Question

Consider a Poisson process with intensity \(\lambda\), and let \(T\) be the time of the last occurrence in the time interval \((0, t]\). If there is no occurrence during \((0, t]\), we set \(T=0\). Compute \(E T\).

Short answer

The expected time of the last occurrence, \(E[T]\), is \(\frac{t}{2}(1 - e^{-\lambda t})\).

Step-by-step solution

Define the Poisson Process

A Poisson process with intensity \( \lambda \) describes events occurring randomly over an interval, such that the number of events in a time period of length \( t \) follows a Poisson distribution with parameter \( \lambda t \). The time between consecutive events is exponentially distributed with parameter \( \lambda \).

Express the Expectation in Terms of Conditional Expectation

To find the expected value of \( T \), use the law of total expectation: \[E[T] = E[T \mid N(t) = 0]P(N(t) = 0) + E[T \mid N(t) \geq 1]P(N(t) \geq 1)\]Where \( N(t) \) is the number of events in \((0, t]\). If \( N(t) = 0 \), we know that \( T = 0 \), and if \( N(t) \geq 1 \), \( T \) follows a certain distribution.

Compute the Probabilities Involved

Compute the probability of no events occurring by noting that the number of events \( N(t) \) in a Poisson process is given by:\[P(N(t) = 0) = e^{-\lambda t}\]Thus, \[ P(N(t) \geq 1) = 1 - e^{-\lambda t}. \]

Determine the Distribution of \(T\) Given Occurrences

Given \( N(t) \geq 1 \), the time \( T \) of the last occurrence follows a Uniform distribution between \( 0 \) and \( t \). This is because all occurrences are equally likely to be the last one. Hence, \[E[T \mid N(t) \geq 1] = \frac{t}{2}.\]

Calculate the Expected Value of \(T\)

Substitute values back into the law of total expectation:\[E[T] = 0 \cdot e^{-\lambda t} + \frac{t}{2} \cdot (1 - e^{-\lambda t}) \]This simplifies to:\[ E[T] = \frac{t}{2}(1 - e^{-\lambda t}). \]

Verify and Interpret the Result

The result makes intuitive sense as it reflects how the expectation of the last event in the time interval depends on both the interval's length \( t \) and the rate \( \lambda \) of occurrences. When \( \lambda \) is very high, the first term dominates, resulting in an expectation close to \( \frac{t}{2} \).

Key concepts

Conditional Expectation

The concept of conditional expectation is like peeling back layers in mathematics. It's how we find the expected value of a random variable, given that we have some additional information. Imagine you're trying to predict an outcome with extra clues in hand.
In the context of a Poisson process, conditional expectation plays a pivotal role. Our original exercise uses conditional expectation to find the expected time of the last occurrence, given certain conditions. It's represented by the formula:

• \[E[T] = E[T \mid N(t) = 0]P(N(t) = 0) + E[T \mid N(t) \geq 1]P(N(t) \geq 1)\]

This formula tells us that the expected value of \(T\) depends on whether there were zero or one/more events during the interval (\(0, t]\). Conditional expectation allows us to break the calculation into manageable parts based on these scenarios.

Exponential Distribution

Exponential distribution is like the heartbeat of a Poisson process. It's the distribution that explains the time between successive events happening in a Poisson process. The exponential distribution is defined with a parameter \( \lambda \), which is the average rate of occurrences.
Think of it in practical terms: If you're waiting for buses arriving randomly, the time between one bus and the next follows an exponential distribution.
Beautifully, the exponential distribution has the 'memoryless' property. This means the future probability of an event doesn’t depend on past events. In our exercise, every gap between events in a Poisson process is exponentially distributed, which simplifies the calculations and helps predict which event might be the last one in any given interval.

Uniform Distribution

Uniform distribution is like drawing lots from a hat. Every outcome has an equal chance of happening. That's exactly how it's used in our problem. When we know that at least one event has occurred in the interval \((0, t]\), the last event is uniformly distributed over this interval.
This means if there's at least one event, the probability is spread evenly across the time interval, making the expected value of \(T\) given the condition, simply the midpoint:

• \[E[T \mid N(t) \geq 1] = \frac{t}{2}\]

In this way, whether the event happened early or late in the interval doesn't influence the calculation. By applying uniform distribution in such scenarios, calculations for expected values become simpler and intuitive.

Total Expectation

The law of total expectation is a powerful tool. Imagine you're calculating an average based on different scenarios. You'd want to weigh each average by how likely it is to occur. The law of total expectation helps achieve this harmony.
This is integral to our Poisson process problem. We used it to calculate the expected time for the last event. By breaking the expectation of \(T\) into two parts—when no event happens and when at least one event does—the total expectation holds all the pieces together.

• It allows us to efficiently calculate overall expectations by averaging across conditions or scenarios.
• With this law, we easily plugged in probabilities and conditional expectations to derive the final solution.

Expectation of Random Variables

Expectation is like the center of gravity of a random variable. It's the weighted average of all possible values. In our problem, the random variable \(T\) (the time of the last occurrence) had its expectation computed using a mix of probability and mean calculations from its conditions.

• For \(N(t) \geq 1\), knowing \(T\) is uniformly distributed makes calculating its expectation straightforward as just \(\frac{t}{2}\).
• The overall expectation calculation also considers the situation where no events occur, assigning \(E[T \mid N(t) = 0] = 0\).

In this way, understanding the concept of expectation is essential for computing possible outcomes in statistical processes like the Poisson process.