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A rotary ATP synthase subunit can exhibit an ATP-dependent exponential distribution of binding waits with rate $k_{1}c$ and an ATP-independent sum of two first-order waits. Convolution gives the latter density $k_a{k}_b(e^{-k_{b}t}-e^{-k_{a}t})/(k_a-k_b)$. The equal-rate limit is the Erlang distribution $k^{2}t{e}^{-kt}$. These fits support one concentration-dependent binding event and two kinetically resolved transitions, but do not uniquely assign the latter to hydrolysis, product release or conformational change. Their rates cannot be ordered from this symmetric density alone.

Under loss $w(\theta )|\theta -a|$, assume $0<\mathbb{E}[w(\theta )\mid x]<\infty$. The posterior expected loss is a positive constant times ordinary absolute-error loss under the probability density proportional to $w(\theta )\pi (\theta \mid x)$. Its minimizers are the medians of that weighted posterior: slopes of the loss change sign where the weighted probability below $a$ crosses one half. In particular, relative absolute loss has $w(\theta )=1/\theta$ for positive parameters. For a Gamma distribution posterior of shape two and rate $b$, the weighted posterior is an exponential distribution of rate $b$, so the estimate is $(\log 2)/b$.

For $a,b>0$, this probability density function is obtained from a beta distribution variable $U$ by $X=U/(1-U)$. Indeed $u=x/(1+x)$ and $du/dx=(1+x{)}^{-2}$, and the change of variables formula gives the displayed density. In particular, the ratio of two independent equal-rate exponential distributions has $a=b=1$, density $(1+x{)}^{-2}$ and cumulative distribution function $x/(1+x)$.

For a continuous uniform distribution on $[0,\theta ]$, the scaled maximum gap $n(\theta -X_{(n)})/\theta$ tends to a unit-rate exponential distribution. An ordinary bootstrap sample includes the unique observed maximum with probability $1-(1-1/n{)}^n\to 1-e^{-1}$. The bootstrap maximum gap therefore retains an atom at zero, whereas the target limit is continuous. This proves failure of the ordinary empirical bootstrap for this endpoint functional; consistency of the endpoint estimate alone does not justify resampling its extreme-value limit.

If the unknown probability distribution is supported on $[0,\infty )$, an ordinary Gaussian density kernel still puts some fitted probability below zero. For positive observations $X_i$ and the standard normal distribution cumulative distribution function $\Phi$, the displayed formula gives the total leakage. For the statistical sample $(0.1,0.4,0.9)$ and smoothing bandwidth $h=0.5$, it isThus about $22.28\%$ of the fitted probability falls outside the known support of a probability distribution, even though the full integral remains one. The leakage panel in smoothing bandwidth shades this area. Simply deleting the negative part leaves area $0.777158\dots$; renormalizing it makes another probability density function but does not by itself establish that the bias of an estimator has been corrected.

The issue also appears in the expectation of the kernel density estimator. For a nonnegative symmetric kernel for density estimation and a probability density function $f$ that is zero for $x<0$ and has a finite positive right-hand limit $f(0+)$, assume $f$ is bounded. Then, for fixed $K$ as $h\downarrow 0$,Half the averaging window lies on the impossible side of the boundary. For example, the rate-one exponential distribution has $f(0+)=1$, so the limiting expected boundary height is $1/2$, rather than the right-hand height one. This refers to estimating the right-hand limit; changing a probability density function at a single point does not change its probability distribution. Decreasing the smoothing bandwidth narrows the affected region but does not remove this boundary-height error. This is why the interior second-order pointwise bias bound for kernel density estimation cannot be applied across a jump at the boundary.

The elapsed-time $h>0$ transition distribution of a driftless square-root diffusion starting at $x$ is the compound Poisson distributionwhere all variables on the right are independent and the empty sum is zero. Here the exponential distribution parameter is its rate. The Laplace transform of this sum iswhich is the Laplace transform of the driftless square-root diffusion; uniqueness of the Laplace transform proves the distributional identity. In particular, for $0\le v<2/h$, the exponential moment isFor $x>0$ this exponential moment is infinite when $v\ge 2/h$, since the event $N=1$ has positive probability and one exponential jump already has an infinite moment there. At $x=0$ the process remains zero and every such moment is one.

For independent exponential distributions with rates ${\lambda }_1>{\lambda }_2>0$, let $T=S_1+S_2$. Dividing the joint density of $(S_1,T)$ by the hypoexponential distribution density of $T$ gives a truncated exponential distribution on $(0,t)$ with rate ${\lambda }_1-{\lambda }_2$. Its displayed conditional expectation supplies an E-step when only pair totals are observed. At equal rates the conditional split is uniform, with mean $t/2$.

// Target: probability-and-statistics.bigb

For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval for the rate transforms to one for the mean by reversing and inverting its endpoints. The monotonic transformation preserves the coverage event; it does not give a prediction interval for one realized holding time.

Suppose $T\ge 0$ has an absolutely continuous probability distribution, with survival function $S(0)=1$. On an interval where $S(t)>0$, its cumulative hazard isHere $h$ is the hazard function; its integral is dimensionless because the units of rate and time cancel. The identity follows by integrating the derivative of $\log S$, so mere continuity of the survival function is not enough: the absolute continuity assumption supplies a probability density function. If $S$ reaches zero, interpret $-\log S$ as $+\infty$. The normalization $S(0)=1$ also matters; an initial probability mass at zero would require an initial term $-\log S(0)$.

Accumulating hazard functions adds on the $H$ scale and multiplies survival factors on the $S$ scale. For $S(a)>0$ and $b>a$,For instance, a cumulative hazard increment of $2$ gives event probability $1-e^{-2}\approx 0.864665$, rather than a probability of $2$. With $h(t)=0.2$ per year, $H(5)=1$ and $S(5)=e^{-1}\approx 0.367879$. With the increasing hazard function $h(t)=0.04t$ per year when $t$ is expressed in years, $H(t)=0.02t^2$, so $H(5)=0.5$ gives $S(5)=e^{-0.5}\approx 0.606531$. Its coefficient has units of inverse years squared.

The constant hazard function produces an exponential distribution; the increasing example is a Weibull distribution. Their survival functions both decrease, although one hazard function stays constant and the other increases. At year ten both have $H(10)=2$ and $S(10)=e^{-2}$ despite different earlier histories. The accumulation identity is described in Lu Tian's Stanford notes, Section 1.1.

For a discrete hazard $q_j$ at each possible event time $t_j$, the exact counterpart is $S(t)={\prod }_{t_j\le t}(1-q_j)$. Taking its negative logarithm sums $-\log (1-q_j)$, rather than $q_j$. Therefore exponentiating a Nelson–Aalen estimator does not give exactly the Kaplan–Meier estimator; small-jump comparison of cumulative hazard estimators quantifies the difference.

Why not accept every delayed pulse? A longer window captures more genuine neutron capture signals, but also admits more accidental coincidences. Consider an original model with a capture waiting time following an exponential distribution of mean $\tau >0$. Assume detection after a prompt signal accepts $0<t<W$, with perfect time measurement and no lower cut. Integrating the normalized density givesThis assumes that moderation, competing capture channels, boundaries and electronic recovery can be represented or neglected in a single effective waiting-time model. It is not a fitted historical capture law.

Let independent prompt-like and delayed-like background streams have Poisson process rates $R_p$ and $R_d$. The accidental coincidence pair rate in a directed window of full width $W$ is $R_p{R}_{d}W$. Electronics producing at most one accepted delayed match per prompt instead give $R_p(1-e^{-R_{d}W})$. These definitions coincide to first order when $R_{d}W\ll 1$. Correlated backgrounds, such as one external neutron producing both a prompt recoil and a later capture, require separate controls; an independent-stream estimate cannot remove them.

For chosen $\tau =20\mu \mathrm{s}$ and $W=50\mu \mathrm{s}$, the accepted fraction is $1-e^{-2.5}\simeq 0.918$. With chosen $R_p=40{\mathrm{s}}^{-1}$ and $R_d=30{\mathrm{s}}^{-1}$, the pair rate is $0.0600{\mathrm{s}}^{-1}$, while the at-most-one rate is approximately $0.05996{\mathrm{s}}^{-1}$. These are teaching parameters, not reported Savannah River rates. A measured uncorrelated background can be tested using displaced time windows; a reactor-off measurement addresses a different question and cannot by itself characterize every reactor-correlated background.

Independent marking is the hypothesis that guarantees the Poisson thinning theorem. Equal marginal retention probabilities do not guarantee it. Consider one Bernoulli random variable $B$ with success probability $p\in (0,1)$, independent of a Poisson process, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with probability $p$, but the marks share the same coin. In an interval whose original count has Poisson distribution of mean $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained mean $1$, but the common-coin rule has variance $2$ and no-arrival probability $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the law of total variance.

An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two independent exponential random variables, hence have the Erlang distribution of shape two and rate $\lambda$, not an exponential distribution. The resulting renewal process and its count variance are analyzed in Every second Poisson arrival has underdispersed counts, and the two dependent streams in alternating split of a Poisson process. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=1_{\{N_t\ge 1\}}$, a Bernoulli distribution at each $t>0$, so it cannot have a positive-mean Poisson distribution. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.

For an independent statistical sample from the beta distribution with parameters $(\theta ,1)$, $\theta >0$, the probability density function is $\theta x^{\theta -1}$ on $(0,1)$. Its raw moments are $\mathbb{E}{X}^r=\theta /(\theta +r)$ for $r>-\theta$, by direct integration. Consequently the method of moments gives $T_1=\bar{X}/(1-\bar{X})$, whereas the maximum-likelihood estimator is $T_2=-n/{\sum }_i\log X_i$.

The second formula follows by setting the score function $n/\theta +{\sum }_i\log X_i$ to zero; its derivative is negative. Also $-\log X_i$ has an exponential distribution of rate $\theta$, since $\mathbb{P}(-\log X_i>u)=e^{-\theta u}$. The central limit theorem and delta method giveThus the asymptotic relative efficiency of $T_1$ relative to $T_2$ is $1-(\theta +1{)}^{-2}$. The transformed sum has a gamma distribution of shape $n$ and rate $\theta$, which also gives $\mathbb{E}{T}_2=n\theta /(n-1)$ for $n>1$ and $\mathrm{Var}(T_2)=n^2{\theta }^2/[(n-1{)}^2(n-2)]$ for $n>2$.

For independent observations $Y_i=\theta +E_i$ with $E_i$ having exponential distribution of known rate $\lambda >0$, the likelihood increases in $\theta$ until the smallest observation, so the maximum-likelihood estimator is $Y_{(1)}$. Its survival function is $\mathbb{P}(Y_{(1)}-\theta >t)=e^{-n\lambda t}$ for $t\ge 0$. Therefore $n\lambda (Y_{(1)}-\theta )$ has exactly the unit-rate exponential distribution for every sample size. This gives statistical consistency and a one-sided nonnormal limit with rate $n$, illustrating how a support endpoint changes usual regular likelihood asymptotics.

For independent and identically distributed random variables $U_i$ with the continuous uniform distribution on $[0,1]$, let $M_n={\max }_{1\le i\le n}{U}_i$. Its cumulative distribution function is $x^n$ on $[0,1]$, so $M_n\stackrel{\mathrm{d}}{\to }1$. The endpoint error has a nonconstant limiting probability distribution: for fixed $z\ge 0$ and $n\ge z$,This survival function gives the displayed exponential distribution limit. More generally, for observations uniform on $[a,b]$, $n(b-M_n)/(b-a)$ has the same limit. The scaling identifies an error of order $1/n$, in contrast to the $1/\sqrt{n}$ fluctuations of a typical sample mean with nonzero finite variance.

Write $a=\nu /2$. Against an exponential distribution proposal of mean $\nu$, the chi-squared distribution density has envelope constant $M=a^a{e}^{1-a}/\Gamma (a)$. The density ratio divided by $M$ is $\exp ((a-1)(\log (y/\nu )-y/\nu +1))$. Thus generate $Y=-\nu \log U$ and accept when $\log V\le (a-1)(\log (Y/\nu )-Y/\nu +1)$ for independent uniform distributions. The inequality $\log t\le t-1$ validates the acceptance probability. The accepted probability density function is the target because rejection sampling weights the proposal by its target-to-envelope ratio.

Let $U_n$ have the uniform distribution on $[0,1]$ and let $E_n$ have an exponential distribution of rate $\lambda >0$, with all these inputs independent. The random affine recursion $X_n=U_n(X_{n-1}+E_n)$ has an invariant exponential law of the same rate. Indeed, for independent exponential $X,E$, beta-gamma independence says that $V=X/(X+E)$ is uniform and independent of $W=X+E$. Thus an independent uniform $U$ gives $UW$ the same law as $VW=X$.

For any deterministic real initial value $c$, couple the recursion to a stationary copy. Synchronous contraction of a random affine recursion gives an expected difference $2^{-n}\mathbb{E}|c-X_0|\le 2^{-n}(|c|+{\lambda }^{-1})$. Hence the one-time laws converge to $\mathrm{Exp}(\lambda )$, independently of $c$. An invariant initialization keeps every one-time law unchanged while the chain continues to fluctuate.

For fixed real $x$, the base $1+x/n$ is positive for sufficiently large $n$. Applying Taylor's theorem to the natural logarithm at one gives $n\log (1+x/n)=x+O(n^{-1})$. Continuity of the exponential function then proves the displayed limit. In particular $(1-z/n{)}^n\to e^{-z}$ for fixed $z\ge 0$, a basic source of exponential distribution limits in probability theory.

For the stated critical geometric offspring law, the conditional upper tail is exactly $[n/(n+1){]}^{\lfloor nx\rfloor }$ for $x\ge 0$. Its logarithm tends to $-x$, proving an exponential distribution limit of rate one without a transform continuity theorem. The unconditional mass at zero tends to one, so survival conditioning changes the relevant limiting population scale.

For an exponential distribution with mean $\theta$, the maximum-likelihood estimator is $\bar{Y}$ and $\sqrt{n}(\bar{Y}-\theta )/\theta$ converges to a standard normal distribution. Inverting the event that this pivot lies between $-z$ and $z$ gives the displayed confidence interval, provided $n>z^2$. Its two-sided coverage probability has error $O(n^{-1})$, because the first Edgeworth expansion term is even and cancels between the two tails.

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.