Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Exponential distribution ▷ Cumulative distribution function

Theorem: Let $X$ be a random variable following an exponential distribution:

$\label{eq:exp} X \sim \mathrm{Exp}(\lambda) \; .$

Then, the cumulative distribution function of $X$ is

$\label{eq:exp-cdf} F_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < 0 \\ 1 - \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; . \end{array} \right.$ $\label{eq:exp-pdf} \mathrm{Exp}(x; \lambda) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < 0 \\ \lambda \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; . \end{array} \right.$

Thus, the cumulative distribution function is:

$\label{eq:exp-cdf-s1} F_X(x) = \int_{-\infty}^{x} \mathrm{Exp}(z; \lambda) \, \mathrm{d}z \; .$

If $x < 0$, we have:

$\label{eq:exp-cdf-s2a} F_X(x) = \int_{-\infty}^{x} 0 \, \mathrm{d}z = 0 \; .$

If $x \geq 0$, we have using \eqref{eq:exp-pdf}:

$\label{eq:exp-cdf-s2b} \begin{split} F_X(x) &= \int_{-\infty}^{0} \mathrm{Exp}(z; \lambda) \, \mathrm{d}z + \int_{0}^{x} \mathrm{Exp}(z; \lambda) \, \mathrm{d}z \\ &= \int_{-\infty}^{0} 0 \, \mathrm{d}z + \int_{0}^{x} \lambda \exp[-\lambda z] \, \mathrm{d}z \\ &= 0 + \lambda \left[ -\frac{1}{\lambda} \exp[-\lambda z] \right]_{0}^{x} \\ &= \lambda \left[ \left( -\frac{1}{\lambda} \exp[-\lambda x] \right) - \left( -\frac{1}{\lambda} \exp[-\lambda \cdot 0] \right) \right] \\ &= 1 - \exp[-\lambda x] \; . \end{split}$
Sources:

Metadata: ID: P48 | shortcut: exp-cdf | author: JoramSoch | date: 2020-02-11, 14:48.