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For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval for the rate transforms to one for the mean by reversing and inverting its endpoints. The monotonic transformation preserves the coverage event; it does not give a prediction interval for one realized holding time.

For the stated critical geometric offspring law, the conditional upper tail is exactly $[n/(n+1){]}^{\lfloor nx\rfloor }$ for $x\ge 0$. Its logarithm tends to $-x$, proving an exponential distribution limit of rate one without a transform continuity theorem. The unconditional mass at zero tends to one, so survival conditioning changes the relevant limiting population scale.

For an event time with an exponential distribution of mean $\mu$ and an independent censoring time, let $X$ be their minimum. Memorylessness of the exponential distribution makes $X+\mu$ the conditional mean of the latent event time when it is censored. Keeping $X$ when the event is observed gives an unbiased imputed event time by the tower property of conditional expectation.

For a mixture distribution of exponential distributions, introduce one convolution state $A_j(u)={\int }_0^u\phi (z)e^{-{\beta }_j(u-z)}dz$ for each rate ${\beta }_j$. It satisfies $A_j^{\prime }=\phi -{\beta }_j{A}_j$. Together with the survival integro-differential equation these form a constant-coefficient first-order system; applying the differential operators $D+{\beta }_j$ eliminates the convolution states. For equal mixing weights at rates one and one half, with $r=\lambda /c$, the result is ${\phi }^{\prime \prime \prime }+(3/2-r){\phi }^{\prime \prime }+(1/2-3r/4){\phi }^{\prime }=0$. The original integral equation supplies initial conditions lost in elimination.

For independent observations from the exponential distribution with rate $\theta$, the log-likelihood is $n\log \theta -\theta {\sum }_i{X}_i$, soThe central limit theorem and delta method give

Here $n(H)dH$ is the expected number of ridge peaks per track length with sea-ice draft in $[H,H+dH]$. If total line density is $\mu$ and mean peak draft is $h_m>h_0$, integration gives $b=1/(h_m-h_0)$ and $B=\mu b{e}^{b{h}_0}$. The normalized peak draft has a shifted exponential distribution. This is a model over a specified draft range, not a universal description of every ridge.

In a classical risk model, before the first claim at time $t$, available capital is $u+ct$. A claim of size $x\le u+ct$ leaves future survival probability $\phi (u+ct-x)$ by the Markov property. Integrating over the independent first-arrival exponential distribution and claim density gives $\phi (u)={\int }_0^{\infty }\lambda e^{-\lambda t}{\int }_0^{u+ct}\phi (u+ct-x)f(x)dxdt$.

A flow-level model treats each document transfer as a flow whose rate depends on the current occupancy and resource capacities. With independent Poisson processes for arrivals and independent unit-mean exponential distributions for sizes, a route's total completion rate equals its allocated aggregate service.

A frozen-age approximation holds an age covariate constant at the start of each observation interval, allowing homogeneous transition probabilities to enter a panel-observed multi-state likelihood. It approximates a model with continuously changing age. Reciprocal rate calibration of mean waiting times is exact for a fixed-age exponential distribution, but not for a continuously ageing Gompertz distribution.

A gamma prior with shape $\alpha$ and rate $\beta$ combined with $n$ independent observations from an exponential distribution of rate $\theta$ gives the posteriorUnder squared-error loss, the Bayes estimator is its posterior mean $(\alpha +n)/(\beta +{\sum }_i{X}_i)$.

If $\Lambda$ has gamma distribution with shape $2$ and rate $p/q$, $N\mid \Lambda$ has Poisson distribution with intensity $\Lambda$, and the independent claim sizes have exponential distribution of mean $\mu$, where $p+q=1$, the aggregate law is $p^2{\delta }_0+2pq\mathrm{Exp}(p/\mu )+q^2\mathrm{Gamma}(2,\text{rate }p/\mu )$. This follows by expanding its moment-generating function as $[p+qp/(p-\mu t){]}^2$.

An observed astronomical colour is the sum of a normal intrinsic colour, a nonnegative dust contribution with an exponential distribution, and normal measurement noise. With latent $C_s\sim N(\mu ,v)$, $E_s\sim \mathrm{Exp}(\text{mean }\tau )$ and $O_s\mid C_s,E_s\sim N(C_s+E_s,r_s)$, the Gibbs sampler conditionals for $C_s$ are normal and those for $E_s$ are truncated normal distributions. A flat prior on $\mu$ and inverse-gamma priors on $v,\tau$ give inverse-gamma conditional updates for the two scales. Log-flat priors on both scales instead make the joint posterior improper when all $r_s>0$, despite formally proper full conditionals at generic latent states: the marginal observed likelihood has a positive limit at either zero-scale boundary. Proper positive-scale priors restore a genuine joint posterior.

For scale $\sigma >0$ and shape $\xi$, the generalized Pareto distribution on nonnegative excesses has cumulative distribution functionThe case $\xi =0$ is the continuous limit $1-e^{-y/\sigma }$. Positive $\xi$ gives a power-law tail, zero gives an exponential distribution, and negative $\xi$ gives a finite upper endpoint $-\sigma /\xi$. Its role in modelling high-threshold excesses is explained by the Pickands-Balkema-de Haan theorem.

Let the summands have gamma distribution with shape two and rate $\beta$, and let the independent positive-support geometric distribution count have parameter $p$. Put $a=\beta (1-\sqrt{1-p})$ and $b=\beta (1+\sqrt{1-p})$. The aggregate moment-generating function factors as $ab/((a-t)(b-t))$, so it is the sum of two independent exponential distributions of rates $a,b$. Its probability density function is $ab(e^{-as}-e^{-bs})/(b-a)$ for $s>0$, nonnegative and integrating to one. This contrasts with the geometric sum of exponential variables, which is itself exponential.

The Gillespie algorithm samples an exact path of a well-mixed stochastic reaction network. At state $x$, let $a_0(x)={\sum }_r{a}_r(x)$; draw a waiting time with exponential distribution of rate $a_0$, choose reaction $r$ with probability $a_r/a_0$, and update $x\leftarrow x+{\nu }_r$.

A positive, exponentially increasing hazard function $h(t)=q{e}^{\beta t}$ gives the survivor functionIts mean is ${\beta }^{-1}{e}^{q/\beta }{E}_1(q/\beta )$, where $E_1$ is the exponential integral. Unlike the exponential distribution, its future mean waiting time is not the reciprocal of its current hazard function.

The holding time of a continuous-time Markov chain is the time spent in its current state before the next jump. In state $i$ it has an exponential distribution of rate equal to the total outgoing rate $q_i$.

A finite mixture distribution of exponential distributions with rates $r_i>0$ and nonnegative weights summing to one. Its expected value is ${\sum }_i{w}_i/r_i$ and its moment-generating function is ${\sum }_i{w}_i{r}_i/(r_i-t)$ on the common finite domain. The mixture chooses one component for each observation; it is not a sum of independent exponential variables.

A hypoexponential distribution is the distribution of a sum of independent variables with exponential distributions. For two positive unequal rates $\lambda ,\mu$, convolution of independent random variables gives the probability density functionIt is zero for negative $s$. Its equal-rate limit is the shape-two gamma distribution with rate $\lambda$.

For two independent rate-$\lambda >0$ exponential distributions, set $U=\min (X,Y)$ and $V=\max (X,Y)$. Summing the two ordering branches gives joint probability density $2{\lambda }^2{e}^{-\lambda (2u+d)}$ for $(U,d=V-U)$ with $u,d>0$. It factors into independent exponential probability density functions. Thus $V$ is distributed as $X+Y/2$, with expected value $3/(2\lambda )$ and variance $5/(4{\lambda }^2)$. This spacing result is the order-statistic manifestation of the memorylessness of the exponential distribution.