Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Exponential distribution ▷ Special case of gamma distribution

Theorem: The exponential distribution is a special case of the gamma distribution with shape $a = 1$ and rate $b = \lambda$.

Proof: The probability density function of the gamma distribution is

\[\label{eq:gam-pdf} \mathrm{Gam}(x; a, b) = \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \; .\]

Setting $a = 1$ and $b = \lambda$, we obtain

\[\label{eq:exp-pdf} \begin{split} \mathrm{Gam}(x; 1, \lambda) &= \frac{\lambda^1}{\Gamma(1)} x^{1-1} \exp[-\lambda x] \\ &= \frac{x^0}{\Gamma(1)} \lambda \exp[-\lambda x] \\ &= \lambda \exp[-\lambda x] \end{split}\]

which is equivalent to the probability density function of the exponential distribution.

Sources:

Metadata: ID: P69 | shortcut: exp-gam | author: JoramSoch | date: 2020-03-02, 20:49.