Index: The Book of Statistical Proofs ▷ Probability 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

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

Proof: The probability density function of the exponential distribution is:

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

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}:

\[\label{eq:exp-cdf-s2b} \begin{split} F_X(x) &= \int_{-\infty}^{0} \mathrm{Exp}(z; \lambda) \, \mathrm{d}z + \int_{0}^{x} \mathrm{Exp}(z; \lambda) \, \mathrm{d}z \\ &= \int_{-\infty}^{0} 0 \, \mathrm{d}z + \int_{0}^{x} \lambda \exp[-\lambda z] \, \mathrm{d}z \\ &= 0 + \lambda \left[ -\frac{1}{\lambda} \exp[-\lambda z] \right]_{0}^{x} \\ &= \lambda \left[ \left( -\frac{1}{\lambda} \exp[-\lambda x] \right) - \left( -\frac{1}{\lambda} \exp[-\lambda \cdot 0] \right) \right] \\ &= 1 - \exp[-\lambda x] \; . \end{split}\]
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Metadata: ID: P48 | shortcut: exp-cdf | author: JoramSoch | date: 2020-02-11, 14:48.