Exponential Distribution

Definition

The probability density function for an exponential random variable is:

\[ f_X(x) = \begin{cases} \begin{alignat}{2} &\lambda e^{-\lambda x} \quad &&x \geq 0, \\ &0 \quad &&\text{else} \end{alignat} \end{cases} \]

where \(\lambda > 0\) is called the rate parameter. An exponentially distributed random variable \(X\) is denoted by \(X \sim \text{Exp} (\lambda)\). The CDF is given by:

\[ F(x; \lambda) = \begin{cases} \begin{alignat}{2} &1 - e^{-\lambda x} \quad &&x \geq 0, \\ &0 && \text{else}. \end{alignat} \end{cases} \]

Figure 1 shows the PDF and CDF for \(\lambda = 0.5, 1.0, 1.5\).

Figure 1 Exponential distribution PDF and CDF

Mean and Variance

Its moment-generating function is:

\[ m(t) = \lambda / (\lambda - t), \]

for \(t < \lambda\). Hence its mean and variance are \(1 / \lambda\) and \(1 / \lambda^2\) respectively. The \(n\)'th moment is \(\mathbb{E} X^n = \frac{n!}{\lambda^n}\).

Failure Rate

Failure rate is defined as:

\[ \lambda_X(t) = \frac{f_X(t)}{1 - F_X(t)}, \]

and is constant and equal to \(\lambda\).

Relationship to Poisson Distribution

The exponential distribution has an important connection to the Poisson distribution. It is the probability distribution of the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. Suppose we observe i.i.d exponential variates \((X_1, X_2, \ldots)\) and define \(S_n = X_1 + \cdots + X_n\). For any positive value \(t\), it can be shown that:

\[ \mathbb{P}(S_n < t < S_{n + 1}) = p_Y(n), \]

where \(p_Y(n)\) is the probability mass function for a Poisson random variable \(Y\) with parameter \(\lambda t\).

Like a geometric random variable, an exponential random variable has the memoryless property:

\[ \mathbb{P}(X \geq u + v | X \geq u) = \mathbb{P}(X \geq v). \]

Examples

Melanoma

The 5-year cancer survival rate in the case of malignant melanoma of the skin at stage 3A is \(78\%\). Assume that the survival time \(T\) can be modeled by an exponential random variable with unknown rate \(\lambda\). Given the 5-year survival rate, compute the probability of a melanoma patient surviving more than 10 years.

The rate of surival for 5 years is given as 0.78. Assuming that the survival rate random variable is an exponential distribution and using the CDF:

\[ \mathbb{P}(T > t) = \exp (-\lambda t), \]

or:

\[ \mathbb{P}(T > 5) = 0.78. \]

Then, we have \(\lambda = -\frac{1}{5} \log(0.78) = 0.05\).

Finally, the probability that the survival time exceeds 10 years is:

\[ \mathbb{P}(T > 10) = 1 - F(x=10, \lambda=0.05) = 1 - (1 - \exp(-0.05 \cdot 10)) = \frac{1}{\sqrt{e}} = 0.6065. \]

Figure 2 shows the survival PDF and CDF as a function of time.

Exponential Lifetimes

Let \(n = 20\) independent components be connected in a serial system; that is, all components need to be operational for the system to work. The lifetime of each component is an exponential random variable with rate of \(1/3\) (in units of 1 per year). What is the probability that the system remains operational for more than one month?

If \(T_i\) are lifetimes of system components, the system's lifetime is \(T = \min_i T_i\). For \(T\) to exceed \(t\), each \(T_i\) has to exceed \(t\):

\[ \mathbb{P}(T > t) = \mathbb{P}(T_1 > t, T_2 > t, \ldots, T_n > t). \]

If \(T_i\) are independent exponentials with rates \(\lambda_i\):

\[ \prod^n_{i = 1} \mathbb{P}(T_i > t) = \prod^n_{i = 1} \exp\left\{-\lambda_i t \right\} = \exp \left\{ -t \sum^n_{i = 1} \lambda_i \right\}, \]

which shows that \(T \sim \text{Exp}(\sum^n_{i = 1} \lambda_i)\). Then we have:

\[ \mathbb{P}(T > 1/12) = \exp \left\{-10/12 \right\} = 0.4346. \]

Appendix

Plotting Script

from scipy.stats import expon
import numpy as np
from matplotlib import pyplot as plt

## Expnential distribution
lambda_1, lambda_2, lambda_3 = 0.5, 1.0, 1.5
scale_1, scale_2, scale_3 = 1 / lambda_1, 1 / lambda_2, 1 / lambda_3
x = np.linspace(0, 6)

fig, ax = plt.subplots(1, 2)
ax[0].set_title("Exponential Distribution PDF")
ax[0].set_xlim((-0.5, 5.5))
ax[1].set_ylim((-0.5, 1.5))
ax[0].plot(x, expon.pdf(x, scale=scale_1), "-", label="lambda = {}".format(lambda_1), alpha=1.0)
ax[0].plot(x, expon.pdf(x, scale=scale_2), "-", label="lambda = {}".format(lambda_2), alpha=0.7)
ax[0].plot(x, expon.pdf(x, scale=scale_3), "-", label="lambda = {}".format(lambda_2), alpha=0.7)
ax[0].set_box_aspect(1)
ax[0].set_xlabel("x")
ax[0].set_ylabel("P(x)")

ax[0].grid(True)
ax[0].legend()

ax[1].set_title("Exponential Distribution CDF")
ax[1].set_xlim((-0.5, 5.5))
ax[1].set_ylim((-0.5, 1.5))
ax[1].plot(x, expon.cdf(x, scale=scale_1), "-", label="lambda = {}".format(lambda_1), alpha=1.0)
ax[1].plot(x, expon.cdf(x, scale=scale_2), "-", label="lambda = {}".format(lambda_2), alpha=0.7)
ax[1].plot(x, expon.cdf(x, scale=scale_3), "-", label="lambda = {}".format(lambda_2), alpha=0.7)
ax[1].set_box_aspect(1)
ax[0].set_xlabel("P(X < x)")
ax[1].set_ylabel("Probability")
ax[1].grid(True)
ax[1].legend()

fig.tight_layout()
fig.savefig("exponential_distribution.png", dpi=800, bbox_inches="tight")