4.3 Exponential Distribution
============================

4.3.1 Main Idea
---------------

The **exponential distribution** models a continuous waiting time until the
next event occurs when events arise independently at a constant average rate.

It is closely connected to the Poisson process:

- the **Poisson distribution** counts how many events occur in a fixed interval;
- the **exponential distribution** measures how long we wait until the next event.

Typical examples include:

- time until the next customer arrives;
- time between telephone calls;
- distance until the next flaw appears;
- lifetime of a component when its failure rate is constant.

Let

.. math::

   \lambda>0

be the event rate per unit time or space. If :math:`X` denotes the waiting time
until the next event, then

.. math::

   X\geq0.

.. image:: ../_static/figures/continuous_distributions/figure_4_3_1_poisson_exponential_timeline.png
   :width: 86%
   :align: center

The same event process can therefore be viewed in two ways. Over a fixed
interval, we may count events. Starting from one event, we may instead measure
the continuous distance or time to the next event.


4.3.2 Rate and Mean-Waiting-Time Parameterizations
--------------------------------------------------

Two equivalent parameterizations are commonly used.

If :math:`\lambda` is the event rate, then the mean waiting time is

.. math::

   \mu=\frac1\lambda.

Equivalently,

.. math::

   \lambda=\frac1\mu.

Some references use a symbol such as :math:`m` for this rate or **decay
parameter**. In this handout, we use :math:`\lambda` consistently.

Thus the two equivalent forms of the density are

.. math::

   f(x)=\lambda e^{-\lambda x}

and

.. math::

   f(x)=\frac1\mu e^{-x/\mu}.

The units are important:

- :math:`\lambda` has units of events per unit time;
- :math:`\mu` and :math:`X` have units of time.

For example, if customers arrive at an average rate of 30 per hour, then

.. math::

   \lambda=\frac{30}{60}=0.5\text{ per minute},

so the mean waiting time is

.. math::

   \mu=\frac1{0.5}=2\text{ minutes}.


4.3.3 Probability Density Function
----------------------------------

A random variable :math:`X` has an exponential distribution with rate
:math:`\lambda` if

.. note::

   **Exponential PDF**

   .. math::

      f(x)
      =
      \begin{cases}
      \lambda e^{-\lambda x}, & x\geq0,\\
      0, & x<0.
      \end{cases}

The density is largest at :math:`x=0`:

.. math::

   f(0)=\lambda.

It then decreases continuously toward zero as :math:`x` increases. This is why
small waiting times are more common than very large waiting times.

As with every probability density function, the total area must equal 1:

.. math::

   \int_{-\infty}^{\infty}f(x)\,dx
   =
   \int_0^\infty \lambda e^{-\lambda x}\,dx.

Since

.. math::

   \int \lambda e^{-\lambda x}\,dx
   =
   -e^{-\lambda x},

we obtain

.. math::

   \int_0^\infty \lambda e^{-\lambda x}\,dx
   =
   \left[-e^{-\lambda x}\right]_0^\infty
   =
   0-(-1)
   =1.

The value :math:`f(x)` is a **density**, not the probability of one exact
waiting time. Because :math:`X` is continuous,

.. math::

   P(X=x)=0.


4.3.4 CDF and Survival Function
-------------------------------

The cumulative distribution function is

.. math::

   F(x)=P(X\leq x).

For :math:`x\geq0`, integrate the density from 0 to :math:`x`:

.. math::

   F(x)
   =
   \int_0^x \lambda e^{-\lambda t}\,dt.

Therefore,

.. math::

   F(x)
   =
   \left[-e^{-\lambda t}\right]_0^x
   =
   1-e^{-\lambda x}.

Hence,

.. note::

   **Exponential CDF**

   .. math::

      F(x)
      =
      \begin{cases}
      0, & x<0,\\
      1-e^{-\lambda x}, & x\geq0.
      \end{cases}

The complementary probability

.. math::

   P(X>x)

is called the **survival function**. It is

.. math::

   P(X>x)
   =
   1-F(x)
   =
   e^{-\lambda x}.

This simple exponential tail is one of the most useful properties of the
distribution.


4.3.5 Probability Calculations
------------------------------

For :math:`x\geq0`,

.. math::

   P(X\leq x)
   =
   1-e^{-\lambda x}.

For waiting **more than** :math:`x`,

.. math::

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

For an interval :math:`0\leq a<b`,

.. math::

   P(a<X<b)
   =
   F(b)-F(a).

Substituting the exponential CDF gives

.. math::

   P(a<X<b)
   =
   e^{-\lambda a}-e^{-\lambda b}.

.. image:: ../_static/figures/continuous_distributions/figure_4_3_2_exponential_probability_area.png
   :width: 78%
   :align: center

The shaded area under the density between two time points represents the
probability that the event occurs during that waiting-time interval.

Because :math:`X` is continuous, endpoint inclusion does not matter:

.. math::

   P(a<X<b)
   =
   P(a\leq X\leq b).


4.3.6 Main Example: Time Between Customer Arrivals
--------------------------------------------------

Suppose customers arrive according to a Poisson process at an average rate of
30 customers per hour.

Express the rate in minutes:

.. math::

   \lambda=\frac{30}{60}=0.5\text{ per minute}.

Let

.. math::

   X=\text{time in minutes until the next customer arrives}.

Then

.. math::

   X\sim\operatorname{Exp}(0.5).

The mean waiting time is

.. math::

   E(X)=\frac1{0.5}=2\text{ minutes}.

**Probability that the next customer arrives within one minute**

.. math::

   P(X<1)
   =
   1-e^{-0.5(1)}
   \approx0.3935.

**Probability that the wait exceeds five minutes**

.. math::

   P(X>5)
   =
   e^{-0.5(5)}
   =
   e^{-2.5}
   \approx0.0821.

**Probability that the waiting time is between one and three minutes**

.. math::

   P(1<X<3)
   =
   e^{-0.5(1)}-e^{-0.5(3)}
   \approx0.3834.

These three calculations illustrate the three common forms of exponential
probability questions: left-tail, right-tail, and interval probabilities.


4.3.7 Deriving the Mean
-----------------------

For a continuous random variable,

.. math::

   E(X)
   =
   \int_{-\infty}^{\infty}x f(x)\,dx.

For an exponential random variable,

.. math::

   E(X)
   =
   \int_0^\infty x\lambda e^{-\lambda x}\,dx.

Use integration by parts. Let

.. math::

   u=x,
   \qquad
   dv=\lambda e^{-\lambda x}\,dx.

Then

.. math::

   du=dx,
   \qquad
   v=-e^{-\lambda x}.

Therefore,

.. math::

   E(X)
   =
   \left[-xe^{-\lambda x}\right]_0^\infty
   +
   \int_0^\infty e^{-\lambda x}\,dx.

The boundary term is zero, so

.. math::

   E(X)
   =
   \left[-\frac1\lambda e^{-\lambda x}\right]_0^\infty
   =
   \frac1\lambda.

Thus,

.. note::

   **Mean of an Exponential Distribution**

   .. math::

      \boxed{E(X)=\mu=\frac1\lambda}.

A larger event rate therefore produces a shorter mean waiting time.


4.3.8 Deriving the Variance and Standard Deviation
--------------------------------------------------

Use

.. math::

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

First compute

.. math::

   E(X^2)
   =
   \int_0^\infty x^2\lambda e^{-\lambda x}\,dx.

Again use integration by parts with

.. math::

   u=x^2,
   \qquad
   dv=\lambda e^{-\lambda x}\,dx.

Then

.. math::

   du=2x\,dx,
   \qquad
   v=-e^{-\lambda x}.

Therefore,

.. math::

   E(X^2)
   =
   \left[-x^2e^{-\lambda x}\right]_0^\infty
   +
   2\int_0^\infty x e^{-\lambda x}\,dx.

The boundary term is zero. Also,

.. math::

   \int_0^\infty x e^{-\lambda x}\,dx
   =
   \frac1{\lambda^2}.

Hence,

.. math::

   E(X^2)=\frac2{\lambda^2}.

Now

.. math::

   \operatorname{Var}(X)
   =
   \frac2{\lambda^2}
   -
   \left(\frac1\lambda\right)^2
   =
   \frac1{\lambda^2}.

Therefore,

.. note::

   **Variance and Standard Deviation**

   .. math::

      \boxed{\operatorname{Var}(X)=\frac1{\lambda^2}}

   and

   .. math::

      \boxed{\sigma=\frac1\lambda}.

Thus, for an exponential distribution,

.. math::

   \boxed{\sigma=\mu}.

The mean and standard deviation have the same numerical value and the same
units.


4.3.9 Percentiles and the Median
--------------------------------

Suppose :math:`k_r` is the :math:`r`-th percentile, where :math:`0<r<1`.
Then

.. math::

   P(X\leq k_r)=r.

Using the CDF,

.. math::

   1-e^{-\lambda k_r}=r.

Therefore,

.. math::

   e^{-\lambda k_r}=1-r.

Take natural logarithms:

.. math::

   -\lambda k_r
   =
   \ln(1-r).

Hence,

.. note::

   **Exponential Percentile**

   .. math::

      \boxed{
      k_r=-\frac{\ln(1-r)}{\lambda}
      }.

For the median, :math:`r=0.50`, so

.. math::

   k_{0.50}
   =
   \frac{\ln2}{\lambda}
   \approx
   \frac{0.6931}{\lambda}.

Because

.. math::

   \frac{\ln2}{\lambda}
   <
   \frac1\lambda,

we have

.. math::

   \text{median}<\text{mean}.

This is consistent with the right-skewed shape of the exponential
distribution.

For the customer-arrival example, :math:`\lambda=0.5`. The 70th percentile is

.. math::

   k_{0.70}
   =
   -\frac{\ln(0.30)}{0.5}
   \approx2.41\text{ minutes}.

Thus, 70% of the waiting times are at most about 2.41 minutes.


4.3.10 Memoryless Property
--------------------------

The exponential distribution is **memoryless**. For :math:`s,t\geq0`,

.. math::

   P(X>s+t\mid X>s)
   =
   P(X>t).

To see why, use conditional probability:

.. math::

   P(X>s+t\mid X>s)
   =
   \frac{P(X>s+t)}{P(X>s)}.

Using the survival function,

.. math::

   P(X>s+t\mid X>s)
   =
   \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}
   =
   e^{-\lambda t}.

But

.. math::

   e^{-\lambda t}=P(X>t).

Therefore,

.. math::

   \boxed{
   P(X>s+t\mid X>s)=P(X>t)
   }.

.. image:: ../_static/figures/continuous_distributions/figure_4_3_3_exponential_memoryless.png
   :width: 82%
   :align: center

For the customer-arrival example, suppose 5 minutes have already passed with no
arrival. The probability of waiting more than one **additional** minute is
still

.. math::

   P(X>1)
   =
   e^{-0.5}
   \approx0.6065.

The previous five minutes do not make an arrival “due.”


4.3.11 Connection to the Poisson Distribution
---------------------------------------------

The Poisson and exponential distributions describe the same underlying event
process from two different perspectives.

Suppose events follow a Poisson process with rate :math:`\lambda`, and let

.. math::

   N(t)=\text{number of events during an interval of length }t.

Then

.. math::

   N(t)\sim\operatorname{Poisson}(\lambda t).

Let

.. math::

   T=\text{waiting time until the first event}.

The event

.. math::

   T>t

means that **no event has occurred by time** :math:`t`. Therefore,

.. math::

   P(T>t)
   =
   P(N(t)=0).

Using the Poisson PMF,

.. math::

   P(N(t)=0)
   =
   \frac{e^{-\lambda t}(\lambda t)^0}{0!}
   =
   e^{-\lambda t}.

Thus,

.. math::

   P(T>t)=e^{-\lambda t}.

Therefore,

.. math::

   F_T(t)
   =
   1-e^{-\lambda t}.

Differentiating gives

.. math::

   f_T(t)
   =
   \lambda e^{-\lambda t}.

This is exactly the exponential distribution.

So:

    **Poisson** -> number of events in a fixed interval.

    **Exponential** -> waiting time until the next event.


4.3.12 Reliability Interpretation and Constant Hazard
-----------------------------------------------------

The exponential distribution is often used in reliability when the failure
rate is approximately constant.

The hazard rate is

.. math::

   h(x)
   =
   \frac{f(x)}{P(X>x)}.

For an exponential random variable,

.. math::

   h(x)
   =
   \frac{\lambda e^{-\lambda x}}{e^{-\lambda x}}
   =
   \lambda.

Thus the hazard is constant over time.

For example, suppose a component lifetime is exponential with mean 10 years.
Then

.. math::

   \lambda=\frac1{10}=0.1\text{ per year}.

The probability that it survives more than 7 years is

.. math::

   P(X>7)
   =
   e^{-0.1(7)}
   \approx0.4966.

This model is appropriate only when a constant failure rate is plausible. It
is not suitable for strong wear-out or aging behavior in which failure risk
increases with age.


4.3.13 Exponential, Poisson, and Geometric
------------------------------------------

.. list-table::
   :header-rows: 1
   :widths: 28 24 24 24

   * - Feature
     - Exponential
     - Poisson
     - Geometric
   * - Type
     - Continuous
     - Discrete
     - Discrete
   * - Random variable
     - Waiting time
     - Event count
     - Number of trials until success
   * - Typical setting
     - Continuous event process
     - Continuous event process
     - Repeated Bernoulli trials
   * - Memoryless
     - Yes
     - Not applicable in this form
     - Yes

The exponential distribution can be viewed as the continuous analogue of the
geometric waiting-time model. Both are memoryless, but one measures continuous
time while the other counts discrete trials.


4.3.14 A Simple Method
----------------------

For an exponential problem:

1. Define what :math:`X` measures and identify its units.
2. Identify the event rate :math:`\lambda`, or convert from the mean using
   :math:`\lambda=1/\mu`.
3. Make the units of :math:`x` compatible with the units of :math:`\lambda`.
4. Translate the probability statement:

   - less than -> use :math:`1-e^{-\lambda x}`;
   - more than -> use :math:`e^{-\lambda x}`;
   - between -> subtract two CDF values.

5. Interpret the result in the original context.


4.3.15 Quick Practice
---------------------

Calls arrive at a service desk according to a Poisson process at an average
rate of 4 calls per hour.

Let :math:`X` be the waiting time in hours until the next call.

Find:

(a) :math:`P(X>0.5)`;

(b) :math:`P(X<0.25)`;

(c) :math:`E(X)`; and

(d) the median waiting time.

**Answer**

Here,

.. math::

   \lambda=4.

For part (a),

.. math::

   P(X>0.5)
   =
   e^{-4(0.5)}
   =
   e^{-2}
   \approx0.1353.

For part (b),

.. math::

   P(X<0.25)
   =
   1-e^{-4(0.25)}
   =
   1-e^{-1}
   \approx0.6321.

For part (c),

.. math::

   E(X)=\frac14\text{ hour}=15\text{ minutes}.

For part (d),

.. math::

   k_{0.50}
   =
   \frac{\ln2}{4}\text{ hour}
   \approx0.1733\text{ hour}
   \approx10.4\text{ minutes}.


4.3.16 Common Mistakes
----------------------

A common mistake is to confuse the rate with the mean waiting time.

.. math::

   \mu=\frac1\lambda,

not :math:`\mu=\lambda`.

Another mistake is to use inconsistent units. If :math:`\lambda` is measured
per hour, then the waiting time inserted into the formula must be measured in
hours.

A third mistake is to treat :math:`f(x)` as a point probability. For a
continuous random variable,

.. math::

   P(X=x)=0.

Another mistake is to use a Poisson PMF when the question asks **how long**
until an event. Poisson counts events; exponential measures waiting time.

Finally, do not use the exponential model automatically for every lifetime.
Its constant hazard rate excludes many aging and wear-out mechanisms.


4.3.17 Summary
--------------

For

.. math::

   X\sim\operatorname{Exp}(\lambda),
   \qquad \lambda>0,

with :math:`x\geq0`,

.. math::

   f(x)=\lambda e^{-\lambda x},

.. math::

   F(x)=1-e^{-\lambda x},

and

.. math::

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

The mean, variance, and standard deviation are

.. math::

   E(X)=\frac1\lambda,

.. math::

   \operatorname{Var}(X)=\frac1{\lambda^2},

and

.. math::

   \sigma=\frac1\lambda.

The :math:`r`-th percentile is

.. math::

   k_r=-\frac{\ln(1-r)}{\lambda}.

The distribution is memoryless:

.. math::

   P(X>s+t\mid X>s)
   =
   P(X>t).

Its main connection to the Poisson distribution is:

    Poisson counts events; exponential measures the waiting time between them.