Poisson Proccess

The Poisson distribution is a distribution of the number of events that occur within a given time interval \(t\). PThe probability mass function (PMF) of the Poisson distribution is:

\[ P(\mathbf{X}=k)=\frac{\lambda^ke^{-k}}{k!} \]

where \(\lambda\) is the expected number of events over the time interval, or the rate times the time interval.

A Poisson process is a process that determines the distribution of the time elapsed between the \((n-1)\)-th and \(n\)-th event. To calculate this, we use the exponential distribution with mean \(1/\lambda\).

Poisson process has the following properties:

• The number of events that occur in different time intervals are independent.
• The distribution of the number of events that occur in a given interval depends only on the length of the interval, not the location in the time domain.
• The probability of two events occurring at the exact same time is 0.

Example, If the rate is 5 orders per hour, compute the probability of the number of events in a 2-hour interval. This is a Poisson distribution with mean \(\lambda \times t\), given \(\lambda = 5\) and \(t=2\) :

\[ \mathbf{X}(2) \sim \text{Poisson}(10) \]

Then, the probability of receiving 8 orders in 2 hours is:

\[ P(\mathbf{X}=8) = \frac{10^8e^{-10}}{8!} \approx 0.1126 \]

To compute the probability of the time interval between order arrivals, we use the exponential distribution. The average time between intervals is:

\[ \text{E}(T) = \frac{1}{\lambda} = 0.2 \text{ hour} \]

Next, we compute the probability that the next order arrives in the next 15 minutes using the CDF of the exponential distribution:

\[ 15 \text{ minutes} = \frac{15}{60} =0.25 \text{ hour}\]

Thus, the probability is:

\[ P(T \leq 0.25) = 1-e^{-5\cdot 0.25} \approx 0.7135 \]