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For a size-$n$ sample from an exponential distribution of mean $\theta$, the likelihood in the maximum-likelihood estimator coordinate $v=\bar{Y}$ has difference $\ell (\theta ;v)-\ell (v;v)=n\log (v/\theta )-nv/\theta +n$, and the fitted observed information is $n/v^2$. Their P-star approximation is the displayed gamma density shape. Its normalization gives exactly the sampling density $n^n{v}^{n-1}{e}^{-nv/\theta }/(\Gamma (n){\theta }^n)$. This illustrates an exact transformation-model case rather than a general claim of exactness.

For independent samples of size $n$ from exponential distributions with rates $\lambda$ and $\psi \lambda$, the maximum-likelihood estimators are $\hat{\lambda }=n/S$ and $\hat{\psi }=S/T$, where $S$ and $T$ are the respective sample sums. Their normalized P-star approximation is exact:Indeed, $S$ and $T$ have independent gamma distributions with shapes $n$ and respective rates $\lambda ,\psi \lambda$. The Jacobian determinant of their inverse transformation is $n^2/({\hat{\psi }}^2{\hat{\lambda }}^3)$, giving the displayed density. Alternatively the square root of the determinant of the fitted observed information is $n/(\hat{\psi }\hat{\lambda })$; multiplying by the exponential of the fitted log-likelihood difference gives the same shape. Its normalizing constant in the P-star approximation is $c_n=n^{2n-1}{e}^{-2n}/\Gamma (n{)}^2$. Integrating over $\hat{\lambda }$ givesso $\hat{\psi }/\psi$ has the F-distribution with degrees of freedom $(2n,2n)$.

For independent Poisson process arrivals and exponential distribution claim sizes of mean $\mu$, minimizing the logarithm of the negative expected exponential utility gives $H^{\prime }(\alpha )=\lambda \beta \mu [(1-\beta \mu \alpha {)}^{-2}-(1+{\theta }_R)]$. Its strictly positive second derivative makes the displayed stationary point unique. On $0\le \alpha \le 1$, the optimum is $\min (1,{\alpha }_{\ast })$. On the strictly open interval $0<\alpha <1$, there is no maximizer when ${\alpha }_{\ast }\ge 1$, only a supremum at full retention. The stationary retention increases with reinsurer loading and is independent of the direct insurer's loading.

If $X,Y$ are independent with density $2e^{-x^2}/\sqrt{\pi }$ on the positive axis, the Jacobian matrix of the inverse transformation has determinant $1/[2(1+v^2)]$. The joint density is $2e^{-u}/[\pi (1+v^2)]$ for $u,v>0$, factoring into an exponential distribution for $U$ and a unit-scale half-Cauchy distribution for $V$. The underlying radial symmetry makes radius and angle independent.

Suppose the recorded durations retain an exponential distribution of rate $\theta$, but an independent fraction $q$ are relabelled as failures and the rest as right-censored. Applying the ordinary survival likelihood gives $\hat{\theta }=d/{\sum }_i{y}_i\to q\theta$, rather than $\theta$. A censoring flag chosen independently of the recorded duration does not establish independent censoring of a latent failure time. Genuine simulation instead draws an event time and an independent censoring time and records their minimum.

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 $a_r(x)$.

Three nodes have integer link capacities $C_{12},C_{13},C_{23}$. Calls between a node pair use either its direct link or the two-link alternative. With unrestricted instantaneous rearrangement of existing calls, counts $n_{12},n_{13},n_{23}$ are feasible exactly when $n_{12}+n_{13}\le C_{12}+C_{13}$, $n_{12}+n_{23}\le C_{12}+C_{23}$ and $n_{13}+n_{23}\le C_{13}+C_{23}$. Necessity follows by counting calls crossing each node cut. For sufficiency, the inequalities allow at most one count to exceed its direct-link capacity. Divert precisely that excess to the other two links; the same inequalities ensure capacity there. The counts thus have the same feasible set as a fixed routing loss network with one virtual resource at each node, capacity equal to the sum of its two incident physical-link capacities, and each call using its two endpoint resources. With independent Poisson processes for arrivals and independent routing-independent exponential distributions for holding times, this gives the usual product-form stationary distribution of a loss network and exact Poisson arrivals see time averages blocking probabilities.

In a continuous-time epidemic model, a constant recovery rate $\gamma$ gives an exponential distribution of infectious duration with mean $1/\gamma$. Other duration distributions require additional stages or infection-age structure.

If every interspike interval is at least $t_{\mathrm{abs}}>0$, a sustained spike train has mean firing rate at most $1/t_{\mathrm{abs}}$. Across a window of duration $T$, the event count satisfies $N(T)\le 1+T/t_{\mathrm{abs}}$; dividing by $T$ and taking the long-time limit gives the bound. The extra one accounts for a possible first event. The bound permits much slower firing and says nothing about whether stronger drive can maintain the limiting rate in a real neuron.

For the chosen phenomenological paired-pulse recovery, the hold alone gives $500\mathrm{Hz}$. Add the further assumption that every successful event resets precisely the same recovery curve, with no accumulation from earlier events. Then $0.35\mathrm{nA}$ onset-timed pulses can sustain at most $1000/4.942488\approx 202.327\mathrm{Hz}$ in that model. This repeatable reset is an extra assumption: a two-pulse result alone cannot establish it. A real neuron's sustained rate also depends on neural adaptation, input delivery and evolving ion-channel inactivation. The existing leaky integrate-and-fire model similarly adds integration time to its imposed hold.

A homogeneous Poisson process of rate $\lambda$ has interspike intervals with an exponential distribution, soWith the purely illustrative choices $\lambda =100\mathrm{Hz}$ and $t_{\mathrm{abs}}=0.002\mathrm{s}$, this is $1-e^{-0.2}\approx 0.181269$. Thus about $18.13\%$ of intervals predicted by that model violate this chosen hold. Matching a mean spike train count does not make a stimulus-only Poisson process respect recent firing.

A minimal alternative is a renewal process with independent and identically distributed intervals $S=t_{\mathrm{abs}}+X$, where $X$ has an exponential distribution of rate $\lambda$. Its hazard function is zero during the hold and $\lambda$ afterwards. Its mean interval is $t_{\mathrm{abs}}+1/\lambda$ and long-run firing rate isThe same $100{\mathrm{s}}^{-1}$ rate after a $2\mathrm{ms}$ hold gives $83.333\mathrm{Hz}$, not $100\mathrm{Hz}$. This separate stochastic example has no gradual relative recovery; it illustrates why an available-state rate differs from an observed mean firing rate. Smooth dependence on time since the last event can extend the hazard function, while longer histories require more than this simple renewal process. Time-rescaled spike-train diagnostics can test the resulting history-dependent predictions against recorded spike trains.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.

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 sample median is a central value of an ordered statistical sample. The usual numerical convention selects the middle order statistic for odd sample size and the arithmetic mean of the two central order statistics for even sample size:NIST uses this convention. ^[ref]. For $(0.7,0.1,0.9,0.4,0.2)$, the sample median is $0.4$; for $(0.1,0.2,0.4,0.7)$, it is $(0.2+0.4)/2=0.3$, which need not be an observed value.

More generally, every value between the two central order statistics in an even statistical sample is a median of its empirical distribution function. The generalized-inverse sample quantile at $p=1/2$ selects the lower central order statistic in that case, so it can differ from the midpoint convention. Ties are allowed in all these definitions.

The sample median measures the center using sorted positions. Changing the largest observation from $0.9$ to $9$ in the five-value example leaves the sample median at $0.4$, while the sample mean rises from $0.46$ to $2.08$. Its sampling distribution still depends on the population probability distribution: the mean $1/2$ found for five observations from a uniform distribution is a special calculation, and a sample median's expected value need not equal the population median in general.

For a concrete counterexample, three independent and identically distributed random variables with rate-one exponential distribution have sample median $X_{(2)}$ with expected value $1/3+1/2=5/6$, by independent exponential order-statistic spacings. The population median instead solves $1-e^{-m}=1/2$, giving $m=\log 2\approx 0.693147$. Thus even the middle order statistic can have mean different from the population median.

A probability distribution is self-decomposable if, for every $0<c<1$, an independent remainder $R_c$ can be chosen so that the displayed equality in distribution holds. A normal distribution $\mathcal{N}(\mu ,{\sigma }^2)$ admits a normal distribution remainder with mean $(1-c)\mu$ and variance $(1-c^2){\sigma }^2$. For an exponential distribution of rate $\lambda$, choose $R_c=0$ with probability $c$ and otherwise an independent exponential distribution of rate $\lambda$. Its moment-generating function is $c+(1-c)\lambda /(\lambda -t)=(\lambda -ct)/(\lambda -t)$; multiplication by $M_X(ct)$ proves the claim using the uniqueness theorem for moment-generating functions.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

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 frequency-dependent transmission, the closed SIR model is a system of ordinary differential equations for the fractions $s,i,r$:Here $\beta >0$ is the coefficient of frequency-dependent transmission, and $\gamma >0$ is the recovery rate, interpreted more generally as the removal rate. Both have units of inverse time. With a constant recovery rate, the associated individual time in the infectious compartment has an exponential distribution with mean $1/\gamma$; the SIR model does not require every person to spend exactly that long there. The fractions themselves have no units.^[ref]

For counts $S=Ns$, $I=Ni$, $R=Nr$, the same SIR model equations readThe factor $1/N$ keeps the same $\beta$ in both descriptions. Dropping it while keeping these units would fail dimensional analysis. Some existing SIR model variants use the unnormalized mass-action infection convention instead, including the SIR susceptible-recovered identity; their coefficient is $b=\beta /N$ in the notation used here.

Adding the SIR model equations gives $d(s+i+r)/dt=0$, a short proof of conservation of population. Nonnegative starting fractions stay nonnegative: the outflow from either the susceptible compartment or the infectious compartment vanishes when that group is empty, while $dr/dt\ge 0$. Thus the susceptible compartment can only shrink and the removed compartment can only grow.

For a concrete initial value problem, stipulateAt the start, the flow into the infectious compartment is $0.3\times 0.99\times 0.01=0.00297$ of the total per day, and the flow into the removed compartment is $0.1\times 0.01=0.00100$ of the total per day. Hence $di/dt=0.00197$ per day. If $N=1000$, these correspond to model rates of $2.97$ and $1.00$ people per day, and net growth $1.97$ people per day. A rate need not be an integer count of events on any particular day.

For this SIR model, the basic reproduction number isIt describes one introduced member of the infectious compartment in an otherwise wholly susceptible compartment. The instantaneous reproduction number at the actual starting state is $R_0{s}_0=2.97$, obtained by freezing the current conditions and the current susceptible compartment's fraction. More generally it is $R_{0}s(t)$ in this SIR model. The basic reproduction number is a ratio of two rates, so it has no units; $r_0$ is a different quantity, the initial removed compartment's fraction.

While $i>0$, its instantaneous relative growth isInitially this equals $0.3\times 0.99-0.1=0.197{\mathrm{day}}^{-1}$. Freezing $s\approx s_0$ gives the early exponential growth approximation $i(t)\approx 0.01e^{0.197t}$ with $t$ measured in days. It must fail as the susceptible compartment shrinks. An epidemic invasion threshold occurs at $R_0{s}_0=1$; if $i_0=0$, the deterministic SIR model stays at $i=0$ even above that threshold.

In a closed population of size $n$, a frequency-dependent SIRS epidemic has the displayed transitions at rates $\lambda SI/n$, $\gamma I$, and $\nu R$. The exponential distribution of infectious lifetimes and the constant immunity-loss hazard make the compartment counts a continuous-time Markov chain. Each jump preserves $S+I+R=n$.

Squaring a Rayleigh distribution variable produces an exponential distribution. For $w\ge 0$,where the exponential distribution parameter is a rate. Equivalently, $R^2/{\sigma }^2$ has chi-squared distribution with two degrees of freedom (statistics). An amplitude and its square therefore have different probability density functions, even though one determines the other.

This makes the Rayleigh-normal scale mixture transparent: at unit scale, $R^2$ is exponential, allowing a random normal scale to produce a Laplace distribution. Another equivalent parametrization uses a Weibull distribution with shape two and scale $\sqrt{2}\sigma$. Conventions that call this latter scale the Rayleigh scale differ by a factor of $\sqrt{2}$; check the probability density function before comparing numerical parameter values.^[ref]