2.2c.4 Exponential Distribution

Exponential Distribution

A non-negative continuous random variable $X\ge 0$ follows the Exponential distribution $X\sim \text{Exp}(\lambda )$ with parameter $\lambda >0$, defined by the probability density function
$f_X(x)=\{\begin{array}{ll}\lambda e^{-\lambda x}& \text{if}x\ge 0\\ 0& \text{otherwise.}\end{array}$
The CDF for the Exponential distribution is $F_X(x)=1-e^{-\lambda x}$.

The Exponential distribution represents the waiting time until the first occurrence of an event in a process where events occur continuously, independently, and at a constant average rate $\lambda$.

Additionally, akin to the discrete Geometric distribution, the exponential distribution is the only continuous PDF that has the memoryless property, which states that the probability of needing more than $m$ additional trials to get the first success is independent of the $k$ trials that have already been conducted, such that
$\mathbb{P}[X>k+m\mid X>k]=\mathbb{P}[X>m],$
showing that whether or not you start fresh or you have already had $k$ failures, the chance of needing more than $m$ additional trials remains the same.

Properties

(Mean) $\mathbb{E}(X\sim \text{Exp}(\lambda ))=\frac{1}{\lambda }$.

(Variance) $\text{Var}(X\sim \mathrm{Exp}(\lambda ))=\frac{1}{{\lambda }^2}$.

(Discretization) If $X\sim \text{Exp}(\lambda )$, $\lfloor X\rfloor +1\sim \mathrm{Geom}(1-e^{-\lambda })$.

We know that the CDF of $W$ is $F_W(w)=1-e^{-\lambda w}$. Then, $F_X(x)=1-e^{-\lambda x}$ for $x\in \mathbb{N}$. We are trying to find the probability of W falling in the interval $[x,x+1)$ (since this continuous interval represents the floor of values of $x$). As such,

f_{X}(x)&=F_{X}(x+1)-F_{X}(x) \\ &=1-e^{-\lambda(x+1)}-1+e^{-\lambda x} \\ &=e^{-\lambda x}(1-e^{-\lambda})\\ &=(e^{-\lambda})^x(1-e^{-\lambda}), \end{align}$$ which represents a geometric distribution with $p=1-e^{-\lambda}$ by definition. # Examples >[!example] Exponential Order Statistics > >Let $X_{1}, \ldots, X_{n} \sim \operatorname{Exp}(\lambda_{i})$ be mutually independent. >>[!example]- What is the distribution of $\min(X_{1}, \ldots, X_{n})$? >>Observe that since each variable $X_{i}$ denotes the time at which the $i$-th event occurs, then $\min(X_{1}, \ldots, X_{n})$ denotes the time until an event occurs. As such, the minimum will be much smaller than just a single $X_{i}$. >> >>Think about this in survival time terms. The survival time of $\min(X_{1}, \ldots, X_{n})$ is >>$$\begin{align} >>P(\min(X_{1}, \ldots, X_{n}) > t) &= P(X_{1} > t, \ldots, X_{n} > t) \\ >>&= P(X_{1}>t) \times \ldots \times P(X_{n} > t) \\ >>&= \prod_{i=1}^{n}e^{-\lambda_{i} t} \\ >>&= e^{-t \sum^{n}_{i=1}\lambda_{i}}, >>\end{align}$$ >>and therefore $\min(X_{1}, \ldots, X_{n}) \sim \operatorname{Exp}\left( \sum\limits_{i=1}^{n} \lambda_{i} \right)$. >> >>Think of it like $n$ independent phone lines, each getting random calls at their own rate $\lambda_i$. The time until the very first phone rings is faster because all the lines are “competing”–together, they get calls at the combined rate $\lambda_1 + \lambda_2 + \dots + \lambda_n$. > >Now let $X_{1}, X_{2}, X_{3} \sim_{\text{iid}} \operatorname{Exp}(\lambda)$ be mutually independent. >>[!example]- Given that $\min(X_{1}, X_{2}, X_{3}) = m$, what is the probability that the difference between the median and the smallest is at least $s$?