The exponential [[Random Variable|random variable]] represents the time it takes for an event to occur in a [[Poisson Distribution|Poisson]] process. $\lambda$ is the same parameter as in the [[Poisson Distribution|Poisson distribution]], representing the average number of times that the event occurs over a time period. It has the following [[Probability Distribution Function|probability distribution function]]: $ \begin{align*} f(x) &= \lambda e^{-\lambda x} \text{ if } x \ge 0 \\ P(X < x) &= \int_{0}^{x}\lambda e^{-\lambda t} dt = 1 - e^{-\lambda t} \\ P(X > x) &= e^{-\lambda t} \end{align*} $ The [[Expectation|mean]] waiting time until the first occurrence is the reciprocal of the event's frequency: $\mu = E(X) = \frac{1}{\lambda}$. The [[Variance|variance]] of the distribution is $\frac{1}{\lambda^2}$. $ P(X > b|X > a) = \frac{P(X > b)}{P(X > a)} = \frac{e^{-\lambda b}}{e^{-\lambda a}} = e^{-\lambda(b - a)} = P(X > b - a) $ The Poisson process has no memory, and prior events do not affect the likelihood of the current event. ### Derivation During the time it takes for the event to occur, no events happen, which means that the [[Probability|probability]] of one unit of such interval to occur is $Po(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$. The probability for this interval to last $t$ times as long or longer is equal to $P(T > t) = Po(X = 0)^{t} = e^{-\lambda t}$, translating to $P(T > t) = 1 - e^{-\lambda t}$. Differentiating the latter returns $\frac{d}{dx}(1 - e^{-\lambda t}) = \lambda e^{-\lambda t}$.