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Use the exponent convention $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$. A real Lévy process starts at $0$, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Its characteristic exponent of a Lévy process satisfies $\psi (0)=0$ and is continuous. This convention has the opposite sign from the alternative $e^{-t\Psi (u)}$, so $\psi =-\Psi$.

For real $u$, the exponential martingale of a Lévy process is integrable because $|M_t^u|=e^{-t\mathrm{Re}\psi (u)}$ is deterministic and finite. For $s\le t$, its conditional expectation with respect to the natural filtration isThe second equality uses both independent increments and stationary increments. Thus this is a complex martingale, meaning that its real and imaginary parts are martingales.

A probability distribution is an infinitely divisible distribution when, for every positive integer $n$, it is the distribution of a sum of $n$ independent and identically distributed random variables. Their common distribution is allowed to depend on $n$. For a Lévy process,and these summands are independent with the common law of $X_{1/n}$. This proves the required infinite divisibility directly.

The Lévy–Khintchine theorem states that a real infinitely divisible distribution has characteristic function $e^{\psi (u)}$, where, for a unique triplet with the following fixed truncation,with $b\in \mathbb{R}$, $a\ge 0$, $K$ a positive measure, $K(\{0\})=0$, and $\int (1\wedge y^2)K(dy)<\infty$. Conversely every such triplet gives an infinitely divisible distribution and a Lévy process, unique in law, whose time-$t$ characteristic function is $e^{t\psi (u)}$. The Lévy measure $K$ need not be finite near zero. Taylor expansion makes the compensated integrand $O(y^2)$ there, while the measure is finite away from zero, so the integral is well defined. The drift parameter depends on the truncation: using $|y|\le 1$ instead can alter $b$ by the contribution of atoms at $\pm 1$. Here $a$ denotes Gaussian variance, not drift.

We now prove that continuity forces $K=0$, using the suggested truncation rather than merely quoting a path classification. For $\epsilon >0$, ${\lambda }_{\epsilon }=K(|y|\ge \epsilon )$ is finite. The truncated exponent is that of the Compound Poisson processwhere $N$ has rate ${\lambda }_{\epsilon }$ and the marks have law $K(dy)1_{\{|y|\ge \epsilon \}}/{\lambda }_{\epsilon }$ when the rate is positive. The remaining exponent $\psi -{\psi }_{\epsilon }$ also has an admissible Lévy–Khintchine formula triplet. Hence construct independently a Lévy process $Z^{\epsilon }$ with that exponent. The sum $Y^{\epsilon }+Z^{\epsilon }$ has the same finite-dimensional distributions as $X$, and hence the same law on càdlàg path space. Continuity is therefore an almost-sure property of the sum if it is one of $X$.

But independent compound Poisson jumps cannot cancel: conditional on the path of $Z^{\epsilon }$, its jump times are countable; the independent Poisson jump times have continuous distributions and avoid them almost surely. At each jump of $Y^{\epsilon }$, the sum has the same nonzero jump, of size at least $\epsilon$. If ${\lambda }_{\epsilon }>0$, there is positive probability of such a jump on any fixed positive-length time interval, contradicting almost-sure continuity. Thus $K(|y|\ge \epsilon )=0$ for every $\epsilon >0$, and taking $\epsilon =1/n$ gives $K=0$.

The characteristic function is now $\exp [t(ibu-a{u}^2/2)]$. If $a>0$, the process $B_t=(X_t-bt)/\sqrt{a}$ has continuous paths and independent centered normal increments of variance $t-s$, so it is standard Brownian motion on the original probability space. If $a=0$, the increments are deterministic, and continuity together with equality at all rational times gives $X_t=bt$ simultaneously for all $t$ almost surely. A Brownian motion on an enlarged space can be used in the zero-variance representation if necessary. Therefore

For real $\lambda$, the normal moment-generating formula gives $\mathbb{E}{e}^{\lambda (B_t-B_s)}=e^{{\lambda }^2(t-s)/2}$. Using independent increments and conditioning on the Brownian filtration,All terms are integrable by the same normal formula. Thus the displayed process is an Exponential martingale for Brownian motion.

For the interval exit time, we first show both almost-sure finiteness and finite expectation. On $\{T>n\}$, the position at time $n$ lies in $(-b,a)$. If $X_{n+1}-X_n\ge a+b$, the continuous path must exit by time $n+1$. This increment is independent of ${\mathcal{F}}_n$ and has law $N(\mu ,1)$, so the conditional probability of this event is a fixed $c>0$. ThereforeThis geometric tail bound from a uniform escape probability implies $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Take $\lambda =-2\mu$. Then $e^{-2\mu X_t}=e^{-2\mu B_t-2{\mu }^{2}t}$ is the exponential martingale just proved. Its stopped values lie between $e^{-2\mu a}$ and $e^{2\mu b}$. The optional stopping theorem at $T\wedge t$, followed by bounded convergence, gives $\mathbb{E}{e}^{-2\mu X_T}=1$. With $v=\mathbb{P}(X_T=a)$ and the two possible boundary values,Solving yieldsNext apply bounded-time optional stopping to $B_t$. Since $B_{T\wedge t}=X_{T\wedge t}-\mu (T\wedge t)$, we obtain $\mathbb{E}{X}_{T\wedge t}=\mu \mathbb{E}(T\wedge t)$. The stopped position is bounded, and the stopped time increases to an integrable $T$. Passing to the limit gives $\mu \mathbb{E}T=av-b(1-v)$. ThereforeAs a check, letting $\mu \downarrow 0$ gives the driftless formulas $v\to b/(a+b)$ and $\mathbb{E}T\to ab$. The expectation calculation uses a justified stopping limit, rather than assuming that optional stopping holds at every unbounded time.

For a fixed environment, use two independent copies of the walk to express $Z_t^2=⟨{\kappa }_t^{(1)}{\kappa }_t^{(2)}⟩$. The product is finite and nonnegative, so we can interchange the two expectations. At each time $j$, if the walks occupy different sites, the two environment signs are independent and their factors have product expectation $1$. If they coincide, both factors use the same sign andDifferent time layers are independent even when a site is visited again. Therefore, with the simultaneous collision count $I_t={\sum }_{j=1}^t{1}_{\{{\xi }_j^{(1)}={\xi }_j^{(2)}\}}$,This is the collision-count second moment for independent path weights. It involves equal-time intersections, not arbitrary intersections of the two spatial ranges.

The difference walk $D_j={\xi }_j^{(1)}-{\xi }_j^{(2)}$ has independent bounded increments symmetric about zero and a full-dimensional step distribution. It may hold at zero for one step; such a hold counts as a positive-time return. The allowed transience fact givesMoreover $\rho >0$ since the two first steps can be equal. Restarting the independent increments after every return to zero shows that the total positive-time return count $I_{\infty }$ has the zero-based geometric distributionIndeed the probability of at least $k$ returns is ${\rho }^k$, and subtracting successive tails gives this law. Consequently, whenever $\rho (1+{\epsilon }^2)<1$,A sufficient small-noise condition is $0<\epsilon <\min (1,\sqrt{(1-\rho )/\rho })$.

The L2 martingale convergence theorem now gives $Z_t\to \zeta$ in $L^2$ as well as almost surely. One can see the mean preservation directly: for $t\ge s$, the martingale identity gives $\mathbb{E}(Z_t-Z_s{)}^2=\mathbb{E}{Z}_t^2-\mathbb{E}{Z}_s^2$. The second moments are increasing and bounded, so the sequence is Cauchy in $L^2$ and its limit is the already identified almost-sure limit. Thus

Assume the displayed Gaussian forward-rate covariance drift restriction. Its diagonal instance givesUsing it on the two parts of the deterministic discounted exponent yieldsThe last equality uses covariance symmetry to reflect the triangular integration region across its diagonal. The independent increments of the integrated Gaussian forward-rate process and its conditional Gaussian moment-generating function then give, for $r\le s\le T$,All these prices are integrable because their exponents are Gaussian random variables with finite variance. Thus the mean restriction implies the discounted bond martingale condition, completing the equivalence of the first two descriptions.

Use $T$ for the fixed maturity and $s\le T$ for the observation time. The original PDF's discounting uses the instantaneous forward rate over $[s,T]$ and the short rate over $[0,s]$. Let $X(s,u)=F_{s,u}-{\mu }_{s,u}$ be the centered Gaussian forward-rate field, and defineAssume the covariance and mean have the regularity needed for these mean-square integrals and the maturity differentiation below. Splitting $Y$ at $u=s$ shows that the discounted zero-coupon bond price isThis integrated Gaussian forward-rate process has deterministic varianceIts independent increments hold relative to the entire specified filtration. Indeed, if $r\le s$ and $z\le r$, the covariance between $Y(s,T)-Y(r,T)$ and any earlier $X(z,w)$ is the integral ofSince uncorrelated jointly Gaussian variables are independent, the increment is independent of every finite family of earlier observations and hence of their generated sigma-algebra. The increment has normal distribution with mean zero and variance $v(s,T)-v(r,T)$. Also $Y(0,T)=0$ because $c(0,u,w)=0$.

The Gaussian moment-generating function now givesTherefore the Gaussian forward-rate covariance drift restriction is exactly the deterministic conditionIt turns $Z_{s,T}/Z_{0,T}$ into a Gaussian exponential martingale with deterministic variance. The subparts establish the requested equivalent descriptions from this common calculation.

Disjoint time strips of the Poisson random measure are independent, and its intensity is translation invariant in time. The construction and its limits therefore give independent increments and stationary increments, withFor the scaling actually printed in the original PDF,For example, $\alpha =2$, $t=1$, $u=1$ gives $e^{-4}$ rather than $e^{-1}$. Thus the requested equality of process laws is false unless $\alpha =1$.

The intended self-similarity of a Cauchy process rescales space and time in opposite directions:Indeed, for $0\le s<t$,The transformed process retains independent increments. For any ordered times, the joint characteristic function of its successive increments is therefore the same product as for $X$; cumulative summation proves equality of all finite-dimensional distributions. Both constructions have càdlàg paths, so these distributions also determine their laws on the usual path space. This derives the corrected invariance and explicitly diagnoses the false printed assertion.

For the amount of work $A_j(0,t]$ offered by one flow in a time interval of length $t$, its effective bandwidth at parameter $\theta >0$ and timescale $t>0$ isIts units are work per unit time. The parameter $\theta$ measures the price placed on bursts: a stringent rare-overflow target generally requires a larger value of $\theta$ than a loose target. The timescale $t$ describes the duration of the congestion episode. When the limit exists, writeThis is the long-time effective bandwidth. It includes a burstiness cost, rather than just the mean offered rate. For a finite-time flow with the required moments, ${\lim }_{\theta \downarrow 0}{\alpha }_j(\theta ,t)=\mathbb{E}{A}_j(0,t]/t$, and effective bandwidth is increasing in $\theta$. If $A_j(0,t]/t$ is bounded, its limit as $\theta \to \infty$ is the essential supremum of that interval's input rate, the peak-rate limit. The analogous long-time mean limit requires the usual interchange or regularity assumptions.

For independent flows their cumulant-generating functions add, soThe Chernoff bound therefore gives, for any $\theta >0$ in the common domain,Optimizing over $\theta$ produces the relevant Legendre-Fenchel transform. The Cramér theorem states that sample means of independent and identically distributed inputs whose moment-generating function is finite near zero have a large deviation principle, with rate function equal to the Legendre-Fenchel transform of their cumulant-generating function. More generally, the Gärtner–Ellis theorem applies to a convergent scaled log moment-generating function when zero lies inside its domain and the limit is lower semicontinuous and essentially smooth. It gives the same transform as a good rate function. A sample-path version, together with the contraction principle for large deviations for the queue workload, translates input deviations into overflow deviations. The compact-time and infinite-horizon continuity/tail hypotheses must be checked for the chosen traffic model.

A particularly useful rigorous queue bound holds for inputs with stationary increments and independent increments, such that $\mathbb{E}{e}^{\theta A(t)}=e^{t\Lambda (\theta )}$. If $\Lambda (\theta )\le C\theta$, thenis a nonnegative supermartingale, since its conditional increment factor is $e^{(t-s)(\Lambda (\theta )-C\theta )}\le 1$. The maximal inequality for a nonnegative supermartingale on $[0,T]$, followed by $T\to \infty$, gives the continuous-time exponential workload boundHere $W$ is the stationary queue workload of the stable infinite-buffer queue, represented by input from the past. This directly bounds the supremum over time; applying a one-time Chernoff bound alone would not establish it.

For a nonzero Compound Poisson process with bounded positive packet sizes and mean input rate less than $C$, a positive root ${\theta }_{\ast }$ of $\Lambda (\theta )=C\theta$ exists and is unique. Here the root theorem can be made explicit:The upper bound was just proved. For the lower bound, use exponential tilting with density $e^{{\theta }_{\ast }(A(t)-Ct)}$ at time $t$. Under the tilted law the net input has positive mean drift ${\Lambda }^{\prime }({\theta }_{\ast })-C>0$, by strict convexity and the positive-root equation. The strong law of large numbers then makes the first crossing time ${\tau }_b$ finite almost surely under that law. Its overshoot is at most $s_{\max }$. Changing measure at ${\tau }_b$, first on $\{{\tau }_b\le T\}$ and then letting $T\to \infty$, givesDividing the logarithmic bounds by $b$ proves ${\lim }_{b\to \infty }{b}^{-1}\log \mathbb{P}(W\ge b)=-{\theta }_{\ast }$.

Thus the familiar admission rule compares ${\sum }_j{a}_j(\theta )$ with $C$, choosing $\theta$ from the desired buffer-tail exponent. For more general inputs, a logarithmic root theorem requires its own light-tail and renewal or sample-path hypotheses; the displayed supermartingale upper bound needs only the moment and increment assumptions stated above.

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.

For standard Brownian motion $B$ and $h>0$, let $T_h$ be its first hitting time of $h$. Define the reflected path byleaving a path unchanged if $T_h=\infty$. The Brownian reflection principle says that $\hat{B}$ has the same path probability distribution as $B$. In its endpoint form, for $t>0$ and every Borel set $A\subset (-\infty ,h)$,Here is a direct proof of preservation of the path law. For a fixed integer horizon $L$, let $T^{(m)}=2^{-m}\lceil 2^m{T}_h\rceil \wedge L$, with an infinite first term interpreted as infinity. This is a stopping time taking finitely many deterministic grid values. On $\{T^{(m)}=j{2}^{-m}\}$, the history determining that event is measurable at $j{2}^{-m}$, and subsequent Brownian motion increments are independent of that history. Those increments and their negatives have the same joint law, by symmetry of centered normal distributions and independent increments. Thus reflecting after $T^{(m)}$ around $B_{T^{(m)}}$ preserves the full path law. For any fixed horizon $t<L$, the reflected paths converge uniformly on $[0,t]$ to the displayed reflected path, by continuity as $T^{(m)}$ decreases to $T_h\wedge L$. This includes $T_h=\infty$, since reflection after $L$ has no effect before $t$. Passing to all finite-dimensional probability distributions proves the assertion; no prior recurrence claim is needed. Equivalently, the Strong Markov property gives the same proof by symmetry of the fresh increments after the hit. Reflection is an involution preserving the hitting event and mapping the indicated endpoint $y$ to $2h-y$. An endpoint above $h$ forces a hit by continuity; this proves the endpoint identity, rather than assuming it.

For $S_t={\sup }_{0\le s\le t}{B}_s$, split $\{T_h\le t\}$ according to whether $B_t$ is above or below $h$. Since $B_t$ has no atom at $h$, reflection givesBoth variables are nonnegative, so their positive tails identify the law:For $t>0$ this is the half-normal distribution of scale $\sqrt{t}$; at $t=0$ both variables vanish.

For the two separated intervals, putHere $A_0$ is measurable for ${\mathcal{F}}_b$, $Z$ has the normal distribution $N(0,c-b)$, and $H$ is a function of increments after $c$. The independent increments make these three objects mutually independent. The later absolute maximum is $B_b+Z+H$, so equality of the two maxima means $Z=A_0-H$. Conditional on $(A_0,H)$, this specifies a single value of the nondegenerate normal distribution $Z$, and hence has probability zero. ThereforeThis proves the atomless maxima on separated Brownian intervals without incorrectly treating the two absolute maxima as independent.

The exit time $T$ is finite almost surely. Indeed, if the Brownian motion is still in $(-b,a)$ at an integer time, its next unit increment exceeds $a+b$ with a fixed probability $p>0$ independent of the past; on this event it must exit before the next integer time. Consequently $\mathbb{P}(T>n)\le (1-p{)}^n\to 0$. Path continuity gives $B_T\in \{a,-b\}$, and the two exit events partition the probability space up to null sets.

For each real $\lambda$, the exponential Brownian martingale $Z_t^{\lambda }=\exp (\lambda B_t-{\lambda }^{2}t/2)$ has expectation one. This follows directly from the moment-generating function of a normal distribution and independent increments. At $t\wedge T$ its value is bounded by $e^{|\lambda |\max (a,b)}$. Bounded-time optional stopping and then dominated convergence therefore giveWrite $p_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_a\}}]$ and $q_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_{-b}\}}]$. Applying the identity to $\lambda$ and $-\lambda$ yieldsFor $\lambda \ne 0$, multiply the first equation by $e^{\lambda b}$, the second by $e^{-\lambda b}$, and subtract. Solving the resulting system givesAdding and using the sum formula for the hyperbolic sine gives the asymmetric Brownian interval-exit transform:At $\lambda =0$, the printed sine ratio is $0/0$ and needs its removable extension. Dominated convergence as $\lambda \to 0$ givesThus the formulas hold for every real parameter with this endpoint convention. The expectation on the left of the first formula is legible in the original PDF; the converted TeX's nested exponential is an OCR error.

In the classical risk model, the insurer's surplus iswhere $u\ge 0$ is initial capital, $N(t)$ is a Poisson process of rate $\lambda >0$, and the claim sizes are positive independent and identically distributed random variables with expected value $\mu$, independent of the claim-arrival process. Premium flows in at constant rate $c$. A positive relative safety loading $\theta$ meansso premium income exceeds the expected claim cost per unit time. The ruin stopping time is $\tau =\inf \{t\ge 0:U(t)<0\}$, with $\tau =\infty$ if ruin never occurs, and the ultimate ruin probability is $\psi (u)=\mathbb{P}(\tau <\infty )$.

Assume the stated positive adjustment coefficient $R$ exists and that the claim moment-generating function is finite at $R$. The Lundberg inequality isHere is a proof, including the stopping argument. By conditioning on the Poisson distribution of $N(t)$, the random-sum transform identity givesThe adjustment coefficient equation is $\lambda (M(R)-1)=cR$. Consequently the exponential surplus martingalehas expected value one at every finite time. More strongly, the claim process has independent increments, so for $s\le t$,Thus $Z$ is a nonnegative continuous-time martingale for the filtration generated by arrivals and claim sizes. For a fixed finite horizon $T$, the optional stopping theorem at the bounded stopping time $\tau \wedge T$ gives $\mathbb{E}{Z}_{\tau \wedge T}=1$. This bounded use is legitimate: the stopped values belong to the closed martingale on $[0,T]$ given by conditional expectations of the integrable variable $Z_T$.

On $\{\tau \le T\}$, the surplus is negative, and therefore $C(\tau )-c\tau =u-U(\tau )>u$. By nonnegativity on the complementary event,The ruin events increase to $\{\tau <\infty \}$ as $T$ increases, so continuity of probability gives the claimed Lundberg inequality. No expectation identity at the unbounded stopping time $\tau$ is needed. Positive loading by itself does not guarantee a positive adjustment coefficient for every claim law; the existence assumption is used here.

For the exponential distribution of claim sizes, $M(r)=1/(1-\mu r)$ for $r<1/\mu$. Substitution in the adjustment coefficient equation givesDiscard the zero root and divide by $\mu r$. The unique positive solution iswhich lies strictly inside the finite-transform domain.

For deterministic claim sizes, the claim moment-generating function is $e^{\mu r}$. Define $h(r)=e^{\mu r}-1-(1+\theta )\mu r$. It satisfies $h(0)=0$, $h^{\prime }(0)=-\theta \mu <0$, $h^{\prime \prime }(r)={\mu }^2{e}^{\mu r}>0$, and $h(r)\to \infty$ as $r\to \infty$. Thus its strict convexity gives exactly one positive zero $R_{\mu }$, with $h(r)<0$ for $0<r<R_{\mu }$.

Put $x=\mu R=\theta /(1+\theta )\in (0,1)$. Since $-\log (1-x)>x$, we have $e^x<1/(1-x)$. HenceIt follows thatThus the deterministic-claim Lundberg inequality provides a smaller upper bound at positive capital; at zero capital both bounds equal one. This is a special case of deterministic claims maximize the adjustment coefficient at fixed mean and loading: fixing the mean and loading, variability lowers the exponential decay coefficient relative to deterministic claims. The ordering of these upper bounds alone is not a proof that the actual ultimate ruin probabilities are ordered.

The surplus in the classical risk model is $U(t)=u+ct-S(t)$. With ruin defined as a strictly negative surplus, its ultimate ruin probability isLet ${\overline{F}}_X(v)={\int }_v^{\infty }{f}_X(x)dx$. The first arrival time has exponential distribution of rate $\lambda$, independent of the first claim. Before that claim no ruin occurs, because the premium rate is positive. Given its arrival at $t$ and size $x$, available capital just before it is $u+ct$. If $x>u+ct$ ruin is immediate; otherwise the independent increments and fresh claim sizes make subsequent ruin probability $\psi (u+ct-x)$. Thus first-claim conditioning givesThis is the ruin counterpart of the first-claim decomposition for survival probability. DefineThe convolution term is continuous: extend the bounded function $\psi$ by zero to negative arguments and use continuity of translations in $L^1$ for the integrable density $f_X$. The tail function is also continuous. Changing variables to $v=u+ct$ yieldsDifferentiating this exponentially weighted integral establishes differentiability rather than presuming it. The fundamental theorem of calculus givesReversing the integration variable in the convolution therefore proves the ruin integro-differential equation:At zero the derivative is the right derivative; the same representation supplies it.

For the exponential claim law, introduceDifferentiating the explicit convolution gives $A^{\prime }=\alpha \psi -\alpha A$. Henceusing the proved integral equation. Differentiating that equation once more givesso the required ordinary differential equation isPositive relative safety loading means $c>\lambda /\alpha$, so $\kappa =\alpha -\lambda /c>0$. The general solution is $a+b{e}^{-\kappa u}$.

The condition at large capital must also be used. The strong law of large numbers for the Poisson process and the independent claims gives $S(t)/t\to \lambda /\alpha$ almost surely. Thus $S(t)-ct\to -\infty$. Its path has only finitely many jumps on each bounded time interval, so its supremum over all time is finite almost surely. Therefore $\psi (u)\to 0$ as $u\to \infty$, forcing $a=0$. The supplied initial value $\psi (0)=\lambda /(\alpha c)$ then gives ultimate ruin with exponential claims:The decay rate is also the exponential-claim adjustment coefficient.

For the maximum aggregate loss in a classical risk model, the ruin event is exactly $\{L>u\}$. Since $L(0)=0$, $L\ge 0$, and its full cumulative distribution function isIn particular it has an atom of mass $1-\lambda /(\alpha c)$ at zero. Conditional on $L>0$, its distribution is exponential with rate $\kappa$; omitting that atom would give an incorrectly normalized law.

For a constant-rate Poisson process, let $T$ be the waiting time from a spike to the next spike. By independent increments, there is no spike in an interval of length $u$ with probability $e^{-ru}$. Hence, for $r>0$,Its density is $r{e}^{-ru}$, so the interspike intervals are independent exponential distributions with mean $1/r$. If $U$ is uniform on $(0,1)$, inverse transform sampling gives $T=-\log U/r$. Repeatedly add independent draws to the current spike time to simulate a constant-rate spike train.

For a prescribed time-dependent nonnegative rate $r(t)$, use an Inhomogeneous Poisson process. Conditional on the present event at $t_i$, its next waiting time has survival probabilityThus its conditional CDF is $1$ minus that expression for $u\ge 0$. Define the cumulative intensity $\Lambda (t)={\int }_0^{t}r(s)ds$. For each new independent uniform draw, set $E_i=-\log U_i$ and solveThis transforms independent unit-rate exponential intervals in integrated-intensity time into physical spike times. Use a generalized inverse when the rate vanishes on an interval: no spikes are placed in a flat part of $\Lambda$. On a finite simulation window, stop when the next intensity target exceeds the available $\Lambda$. If the total future intensity is finite, a target beyond it means that no further spike occurs.

Numerically, integrate the known rate and invert the accumulated integral, or integrate exactly across piecewise-constant rate bins. Drawing an exponential interval using only $r(t_i)$ is not exact when the rate changes during the interval. The Poisson model also omits the neuronal refractory period and other spike-history effects.

From the masses of the Poisson distribution, its probability generating function isAt $z=0$, the term $z^0$ is interpreted as one, so this also gives $\mathbb{P}(Z=0)=e^{-\lambda }$.

One constructive definition of a rate-$\lambda$ Poisson process uses independent and identically distributed exponential random variables $E_j$ of rate $\lambda$. Set $T_0=0$, $T_k={\sum }_{j=1}^k{E}_j$ andThe strong law of large numbers gives $T_k/k\to 1/\lambda$ almost surely, so the maximum is finite at every finite $t$. The process starts at zero, has right-continuous nondecreasing integer paths, and jumps by one at the arrival times.

For $k\ge 1$, exactly $k$ arrivals by $t$ means $T_k\le t<T_{k+1}$. Integrating the first $k$ exponential gaps and the survival probability of the next gap givesThe last factor is the volume of the $k$-dimensional simplex, obtained for example by iterated integration. For $k=0$, directly $\mathbb{P}(N_t=0)=\mathbb{P}(E_1>t)=e^{-\lambda t}$. ThusAt a fixed time $s$, conditional on the counting history, the residual time until the next arrival has rate-$\lambda$ exponential distribution, independently of that history, by the memoryless property. After that arrival all subsequent gaps are fresh independent exponentials. Thus the process of future counts $N_{s+u}-N_s$ has the same law as $N_u$ and is independent of the past. This proves stationary independent increments, the equivalent usual defining properties of the Poisson process, and gives