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Since $T_i$ has cumulative hazard function ${\varphi }_i{H}_0(t)$,Thus the cumulative-hazard time change $U_i=H_0(T_i)$ has an exponential distribution of rate ${\varphi }_i$.

By Bayes theorem, the frailty distribution among survivors has densityIt is an exponential distribution with rate $1+H_0(t)$.

Generate $U\sim \mathrm{Uniform}[0,1]$ and setFor $x\ge 0$,which is the distribution function of an exponential distribution with rate one. This is inverse transform sampling.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M-M-1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

Under the Cameron-Martin theorem for a linear drift, the probability that Brownian motion with drift $-\mu$ reaches $x$ isUsing the supplied Laplace transform with $\lambda ={\mu }^2/2$ givesand henceThis is the survival function of the exponential distribution with rate $2\mu$.

The minimum of the recovery and death clocks has an exponential distribution with rate $\mu +\gamma$. Its expected value is therefore

The two event clocks have independent exponential distributions with rates $\mu$ and $\gamma$. The competing exponential clocks identity gives

The transformation is $T_k=a_k{U}_k^b$. Because $U_k$ has a unit-rate exponential distribution,Thus $T_k$ has a Weibull distribution, withConsequently $h_2(t)/h_1(t)=(a_1/a_2{)}^{1/b}$ is constant, proving proportional hazards.

Write the molecular copy numbers as $\mathbf{x}=(x_1,x_2,x_3,x_4)$ and use the power-law reaction propensity convention of the paper. The seven propensities and stoichiometric vectors areA channel is assigned zero propensity whenever its update would leave the nonnegative integer lattice.

The Gillespie algorithm starts from $\mathbf{x}=(5,0,0,0)$ and $t=0$. At the current state compute $a_0={\sum }_{r=1}^7{a}_r$. If $a_0=0$, terminate the path. Otherwise draw independent random variables $U_1,U_2$ from the uniform distribution on $(0,1)$, set the next waiting time toand choose the least index $r$ satisfyingThen update $t\leftarrow t+\tau$ and $\mathbf{x}\leftarrow \mathbf{x}+{\nu }_r$, and repeat. The minimum of the seven competing reaction clocks has an exponential distribution of rate $a_0$, while reaction $r$ wins with probability $a_r/a_0$.

Only reaction 1 changes $X_1$. Starting from five molecules, its successive states are $5\to 3\to 1$, with rates $5^2=25$ and $3^2=9$; state one is then absorbing for this channel. ThusThe probability of still being at three is the convolution of the first waiting time with an exponential distribution and survival of the second:Applying the law of total probability to the three possible states givesIt has $g(0)=0$ and tends to one as both reaction waiting times elapse.

If $U\sim \mathrm{Uniform}(0,1)$, the inverse transform sampling formulahas survival function $\mathbb{P}(T>t)=e^{-\lambda t}$ and hence the required exponential distribution.

For unit-rate exponential distribution frailty, the population survival and hazard in a proportional frailty model areThe surviving frailty distribution is exponential with rate $1+H_0(t)$.

At unit scale the probability density function is $r\exp (-r^2/2)$ for $r>0$. Its square has an exponential distribution of rate $1/2$. Thus $\sqrt{-2\log U}$ has this distribution for a uniform distribution variable $U$ on $(0,1)$.

The reaction propensity $a_r(x)$ is the instantaneous rate at which reaction $r$ fires when the copy-number state is $x$. Conditional on the present state, its reaction clock has an exponential distribution of rate $a_r(x)$.

If $S$ has exponential distribution with expected value $\mu$, the tail integral formula for moments gives $\mathbb{E}\min (S,M)=\mu (1-e^{-m})$ and $\mathrm{Var}(\min (S,M))={\mu }^2(1-2m{e}^{-m}-e^{-2m})$, where $m=M/\mu$. At matching retained expected value, the excess variance under quota share reinsurance is $2{\mu }^2{e}^{-m}(m-1+e^{-m})\ge 0$.

The time a customer occupies a server while receiving service. Its mean is a duration; its reciprocal is a service rate. An exponential distribution is convenient for Markov queue models, but Little's law can use a general finite mean.

A shifted exponential distribution has location estimate equal to the sample minimum and rate estimate given above, for a nondegenerate sample of at least two observations. The minimum exceeds the true location on average by $1/(n\lambda )$. The residual sum above the minimum is gamma with shape $n-1$, so the rate estimate has expectation $n\lambda /(n-2)$ for $n>2$ and infinite expectation at $n=2$. The estimated location plus reciprocal rate is the unbiased sample mean.

When the random intensity is $\alpha +Y$ with $Y$ having an exponential distribution of rate $\beta$, the mixed count is the sum of independent Poisson distribution and support-zero geometric distribution counts. The Poisson parameter is $\alpha$ and the geometric success probability is $\beta /(\beta +1)$. At $\alpha =0$ the law is geometric.

For independent samples from an exponential distribution, a test of rate at most ${\lambda }_0$ against rate greater than ${\lambda }_0$ rejects for small sample sums. All simple-alternative likelihood ratios against ${\lambda }_0$ decrease with that sum; its gamma distribution calibrates the size at ${\lambda }_0$. Monotonicity of the rejection probability in the rate proves that this is a uniformly most powerful test for the composite hypotheses.

For exponential distribution claims of mean $\mu$ and compound Poisson distribution count parameter $\lambda$, the total party variance under excess of loss reinsurance is $g(M)=2\lambda {\mu }^2(1-(M/\mu )e^{-M/\mu })$. Its derivative is $2\lambda e^{-M/\mu }(M-\mu )$, so $M=\mu$ is the unique global minimum, with value $2\lambda {\mu }^2(1-e^{-1})$.