3.7 Poisson Distribution
========================

:bolditalic:`Key Terms`

:bolditalic:`Poisson distribution（卜瓦松分配）`
   A discrete probability distribution for the number of events occurring in
   a fixed interval or region.

:bolditalic:`Rate（發生率）`
   The average number of events per unit of time, distance, area, volume, or
   another exposure unit.

:bolditalic:`Interval（區間）`
   The time period or spatial region over which events are counted.

:bolditalic:`Poisson process（卜瓦松過程）`
   A model in which events occur independently at a stable average rate.

:bolditalic:`Rare-event approximation（稀有事件近似）`
   The use of a Poisson distribution to approximate a binomial distribution
   when the number of trials is large and the success probability is small.


3.7.1 Main Idea
---------------

The **Poisson distribution** models the number of events occurring in a fixed
interval of time or space when events occur randomly at a stable average rate.

The interval may represent:

- time, such as calls per hour;
- distance, such as flaws per meter of wire;
- area, such as defects per square meter;
- volume, such as particles in a fixed volume;
- another fixed exposure unit, such as errors per 100 pages.

Let

.. math::

   X=\text{number of events in the interval of interest}.

Then

.. math::

   X\in\{0,1,2,3,\ldots\}.

Unlike a binomial random variable, there is no fixed upper bound on the number
of events that can occur in the interval.

A useful way to see the conceptual change from the previous binomial model is:

- **Binomial:** fix the number of trials and count successes;
- **Poisson:** fix an interval or region and count events occurring in it.

Many engineering processes do not naturally have a fixed number of Bernoulli
trials. Instead, events such as machine breakdowns, incoming calls, surface
flaws, or accidents occur randomly over time or space. In such settings, the
mean event count for the interval is often more natural to specify than a
number of trials and a per-trial success probability.


3.7.2 Conditions for a Poisson Model
-------------------------------------

A standard Poisson model assumes that events occur according to a stable rate.
A useful introductory set of conditions is:

1. events are counted in a fixed interval or region;
2. counts in non-overlapping intervals are independent;
3. the average event rate is constant over the region being modeled;
4. for a very small interval, the chance of one event is approximately
   proportional to the interval size;
5. for a very small interval, the chance of two or more events is negligible.

These conditions are often summarized by saying that events occur
**independently and randomly at a constant average rate**.

The model is not appropriate when events strongly influence one another or
when the rate changes substantially within the interval.


3.7.3 Rate and the Interval of Interest
----------------------------------------

Suppose :math:`\lambda` is the average rate per unit interval and the interval
of interest has size :math:`t`. The expected number of events in that interval
is

.. math::

   \mu=\lambda t.

The units must match. For example, if the rate is given per hour but the
question asks about 15 minutes, first convert 15 minutes to :math:`0.25` hour.

Consider a bakery that places an average of 12 loaves on a shelf every
30 minutes. For a 5-minute interval,

.. math::

   \mu
   =
   12\left(\frac{5}{30}\right)
   =2.

Thus the expected number of loaves in 5 minutes is 2.

.. image:: ../_static/figures/discrete_distributions/figure_3_7_1_rate_scaling.png
   :width: 78%
   :align: center

The figure emphasizes that the Poisson parameter must correspond to the
**actual interval in the probability question**.

.. note::

   A common notation is

   .. math::

      X\sim\operatorname{Poisson}(\mu),

   where :math:`\mu` is the expected count for the interval being modeled.
   Some texts, including the accompanying lecture slides, use
   :math:`\lambda` directly for the **mean count in the interval**. In this
   handout, :math:`\lambda` denotes a rate per unit and
   :math:`\mu=\lambda t` denotes the mean count for the chosen interval.
   Thus the slide notation :math:`\lambda` corresponds to :math:`\mu` here.


3.7.4 Probability Mass Function
--------------------------------

If

.. math::

   X\sim\operatorname{Poisson}(\mu),

then the probability of exactly :math:`x` events is

.. note::

   **Poisson PMF**

   .. math::

      P(X=x)
      =
      \frac{e^{-\mu}\mu^x}{x!},
      \qquad
      x=0,1,2,\ldots.

If :math:`\mu=\lambda t`, the same formula can be written as

.. math::

   P(X=x)
   =
   \frac{e^{-\lambda t}(\lambda t)^x}{x!}.

The formula gives the probability of **exactly** :math:`x` events in the
specified interval. The constant :math:`e` is Euler's number,
:math:`e\approx2.71828`.


3.7.5 Why the PMF Sums to One
------------------------------

A valid PMF must satisfy

.. math::

   \sum_{x=0}^{\infty}P(X=x)=1.

For the Poisson PMF,

.. math::

   \sum_{x=0}^{\infty}P(X=x)
   =
   \sum_{x=0}^{\infty}
   \frac{e^{-\mu}\mu^x}{x!}.

Factor out :math:`e^{-\mu}`:

.. math::

   =
   e^{-\mu}
   \sum_{x=0}^{\infty}\frac{\mu^x}{x!}.

Recall the exponential-series identity

.. math::

   e^{\mu}
   =
   \sum_{x=0}^{\infty}\frac{\mu^x}{x!}.

Therefore,

.. math::

   \sum_{x=0}^{\infty}P(X=x)
   =
   e^{-\mu}e^{\mu}
   =1.


3.7.6 Example 1: Loaves in Five Minutes
----------------------------------------

Continue the bakery example. The expected number of loaves placed on the shelf
in five minutes is

.. math::

   \mu=2.

Let :math:`X` be the number of loaves placed on the shelf in five minutes.
Find the probability of exactly three loaves:

.. math::

   P(X=3)
   =
   \frac{e^{-2}2^3}{3!}.

Since

.. math::

   3!=6,

we obtain

.. math::

   P(X=3)
   =
   \frac{8e^{-2}}{6}
   \approx0.1804.

The entire distribution for :math:`\mu=2` is concentrated around small
nonnegative counts, with the largest probabilities near the mean.

.. image:: ../_static/figures/discrete_distributions/figure_3_7_2_poisson_pmf_mu2.png
   :width: 72%
   :align: center


3.7.7 Cumulative Probabilities and Complements
-----------------------------------------------

For a Poisson random variable,

.. math::

   P(X\leq k)
   =
   \sum_{x=0}^{k}
   \frac{e^{-\mu}\mu^x}{x!}.

There is no simpler elementary closed form for this finite sum, so cumulative
probabilities are usually evaluated by software or a probability table.

The complement rule is especially useful:

.. math::

   P(X>k)
   =
   1-P(X\leq k).

Likewise,

.. math::

   P(X\geq k)
   =
   1-P(X\leq k-1).

Consider a person who receives an average of 6 calls in two hours. For a
15-minute interval,

.. math::

   \mu
   =
   6\left(\frac{15}{120}\right)
   =0.75.

Let :math:`X` be the number of calls in the next 15 minutes. To find the
probability of more than one call,

.. math::

   P(X>1)
   =
   1-P(X\leq1).

Thus,

.. math::

   P(X>1)
   =
   1-
   \left[
   P(X=0)+P(X=1)
   \right].

Substitute the Poisson PMF:

.. math::

   P(X>1)
   =
   1-
   \left[
   \frac{e^{-0.75}(0.75)^0}{0!}
   +
   \frac{e^{-0.75}(0.75)^1}{1!}
   \right].

Therefore,

.. math::

   P(X>1)
   =
   1-e^{-0.75}(1+0.75)
   \approx0.1734.


3.7.8 Deriving the Mean
------------------------

For

.. math::

   X\sim\operatorname{Poisson}(\mu),

start from the definition of expectation:

.. math::

   E(X)
   =
   \sum_{x=0}^{\infty}xP(X=x).

Substitute the PMF:

.. math::

   E(X)
   =
   \sum_{x=0}^{\infty}
   x\frac{e^{-\mu}\mu^x}{x!}.

The :math:`x=0` term is zero, so

.. math::

   E(X)
   =
   \sum_{x=1}^{\infty}
   \frac{x e^{-\mu}\mu^x}{x!}.

Because

.. math::

   \frac{x}{x!}
   =
   \frac{1}{(x-1)!},

we have

.. math::

   E(X)
   =
   \mu
   \sum_{x=1}^{\infty}
   \frac{e^{-\mu}\mu^{x-1}}{(x-1)!}.

Let :math:`j=x-1`. Then

.. math::

   E(X)
   =
   \mu
   \sum_{j=0}^{\infty}
   \frac{e^{-\mu}\mu^j}{j!}.

The sum is 1 because it is the total probability of a Poisson distribution.
Therefore,

.. note::

   .. math::

      \boxed{E(X)=\mu}.

So the Poisson parameter is also the mean count in the interval.


3.7.9 Deriving the Variance
----------------------------

It is convenient to first calculate the factorial moment
:math:`E[X(X-1)]`:

.. math::

   E[X(X-1)]
   =
   \sum_{x=0}^{\infty}
   x(x-1)P(X=x).

Substitute the PMF:

.. math::

   E[X(X-1)]
   =
   \sum_{x=2}^{\infty}
   x(x-1)
   \frac{e^{-\mu}\mu^x}{x!}.

Since

.. math::

   \frac{x(x-1)}{x!}
   =
   \frac{1}{(x-2)!},

we obtain

.. math::

   E[X(X-1)]
   =
   \mu^2
   \sum_{x=2}^{\infty}
   \frac{e^{-\mu}\mu^{x-2}}{(x-2)!}
   =
   \mu^2.

Now use

.. math::

   X^2=X(X-1)+X.

Therefore,

.. math::

   E(X^2)
   =
   E[X(X-1)]+E(X)
   =
   \mu^2+\mu.

Finally,

.. math::

   \operatorname{Var}(X)
   =
   E(X^2)-[E(X)]^2

so

.. math::

   \operatorname{Var}(X)
   =
   (\mu^2+\mu)-\mu^2
   =
   \mu.

Thus,

.. note::

   **Poisson Mean, Variance, and Standard Deviation**

   .. math::

      \boxed{
      E(X)=\mu,
      \qquad
      \operatorname{Var}(X)=\mu,
      \qquad
      \sigma=\sqrt{\mu}
      }.

The equality of the mean and variance is a distinctive property of the basic
Poisson model.


3.7.10 Effect of the Parameter
-------------------------------

The parameter :math:`\mu` controls both the center and spread of the Poisson
distribution.

When :math:`\mu` is small, the distribution is strongly right-skewed and most
probability is concentrated near zero. As :math:`\mu` increases, the center
moves to the right and the distribution becomes less skewed and more
bell-shaped.

Because

.. math::

   \sigma=\sqrt{\mu},

larger values of :math:`\mu` also produce greater absolute spread.


3.7.11 Deriving Poisson from a Binomial Limit
-----------------------------------------------

The lecture slides motivate the Poisson distribution as a limiting case of a
binomial model. The idea is to imagine **many possible opportunities** for an
event, while the probability of an event in each opportunity becomes very
small.

Let

.. math::

   X_n\sim\operatorname{Binomial}(n,p_n).

Keep the expected number of successes fixed at

.. math::

   np_n=\mu.

Therefore,

.. math::

   p_n=\frac{\mu}{n}.

As :math:`n\to\infty`, we have :math:`p_n\to0`, while the average number of
successes :math:`\mu` remains fixed. This is the ``many opportunities, rare
event`` setting emphasized in the slides.

Start with the binomial PMF:

.. math::

   P(X_n=x)
   =
   \binom nx
   p_n^x(1-p_n)^{n-x}.

Substitute :math:`p_n=\mu/n`:

.. math::

   P(X_n=x)
   =
   \binom nx
   \left(\frac{\mu}{n}\right)^x
   \left(1-\frac{\mu}{n}\right)^{n-x}.

To make the limit easier to evaluate, separate the expression into two
factors:

.. math::

   P(X_n=x)=A_nB_n,

where

.. math::

   A_n
   =
   \frac{n!}{x!(n-x)!}
   \frac{\mu^x}{n^x}

and

.. math::

   B_n
   =
   \left(1-\frac{\mu}{n}\right)^{n-x}.

**Part 1: the combinatorial factor**

Rewrite :math:`A_n` as

.. math::

   A_n
   =
   \frac{\mu^x}{x!}
   \frac{n(n-1)\cdots(n-x+1)}{n^x}.

Equivalently,

.. math::

   A_n
   =
   \frac{\mu^x}{x!}
   \prod_{k=0}^{x-1}
   \left(1-\frac{k}{n}\right).

For fixed :math:`x`, each factor satisfies

.. math::

   \lim_{n\to\infty}
   \left(1-\frac{k}{n}\right)
   =1.

Therefore,

.. math::

   \lim_{n\to\infty}A_n
   =
   \frac{\mu^x}{x!}.

**Part 2: the exponential factor**

Rewrite :math:`B_n` as

.. math::

   B_n
   =
   \left(1-\frac{\mu}{n}\right)^n
   \left(1-\frac{\mu}{n}\right)^{-x}.

Use the fundamental exponential limit

.. math::

   \lim_{n\to\infty}
   \left(1-\frac{\mu}{n}\right)^n
   =e^{-\mu}.

Also, for fixed :math:`x`,

.. math::

   \left(1-\frac{\mu}{n}\right)^{-x}
   \to1.

Hence,

.. math::

   \lim_{n\to\infty}B_n=e^{-\mu}.

Combining the two limits gives

.. math::

   \lim_{n\to\infty}P(X_n=x)
   =
   \frac{\mu^x}{x!}e^{-\mu}
   =
   \frac{e^{-\mu}\mu^x}{x!}.

This is exactly the Poisson PMF.

.. note::

   The slide derivation uses :math:`\lambda=np` for the fixed mean count and
   obtains

   .. math::

      P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}.

   This is the same result. In this handout, the corresponding interval mean
   is denoted by :math:`\mu`.

This limit explains why the Poisson distribution can approximate a binomial
model when :math:`n` is large and :math:`p` is small. It does **not** mean that
every Poisson application must literally arise from an identifiable set of
Bernoulli trials; the Poisson model is also used directly for event counts in
continuous time or space.

A common practical rule is:

    large :math:`n`, small :math:`p`, and use :math:`\mu=np`.

For example, suppose

.. math::

   n=200,
   \qquad
   p=0.0102.

Then

.. math::

   \mu=np=2.04.

For exactly 10 successes,

.. math::

   P_{\text{Binomial}}(X=10)
   \approx3.90\times10^{-5},

while the Poisson approximation gives

.. math::

   P_{\text{Poisson}}(X=10)
   =
   \frac{e^{-2.04}(2.04)^{10}}{10!}
   \approx4.47\times10^{-5}.

The two distributions are very close across the region where most of the
probability lies.

.. image:: ../_static/figures/discrete_distributions/figure_3_7_3_binomial_poisson_approximation.png
   :width: 76%
   :align: center

The approximation becomes increasingly useful when individual successes are
rare but there are many opportunities for them to occur.


3.7.12 Poisson Versus Binomial
-------------------------------

.. list-table::
   :header-rows: 1
   :widths: 34 33 33

   * - Feature
     - Binomial
     - Poisson
   * - Basic setting
     - Fixed number of trials
     - Fixed time/space interval
   * - Random variable
     - Number of successes
     - Number of events
   * - Main parameters
     - :math:`n,p`
     - :math:`\mu` or rate :math:`\lambda`
   * - Possible values
     - :math:`0,1,\ldots,n`
     - :math:`0,1,2,\ldots`
   * - Mean
     - :math:`np`
     - :math:`\mu`
   * - Variance
     - :math:`np(1-p)`
     - :math:`\mu`

The distinction is mainly about the data-generating mechanism. A Poisson model
is not chosen merely because the response is a count.


3.7.13 A Simple Modeling Procedure
-----------------------------------

For a possible Poisson problem:

1. Define the event being counted.
2. Identify the interval or region of interest.
3. Identify the average rate and its units.
4. Convert the rate to the required interval using

   .. math::

      \mu=\lambda t.

5. Define :math:`X` as the number of events in that interval.
6. Translate the requested probability into notation.
7. Use the PMF, a cumulative sum, or a complement as appropriate.
8. Interpret the result in the context of the interval being studied.


3.7.14 Quick Practice
----------------------

An emergency room receives an average of 5 patients per hour. Assume a
Poisson model.

Let :math:`X` be the number of patients arriving in the next 30 minutes.
Find:

(a) the mean number of arrivals in 30 minutes;

(b) :math:`P(X=2)`;

(c) :math:`P(X>2)`.

**Answer**

The 30-minute interval is :math:`0.5` hour, so

.. math::

   \mu=5(0.5)=2.5.

Then

.. math::

   P(X=2)
   =
   \frac{e^{-2.5}(2.5)^2}{2!}
   \approx0.2565.

For more than two arrivals,

.. math::

   P(X>2)
   =
   1-P(X\leq2)

.. math::

   =
   1-
   \left[
   P(X=0)+P(X=1)+P(X=2)
   \right]
   \approx0.4562.


3.7.15 Common Mistakes
-----------------------

Do not use the original rate without converting it to the interval in the
question. If the rate is :math:`\lambda` per unit and the interval has size
:math:`t`, use

.. math::

   \mu=\lambda t.

Do not confuse a Poisson model with a binomial model. Binomial problems count
successes in a fixed number of trials; Poisson problems count events in a
fixed interval or region.

Do not forget that the support begins at zero:

.. math::

   X=0,1,2,\ldots.

Do not forget

.. math::

   0!=1.

This is essential when calculating :math:`P(X=0)`.

Do not assume that every count variable is Poisson. The independence and
stable-rate assumptions must still be reasonable.


3.7.16 Summary
--------------

The Poisson distribution models the number of events occurring in a fixed
interval or region at a stable average rate.

If :math:`\lambda` is the rate per unit and :math:`t` is the interval size,

.. math::

   \mu=\lambda t.

For

.. math::

   X\sim\operatorname{Poisson}(\mu),

.. math::

   P(X=x)
   =
   \frac{e^{-\mu}\mu^x}{x!},
   \qquad
   x=0,1,2,\ldots.

Its mean and variance are

.. math::

   E(X)=\mu,
   \qquad
   \operatorname{Var}(X)=\mu,

and

.. math::

   \sigma=\sqrt{\mu}.

The Poisson distribution can also approximate a binomial distribution when
:math:`n` is large, :math:`p` is small, and

.. math::

   \mu=np.