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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 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.

Construct a rate-$\lambda$ Poisson process, for $\lambda >0$, from independent waiting times $E_j$ with exponential distribution $\mathrm{Exp}(\lambda )$. Put $S_k=E_1+\cdots +E_k$, $S_0=0$, and $N_t=\max \{k:S_k\le t\}$. The strong law of large numbers gives $S_k/k\to 1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are càdlàg, and $N_0=0$ almost surely.

At any deterministic time $s$, the memorylessness of the exponential distribution says that the residual waiting time has again exponential distribution $\mathrm{Exp}(\lambda )$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the Poisson process restarts independently at $s$, proving independent increments and stationary increments. Integrating the joint waiting-time densities over $0<s_1<\cdots <s_k\le t<s_{k+1}$ givesso its time values have the expected Poisson distribution.

For $h\downarrow 0$,Together with stationary increments, this gives stochastic continuity at every time, from either side where applicable. All the Lévy process requirements hold, andFor $\lambda =0$, the identically zero process gives the degenerate case.

For an exact failure half an hour after entry, use the conditional density, rather than the probability of an event at one exact time:Again this is independent of $\tau$ by memorylessness of the exponential distribution. Whether successive observed lecture segments came from the same bulb does not affect the conditional likelihood: surviving segments of one bulb multiply into its survival exposure, and replacement starts another memoryless lifetime. Assume no unrecorded use relevant to the exposure calculation or, equivalently, condition on the bulb being alive at each observed lecture's start; unobserved ages themselves need not be known.

The observed working exposure is $12+1/2=12.5$ bulb-hours with one failure. ThusStrict concavity of $\log \lambda -12.5\lambda$ verifies the maximum. Immediate replacement would increase observed working exposure to 13 hours, since the replacement survives the remaining half hour, givingFor the alternative Weibull distribution and unknown entry age in a Weibull survival segment, keep its different inverse-scale convention: $F(t)=\exp [-(\lambda t{)}^p]$. Conditional survival for an additional duration $u$ isand the conditional failure density after $u$ isBoth depend on the unknown age unless $p=1$. The available segment data, with ages and bulb identity unrecorded, do not specify the usual Weibull likelihood needed to estimate $\lambda ,p$. Merely collecting more of the same unknown-age records does not fix the missing age/design information. A larger dataset with documented ages or identities, or a justified model for the entry-age distribution, could permit estimation. If $p=1$ is imposed, the exponential calculation remains valid. The issue is loss of memorylessness and unspecified entry history, not a universal impossibility of learning from any larger bulb experiment.

Condition on the bulb being alive at age $\tau$. The next hour's survival likelihood contribution isIt does not depend on the unknown age $\tau$. This is memorylessness of the exponential distribution: the residual lifetime given survival has the same exponential distribution as a new lifetime.

A feasible per-flow allocation satisfies $x_r>0$ for each active route and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$. It has proportional fairness if every feasible competitor obeysBy the first-order inequality for the strictly concave logarithm, this is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible allocations. Inactive-route rates are immaterial; set them to zero as a convention.

Write $a=n_{\{1\}}$, $b=n_{\{2\}}$, $c=n_{\{1,2\}}$, and $N=a+b+c$. The capacity constraints are $a{x}_1+c{x}_0\le 1$ and $b{x}_2+c{x}_0\le 1$. For $c>0$, put $q=c{x}_0$. Each active local route uses its remaining capacity $1-q$. Omitting constants, the utility isIts maximum has $q=c/N$; when $a+b=0$, the maximum is the endpoint $q=1$. Hence for active routesThese formulas also give $x_1=1/a$ and $x_2=1/b$ when $c=0$. When $N=0$ there is no allocated service. Equivalently, the aggregate route rates for $N>0$ areThis is the proportionally fair allocation on a two-resource linear network for a two-resource linear flow network.

For independent Poisson processes of documents and independent document sizes with exponential distribution of rate ${\mu }_r$, the memorylessness of the exponential distribution makes the population a continuous-time Markov chain. Its transition intensities areThe departure intensity sums the $n_r$ independent completion hazards ${\mu }_r{x}_r$. There is no departure on an inactive route. Set ${\rho }_r={\nu }_r/{\mu }_r$, with ${\mu }_r>0$.

To prove the proposed stationary distribution, define the balance function of a flow-level networkFor an active local route, $\Psi (n-e_r)/\Psi (n)=(a+b)/N$, while for the through route this ratio is $c/N$. Thus ${\Lambda }_r(n)=\Psi (n-e_r)/\Psi (n)$. The weightssatisfy detailed balance on every arrival-departure pair:Each total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$, so the chain is a nonexplosive Markov chain; normalized detailed-balance weights therefore give its stationary distribution, the stationary law of a two-resource linear flow network.

Use the negative binomial series to sum first over $c$:The remaining two geometric series giveAll summands are nonnegative. Consequently the sum is finite exactly under the stability conditions for a two-resource linear flow networkThese say that the offered load on each unit-capacity resource is strictly below one. Equality is insufficient: at least one geometric series then diverges. If all arrival rates are positive, the chain is irreducible and the normalizable law is its unique equilibrium law.

Finally put $q_i={\rho }_i/(1-{\rho }_0)$, $i=1,2$. Summing over $c$ yieldsThus the two local-route counts are independent, each with geometric distribution on the nonnegative integers. The conditional through-route count has negative binomial distributionFor positive traffic rates this depends on the local counts, so all three route counts are not independent. More quantitatively, $\mathrm{Cov}(a,c)={\rho }_0\mathrm{Var}(a)/(1-{\rho }_0)>0$, and likewise for $b$. If zero arrival rates are allowed, the exceptional independent cases are ${\rho }_0=0$, or ${\rho }_1={\rho }_2=0$, when the relevant counts are deterministic zero. This distinction is local-count independence with through-flow dependence.

Put $U=H(T)$ and $V=H(C)$, so $Y=\min (U,V)$ and $0\le \mathbb{E}Y\le \mathbb{E}U=1$. Its expectation is not generally one and depends on censoring. Under independent censoring, condition on $V=v$: the exponential distribution givesWriting $\Delta =1_{T\le C}$ for the observed event indicator, it follows that $\mathbb{E}Y=\mathbb{E}\Delta$. The exponential mean imputation under independent censoring therefore gives a modified Cox–Snell residual with known mean:On censoring, the added one is the expected remaining transformed lifetime, by memorylessness of the exponential distribution. This gives a mean-one variable, not necessarily another exponential variable. Equivalently, $M=\Delta -Y=1-Y^{\ast }$ is a martingale residual with mean zero. Independent censoring, or its appropriate conditional version, is essential for these identities; arbitrary informative censoring does not justify the correction.

Fit a model using the relevant covariates, calculate ${\hat{M}}_i=v_i-{\hat{H}}_i(x_i)$, and plot these martingale residuals against a candidate explanatory variable or against an included variable's value. A smooth systematic trend away from zero can indicate an omitted effect or an unsuitable functional form, such as a nonlinear age effect. After fitting an appropriate effect, the residual trend should diminish. The equivalent plot of ${\hat{Y}}_i^{\ast }$ has mean-one reference rather than zero.

For the exponential distribution with mean $\mu$, $f_A(t)=F_A(t)/\mu$ and $F_A(t)=e^{-t/\mu }$. ThereforeSince ties have probability zero, $\mathbb{P}(X=T_B)=1-\mathbb{E}X/\mu$. Taking expected values in the definition of $U$ givesThere is also an imputation interpretation. If A is observed first, $T_A=X$; if B censors A, memorylessness of the exponential distribution and independence imply that the remaining mean lifetime is $\mu$. Thus $U$ is the conditional expectation of $T_A$ given the observed minimum and censoring indicator. The tower property of conditional expectation gives the same equality. This does not say that $U$ and $T_A$ have the same distribution.

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.