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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 the rate-one exponential distribution, the probability density function is $e^{-t}$ for $t\ge 0$. By integration by parts,Thus

For the exponential distribution with rate $\theta$, the change of variables $u=\theta x$ givesThe integrals are one and two by integration by parts. Hence $\boxed{\mathbb{E}[X]=1/\theta ,\mathrm{Var}(X)=1/{\theta }^2}$.

The printed numerical example reverses the digits and contradicts both definitions. With the stated last-digit definition, $[Z]=17$ gives $X=7$, $Y=1$. Use those definitions throughout.

Put $q=e^{-\lambda }\in (0,1)$ and $N=\lfloor Z\rfloor$. The exponential distribution givesThe Euclidean division $N=10Y+X$ has unique remainder $X\in \{0,\dots ,9\}$ and quotient $Y\in \{0,1,\dots \}$. Thus the joint probability mass function isand zero otherwise. This is decimal digit decomposition of a geometric random variable. Following the erroneous example instead would require different definitions and would not answer the stated question.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

The survival function of this exponential distribution is $\mathbb{P}(X\ge t)=e^{-kt}$ for $t\ge 0$. Since $\{X\ge 3\}\subseteq \{X\ge 1\}$, the conditional probability isThis is the memoryless property: after surviving the first unit, surviving two additional units has the same probability as it did initially.

An integrable nonnegative probability density of this form requires $k>0$. Normalization gives $1=A{\int }_0^{\infty }{e}^{-kt}dt=A/k$, so $\boxed{A=k}$. This is the exponential distribution with rate $k$.

The cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

The survival function of this exponential distribution is $P(T>t)=e^{-\alpha t}$. At the expected value found above,

The probability density function is normalized because ${\int }_0^{\infty }\alpha e^{-\alpha t}dt=1$. Integration by parts gives $\mathbb{E}[T]={\int }_0^{\infty }t\alpha e^{-\alpha t}dt=1/\alpha$ and $\mathbb{E}[T^2]=2/{\alpha }^2$. ThereforeThese are the mean and variance of the exponential distribution with rate $\alpha$.

For $t_0\ge 0$, integration of the exponential distribution gives its cumulative distribution function:For $t_0<0$ the event is empty and its probability is zero.