Relationship between Gamma Distribution and Exponential Distribution
Theorem
$$ \Gamma \left(1, { 1 \over \lambda } \right) \iff \text{exp} (\lambda) $$
Description
If we think about the intuitive definition of the exponential distribution, it’s about the interest in the amount of time it takes for a certain event to occur. If we were to relate this to a discrete probability distribution, the geometric distribution would correspond to this.
In this sense, the generalization of the geometric distribution in terms of the ’number of occurrences’ is the negative binomial distribution. From this perspective, the generalization of the exponential distribution could be considered the gamma distribution. Here, the ’number of occurrences’ corresponds from $\displaystyle \Gamma \left( k, { 1 \over \lambda } \right)$ to $k$ in the gamma distribution, but since the parameter $k$ of the gamma distribution does not necessarily have to be an integer, it is problematic to consider them completely equivalent.
In the gamma distribution, note that $\displaystyle { 1 \over \lambda }$ is taken instead of $\lambda$, which is the parameter of the exponential distribution. Don’t think too hard about it; just knowing this much should be sufficient.
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 $\displaystyle m_{2}(t) := (1-\theta t)^{-k}$. When we substitute $ k = 1$ and $\displaystyle \theta = { 1 \over \lambda }$ into the moment generating function of the gamma distribution, $$ m_{2}(t) = (1 - \theta t)^{-k} = (1- {t \over \lambda})^{-1} =m_{1}(t) $$
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