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Here the gamma distribution uses a rate statistical parameter $\theta$, with known shape $\lambda$. Write $S={\sum }_i{X}_i$ and $a=n\lambda$. For each fixed ${\theta }_1>{\theta }_0$, the alternative-to-null likelihood ratio isstrictly decreasing in $S$. The Neyman-Pearson lemma therefore gives the same lower-tail critical region $S<c_{\alpha }$ for every larger rate. Since $S$ has a continuous gamma distribution under the null hypothesis, there is a unique positive $c_{\alpha }$ satisfyingThus reject when $S<c_{\alpha }$ is a uniformly most powerful test of exact test size $\alpha$. If $G_a$ denotes the unit-rate gamma distribution cumulative distribution function with shape $a$, thenThis is the uniformly most powerful test for a gamma rate.

When $\lambda =1/n$, the sum has shape one and hence an exponential distribution with rate $\theta$. ThereforeIn particular $\pi ({\theta }_0)=\alpha$, and the power function is strictly increasing, exceeding $\alpha$ for $\theta >{\theta }_0$ and tending to one as $\theta \to \infty$.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

At the empty Markov chain state, only the constant-rate production chemical reaction is possible. Thus the first-passage time ${\tau }_1$ has an exponential distribution of rate $2$:Its expected value can be obtained by integrating this survival function, and its second moment by integrating $2t$ times the survival function:Consequently the requested expected value and variance are

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.

Adopt the lower-bound scale of the Heisenberg uncertainty principle, $\Delta r\Delta p\sim \hslash /2$, with the prescribed radial scale from a. Using $\hslash c=197.327$ MeV fm,For a one-sided exponential distribution of a positive momentum magnitude, $f(p)=p_0^{-1}{e}^{-p/p_0}$, both its mean and its standard deviation are $p_0$. Under the additional identification made in this estimate, set the characteristic momentum to $\Delta p$. This identification concerns a positive magnitude; the mean of a signed momentum component in an isotropic stationary state would instead vanish.

Use the relativistic energy-momentum relation, not a nonrelativistic kinetic-energy approximation, because the momentum-energy scale is much larger than the stipulated quark rest energy:This is a characteristic energy estimated at the characteristic momentum. It is not an exact evaluation of $⟨\sqrt{p^2{c}^2+m_q^2{c}^4}⟩$ over a fully specified three-dimensional wavefunction. An order-of-magnitude convention $\Delta p\sim \hslash /\Delta r$ would give an answer about twice as large; the uncertainty principle by itself specifies a bound, not a unique confinement energy.

The exponential distribution prior has rate $\lambda =1/3$ and support $\mu \ge 0$, so its prior density is $(1/3)e^{-\mu /3}{1}_{\{\mu \ge 0\}}$. By Bayes' theorem,where $C$ is a normalizing constant independent of $\mu$.

Completing the square gives a useful exact reference for the numerical sampling. PutThen this normal mean posterior with an exponential prior is a normal distribution with location $m$ and variance $s^2$, truncated below at zero. With $\varphi$ the standard normal distribution density and $\Phi$ its cumulative distribution function,The exponential rate is $1/3$, not three; the prior mean is three. The support restriction must be retained in both the formula and the sampler.

The logarithm is strictly increasing, so $\mathbb{P}(Y\le y)=\mathbb{P}(X\le e^y)$. Since $X>1$ almost surely,Thus the logarithm of a Pareto random variable has an exponential distribution with rate $c$. Its probability density function value at the single endpoint can be chosen arbitrarily without changing the law.

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

The rate-one exponential distribution has density $e^{-t}$ for $t\ge 0$. Integration by parts gives $\mathbb{E}T=1$ and $\mathbb{E}[T^2]=2$, so

Choose $g(t)=e^t$. If $W$ has an exponential distribution with rate $r$, then for $c<r$,while the integral diverges to $+\infty$ when $c\ge r$. Since $Z$ has rate $3$,For the other two rates, $\mathbb{E}[g(V)]=2/(2-1)=2$, whereas $\mathbb{E}[g(U)]={\int }_0^{\infty }1dt=+\infty$. These computations establish exactly the required finite variance, finite expected value, and infinite expected value.

Using the rate parameter convention for the exponential distribution, independence gives the joint probability density functionFor $z\ge 0$, the survival probability of the minimum isThus $Z$ has exponential distribution with rate $3$, andValues of a probability density function at the single endpoint $0$ do not affect the probability distribution.

Because $Z=e^{-Y}>0$, its cumulative distribution function is zero for $z\le 0$. For $z>0$, the transformation is decreasing, and $Y$ has a continuous cumulative distribution function, soThus the exponential transform of a standard Gumbel variable givesThis is the exponential distribution with rate $1$.

Write $M_n={\max }_i{E}_i$. Each exponential distribution has cumulative distribution function $1-e^{-x}$ for $x\ge 0$. Independence therefore gives the maximum of independent exponential random variables formulaFor any fixed $y$, the threshold $y+\ln n$ is nonnegative for all sufficiently large $n$. ThenThis is convergence in distribution to the standard Gumbel distribution.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

The time of the $n$-th event of a Poisson process. At positive rate $\lambda$, the Poisson interarrival times $T_j-T_{j-1}$, with $T_0=0$, are independent and identically distributed random variables with the rate-$\lambda$ exponential distribution. In particular, $T_n$ has expected value $n/\lambda$ and variance $n/{\lambda }^2$. The inverse-count identity is $\{T_n\le t\}=\{N_t\ge n\}$.

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

A Poisson process observed for a fixed duration $H$ has likelihood proportional to ${\lambda }^{N_H}{e}^{-\lambda H}$. Complete independent exponential distribution gaps observed until a fixed event count $n$ have likelihood ${\lambda }^n{e}^{-\lambda T_n}$. Both give count divided by exposure, but the exposure is random in the second design. Under fixed exposure the total count is sufficient: conditional on it, the event locations are independent of the rate, so exact timestamps add no rate information to interval counts. Gap estimates and fixed-horizon estimates should not be compared without distinguishing their stopping rules and any final censored gap.

For independent standard normal random variables $X,Y$, put $X=R\cos \Theta$, $Y=R\sin \Theta$, with $R\ge 0$ and $0\le \Theta <2\pi$. The polar coordinates Jacobian determinant is $r$, so the joint probability density is $r{e}^{-r^2/2}/(2\pi )$ on $r>0$. It factorizes into a Rayleigh distribution density for $R$ and a continuous uniform distribution density for $\Theta$. Consequently the radius and angle are independent. The change of variables for probability densities $S=R^2$ gives $f_S(s)=e^{-s/2}/2$ for $s>0$, the exponential distribution of rate $1/2$, also the chi-squared distribution with two degrees of freedom. Measurable transformation preserves independence, giving $\Theta \perp S$. The angle at $R=0$ is arbitrary on an event of probability zero.

Thus a sector with angular length $\ell$ has probability $\ell /(2\pi )$, independently of any radial restriction. For instance $Y>|X|$ occupies a sector of angle $\pi /2$ and has probability $1/4$.

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

Let $S,T$ be independent random variables with exponential distributions of respective positive rates $\alpha ,\beta$. Condition on $T=t$ and use the survival function of $S$:At $u=0$ this is $P(S>T)$. Dividing by that probability proves that the positive difference conditional on $S>T$ again has an exponential distribution with rate $\alpha$. The memorylessness of the exponential distribution persists after an independent competing event has already occurred. Replacing the difference by a negative sentinel when $S\le T$ adds an atom of a measure of mass $\alpha /(\alpha +\beta )$ at that sentinel without changing the positive tail.