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Use the intended model of independent random variables $A,B$ with exponential distributions of rates $\alpha ,\beta >0$. For $M=\min (A,B)$, independence gives the survival functionThusThe variance is $1/(\alpha +\beta {)}^2$, obtained from the standard first two exponential moments, $EM=1/(\alpha +\beta )$ and $E{M}^2=2/(\alpha +\beta {)}^2$.

For these competing exponential clocks, ties have probability zero. Integrating over Alice's firing time givesLet $H$ denote the event that the next shot hits, and use the stated success probability as $P(H\mid A<B)=1/2$. Bill's corresponding conditional success probability is one. The law of total probability givesThe Bayes' theorem therefore yields

The printed assumptions specify the two marginal exponential distributions but do not explicitly specify their independence. The answers above require that extra assumption; marginals alone do not determine the race. For a concrete counterexample, take one $Z\sim \mathrm{Exp}(1)$ and set $A=Z/\alpha$, $B=Z/\beta$. These have the required marginals, but $M\sim \mathrm{Exp}(\max (\alpha ,\beta ))$; if $\alpha >\beta$, Alice always fires first and $P(A<B\mid H)=1$. Thus the independence qualification changes actual answers, rather than merely their derivation.

For $t\ge 0$, independence factors the survival function of the minimum:The minimum is nonnegative, and this is exactly the survival function of an exponential distribution. HenceThe parameter here is the rate, rather than the mean. The calculation is the simplest example of competing exponential clocks.

An exponential distribution of rate $\lambda >0$ has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ and zero for $x<0$, with survival probability $\mathbb{P}(X>x)=e^{-\lambda x}$. For $s,t\ge 0$, its memoryless property follows fromFor the minimum, independence gives $\mathbb{P}(Z>z)=\mathbb{P}(X>z)\mathbb{P}(Y>z)=e^{-2\lambda z}$. HenceThe density is zero for negative $z$. The comparison probability follows from identical continuous laws and zero tie probability; directly it is ${\int }_0^{\infty }\lambda e^{-2\lambda y}dy=1/2$. These are competing exponential clocks.

For the last density, substitute $y=s^2$ into the normalizing integral to obtain $C=2{\int }_0^{\infty }{e}^{-s^2}ds=\sqrt{\pi }$, the Gaussian integral. Equivalently, each variable has a gamma distribution with shape $1/2$ and rate one. Independence permits convolution of the densities. For $z>0$,Putting $y=z{\sin }^2\theta$ turns the integral into $2{\int }_0^{\pi /2}d\theta =\pi$. ConsequentlyThis is also the additivity of independent gamma distributions with a common rate: their shapes add to one. Density values at the single endpoint zero are immaterial.

Initially, the infection and recovery transition intensities are $\beta N$ and $\gamma$. By competing exponential clocks, the first-event waiting time has distribution

The first event causes extinction precisely when it is recovery of the sole initially infectious individual. An infection would instead give two infectious individuals. The competing exponential clocks calculation therefore gives the first-event extinction probability

The competing exponential clocks calculation givesso $V$ is exponential with rate ${\lambda }_1+{\lambda }_2$. Also

Put $Q={\sum }_{j=1}^n{q}_j$. Since the variables have continuous distributions, the minimum is unique almost surely. Integrating over its possible value givesThis is the joint tail calculation for competing exponential clocks.

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

Integrate over the region $0\le s<t$:ThereforeThis is also the two-clock case of competing exponential clocks. A larger mean time to noise increases this probability, while a larger mean service length decreases it.

Conditioning on the observed time. The native PDF defines ${\tau }_2$ as the first production time and then conditions on that production occurring at ${\tau }_2$. If the intended information is its numerical value ${\tau }_2=t$, the answer depends on $t$. We first compute this conditional probability using a conditional probability density; conditioning on an exact value of a continuous random variable means a density ratio, since $\mathbb{P}({\tau }_2=t)=0$.

Before ${\tau }_2$, no molecules of the second chemical species are present, so the joint-removal chemical reaction cannot occur. Molecules of the first chemical species arrive as a rate-$2$ Poisson process, and each independently starts a clock with a rate-$1/2$ exponential distribution for producing the first molecule of the second chemical species. Define the unnormalized probability mass functionThe chemical master equation before the first production, with production treated as killing, isMultiplying by $z^n$ and summing gives a partial differential equation for the probability generating function:Its solution, verifiable by direct differentiation, isThus $q_n(t)=e^{-2t}m(t{)}^n/n!$, the survival function is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the conditional distribution of $A_1(t)$ given survival to $t$ is a Poisson distribution of mean $m(t)$.

Production in $(t,t+dt)$ from Markov chain state $n$ has joint probability $(n/2)q_n(t)dt+o(dt)$. Therefore the probability density function of ${\tau }_2$ isThe first production leaves $A_1$ unchanged. Taking the density ratio gives a size-biased distribution of the surviving count:In particular, under this reading the requested conditional probability isThe shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production reaction propensity function is larger.

Without observing the numerical time. If the conditioning merely repeats the definition of ${\tau }_2$, the requested probability is instead the unconditional count at that stopping time. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in Markov chain state $n$. The competing exponential clocks implyThe first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with probability one, as is also clear from $S(t)\to 0$. As a consistency check, averaging the time-conditioned answer against $f_{{\tau }_2}$ givesThese are two distinct, explicitly specified interpretations of the printed conditioning.

Compute all three probabilities. From the preceding part, $\mathbb{P}(A)=e^{-6}$. Integrating the joint probability density function over $u>v$ givesOn $A\cap B$ one has $v>2$ and $u>v$, soTherefore $A$ and $B$ are independent events. The minimum waiting time and the identity of the first clock are independent for competing exponential clocks; unequal rates change the winner probabilities but do not destroy that independence.