distribution is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, provided the events occur with a known constant mean rate and independently of the time since the last event. The classic example of a Poisson process is the number of cars passing through a point on a road per unit time.

For a Poisson process with a rate parameter \( \lambda \), the probability of observing exactly \( k \) events in a time interval \( T \) is given by the Poisson probability mass function:
\[ P(X = k) = \frac{e^{-\lambda T}( \lambda T)^k}{k!} \]
This formula is fundamental when dealing with scenarios where events happen at a constant average rate but randomly and independently from each other.

Exponential Distribution

The Exponential distribution is closely linked with the Poisson distribution; it describes the time between consecutive events in a Poisson process. If the number of events per time period follows a Poisson distribution with rate \( \lambda \), then the time between events follows an Exponential distribution with the same rate \( \lambda \).

For an Exponential distribution, the probability density function is defined as:
\[ f_X(x) = \lambda e^{-\lambda x} \] for \( x \geq 0 \).
This relationship is essential for understanding phenomena where we're interested in the time it takes for the next event to occur, like the time until the next car passes in our exercise.

Expected Waiting Time

Expected waiting time is a concept in probability theory that represents the average time one expects to wait for a particular event to occur. In the case of a Poisson process, it’s often linked to the waiting time until the first event. For our textbook example, to calculate the expected waiting time for the woman to safely cross the street, we condition on the time of arrival of the first car.

Her expected waiting time is calculated by integrating over the probability distribution of waiting times, taking care to treat the scenarios differently depending on whether the first car arrives before or after \( T \) time units.
The interplay between the Poisson process and the Exponential distribution allows us to compute this expected value, which turns out to be an integral using the density function of the Exponential distribution.

Probability Mass Function

The probability mass function (PMF) is a function that provides the probabilities of all possible values for a discrete random variable. It is a key concept when dealing with discrete probability distributions like the Poisson distribution. The PMF allows us to calculate how likely it is to observe a certain number of events within a given interval.

For a Poisson distribution with parameter \( \lambda \), the PMF is defined as:
\[ P(X = k) = \frac{e^{-\lambda T}(\lambda T)^k}{k!} \]
where \( k \) is the number of occurrences, and \( T \) is the time interval. This function is crucial for solving parts of our exercise and providing a proper understanding of how the Poisson distribution enables us to model and compute probabilities in situations characterized by randomness and independence.