Question

A Poisson process is observed during \(n\) days. The intensity is, however, not constant, but varies randomly day by day, so that we may consider the intensities during the \(n\) days as \(n\) independent, \(\operatorname{Exp}\left(\frac{1}{\alpha}\right)\) distributed random variables. Determine the distribution of the total number of occurrences during the \(n\) days.

Short answer

The total occurrences follow a Negative Binomial distribution with parameters \( n \) and \( \frac{1}{1+\alpha} \).

Step-by-step solution

Understand the problem

We are asked to find the distribution of the total number of occurrences over \( n \) days for a Poisson process with varying intensity. Each day's intensity is independent and follows an \( \text{Exp}\left(\frac{1}{\alpha}\right) \) distribution.

Identify the random components

Each day's intensity is \( \lambda_i \sim \text{Exp}\left(\frac{1}{\alpha}\right) \). The process over each day is a Poisson process with intensity \( \lambda_i \), yielding a Poisson distribution for the number of occurrences, \( N_i \sim \text{Poi}(\lambda_i) \).

Determine the distribution of occurrences for one day

Since \( \lambda_i \sim \text{Exp}\left(\frac{1}{\alpha}\right) \) and the number of occurrences \( N_i \sim \text{Poi}(\lambda_i) \), the marginal distribution of \( N_i \) is \( \text{Geo}\left(\frac{1}{1+\alpha}\right) \), where \( \text{Geo}(p) \) denotes a Geometric distribution starting from zero.

Determine the distribution for total occurrences over multiple days

The total number of occurrences over \( n \) days is \( N = N_1 + N_2 + \cdots + N_n \). Since each \( N_i \) is independent and identically distributed as \( \text{Geo}\left(\frac{1}{1+\alpha}\right) \), the sum \( N \) follows a \( \text{NegBin}(n, \frac{1}{1+\alpha}) \) distribution.

Find the final answer

The total number of occurrences during \( n \) days follows a Negative Binomial distribution with parameters \( n \) (the number of trials) and \( \frac{1}{1+\alpha} \) (the probability of success on each trial).

Key concepts

Exponential Distribution

The Exponential Distribution is a continuous probability distribution often used to model the time until an event occurs, such as the time between events in a Poisson process. It is defined by a single parameter, \( \alpha \), which is the rate parameter. The probability density function of an exponential distribution is given by: \[ f(x; \alpha) = \alpha e^{-\alpha x} \quad \text{for } x \geq 0 \] In our problem, each day's intensity, \( \lambda_i \), is modeled as an exponentially distributed random variable with parameter \( \frac{1}{\alpha} \). This means we are considering the rate at which events happen to be random. The memoryless property is a key feature of exponential distributions, meaning the future probability of an event occurring is independent of any past events.

Geometric Distribution

The Geometric Distribution models the number of trials needed to get the first success in a series of Bernoulli trials (trials with only two possible outcomes, like success or failure). For this distribution:

• The parameter \( p \) is the probability of success on each trial.
• The support is \( 0, 1, 2, \ldots \) if considering successes in open trials.

The probability mass function is given by: \[ P(X = k) = (1-p)^{k} p \] In the exercise, the number of occurrences for each day, \( N_i \), follows a Geometric distribution with parameter \( \frac{1}{1+\alpha} \). This is derived because each Poisson count is modeled through its random intensity from an Exponential distribution. This setup offers a unique blend where each day's occurrences have a geometric flavor under the Poisson environment.

Negative Binomial Distribution

The Negative Binomial Distribution summarizes the number of trials required to achieve a certain number of successes. Specifically, it extends the geometric model. While the geometric distribution deals with the first success, the negative binomial distribution generalizes this to getting \( r \) successes. It is characterized by:

• \( r \): The number of successful events we want to occur.
• \( p \): The probability of success on each trial.

The probability mass function for the negative binomial distribution is: \[ P(X = k) = \binom{k+r-1}{r-1} (1-p)^k p^r \] In our problem, the total number of occurrences over \( n \) days, \( N \), fits a Negative Binomial distribution with parameters \( n \) (representing the number of days) and \( \frac{1}{1+\alpha} \) (desired outcomes' probability). This situates where each day's Poisson counts contribute to the sum, with geometric distributions summing up to negative binomial outcomes.

Random Intensity

Random intensity implies that the rate at which events occur is not fixed but subject to random variation. In the context of a Poisson process, it means the parameter controlling the average number of events per unit time varies randomly. This concept is pivotal when dealing with real-world scenarios where conditions fluctuate over time.

• In our exercise, random intensities are modeled by the Exponential distribution.
• This randomness in intensity leads to variability in daily occurrences.
• Each day's Poisson process thus behaves differently based on its day's randomized intensity value.

The notion captures the essence of unpredictability in event rates, allowing for realistic modeling of complex systems where deterministic parameters don't suffice.