Exponential distribution ($\mathrm{Exp}(\lambda )$)

The exponential distribution with rate $c>0$ has densityIt is the gamma distribution with shape one.

For independent exponential waits of rates $a$ and $b$, their sum has density $ab(e^{-bt}-e^{-at})/(a-b)$, obtained by convolution on $[0,t]$. The limit at $a=b$ is $a^{2}t{e}^{-at}$ and the mean is $a^{-1}+b^{-1}$.

An exponential clock has an exponentially distributed waiting time. Independent clocks with respective rates $r_i$ have a minimum waiting time with rate ${\sum }_i{r}_i$.

Let $S$ be a fixed sum of independent exponential random variables with smallest rate $\lambda >0$, and let $Z_N$ be the sum of $N-k$ independent rate-$r$ exponential random variables, independent of $S$, with fixed $k$ and $r>\lambda$. At large-deviation speed $N$, $S/N$ has rate function $\lambda y$ for $y\ge 0$; the sample mean has rate function $K_r(z)=rz-1-\log (rz)$ for $z>0$. The product large-deviation principle and contraction principle for large deviations giveIndeed the derivative of the quantity minimized with respect to $z$ is $r-\lambda -1/z$. Its minimizer is $\min (x,(r-\lambda {)}^{-1})$. Beyond that transition, a single slow exponential random variable carries the extra excursion; the rate function becomes affine.

For independent exponential random variables $X,Y$ with the same rate $\lambda >0$, the convolution density of $X+Y/2$ is $2\lambda (e^{-\lambda v}-e^{-2\lambda v})$ for $v>0$. The cumulative distribution function of $\max (X,Y)$ is $(1-e^{-\lambda v}{)}^2$, whose derivative is the same density. This is equality of probability distributions, not an almost-sure equality of the two expressions formed from the same sample.

For exponential rate $\lambda >0$, the integer part has mass $(1-e^{-\lambda })e^{-\lambda m}$ on nonnegative integers. The fractional part has density $\lambda e^{-\lambda u}/(1-e^{-\lambda })$ for $0\le u<1$. The mixed joint density factors into those marginals, proving independence. The integer part is a geometric distribution on zero-based support, with expectation $1/(e^{\lambda }-1)$.

If $X\sim \mathrm{Exp}(\lambda )$, then direct integration gives

If $T\sim \mathrm{Exp}(\mu )$, thenThus an exponential remaining lifetime has the same law regardless of the elapsed lifetime.

If $A$ and $B$ are independent exponential variables with the same rate, then $A/(A+B)$ is uniform on $(0,1)$ and is independent of $A+B$.

For independent exponential variables $S_i$ of rates $q_i$ and $Q={\sum }_i{q}_i$, let $T={\min }_i{S}_i$ and let $K$ be the minimizing index. ThenConsequently $T$ is exponential with rate $Q$, $\mathbb{P}(K=k)=q_k/Q$, and $K$ and $T$ are independent.

Let $S,T$ be independent random variables with exponential distributions of respective positive rates $\alpha ,\beta$. Condition on $T=t$ and use the survival function of $S$:At $u=0$ this is $P(S>T)$. Dividing by that probability proves that the positive difference conditional on $S>T$ again has an exponential distribution with rate $\alpha$. The memorylessness of the exponential distribution persists after an independent competing event has already occurred. Replacing the difference by a negative sentinel when $S\le T$ adds an atom of a measure of mass $\alpha /(\alpha +\beta )$ at that sentinel without changing the positive tail.

If $N$ is geometric on $\{1,2,\dots \}$ with parameter $p$, independently of unit-rate exponential variables $S_i$, thenis exponential with rate $p$. Its moment-generating function is $p/(p-\theta )$ for $\theta <p$.