Relationship Between the Gamma Distribution and the Exponential Distribution
Theorem
$$ \Gamma \left(1, { 1 \over \lambda } \right) \iff \text{exp} (\lambda) $$
Explanation
Considering the intuitive definition of the exponential distribution, it is concerned with the time it takes until some event occurs. In terms of discrete probability distributions, the geometric distribution corresponds to this.
Here, the generalization of the geometric distribution with respect to the ’number of occurrences’ of an event is the negative binomial distribution. In this sense, one could say that the generalization of the exponential distribution is the gamma distribution. In this case, the ’number of occurrences’ corresponds to $k$ in the gamma distribution $\displaystyle \Gamma \left( k, { 1 \over \lambda } \right)$, but since there is no particular reason why the parameter $k$ of the gamma distribution must be an integer, it would be problematic to regard them as completely equivalent.
Also note that in the gamma distribution, one takes $\displaystyle { 1 \over \lambda }$ rather than the parameter $\lambda$ of the exponential distribution. Do not think about it too hard; it is enough to just know that it can be thought of in this way.
Proof
Strategy: We show that the moment generating functions of the two distributions can be expressed in the same form.
The moment generating function of the exponential distribution $\text{exp} (\lambda)$ is $\displaystyle m_{1}(t) := (1- {t \over \lambda})^{-1}$, and the moment generating function of the gamma distribution $\Gamma (k, \theta)$ is $\displaystyle m_{2}(t) := (1-\theta t)^{-k}$. Substituting $ k = 1$ and $\displaystyle \theta = { 1 \over \lambda }$ into the moment generating function of the gamma distribution gives $$ m_{2}(t) = (1 - \theta t)^{-k} = (1- {t \over \lambda})^{-1} =m_{1}(t) $$
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