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For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property gives independent increments and stationary increments; the strict inverse of the continuous running maximum has càdlàg paths. The Brownian first-passage Laplace transform gives Laplace exponent $\Phi (\lambda )=\sqrt{2\lambda }$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the fixed-level versus simultaneous Brownian passage-time equality.

A càdlàg modification of a stochastic process is a modification of a stochastic process whose paths are càdlàg on one event of probability one. Equality almost surely at every separately fixed time does not imply indistinguishability of stochastic processes when the original process lacks path regularity. A stochastic process starting at zero, with independent increments, stationary increments, and stochastic continuity, has a càdlàg modification and is a Lévy process under the intrinsic definition.

Suppose Lévy processes $X^n$ converge to $X$ in probability at every fixed time, andThen $X$ starts at zero, has independent increments and stationary increments, and has stochastic continuity. Finite increment vectors inherit their independent laws by convergence in distribution; the near-zero estimate supplies continuity at zero. Hence $X$ has a càdlàg modification that is a Lévy process. The hypotheses alone cannot assert that the original version has càdlàg paths.

Conditionally on an ${\mathrm{SLE}}_{\kappa }$ initial segment through time $t$, mapping out that segment by $g_t-U_t$ turns the future into an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. This follows because its driving function is $U_{t+s}-U_t$, and Brownian motion has stationary increments and independent increments.

For a nondecreasing process $A$ with stationary increments and independent increments, suppose $\mathbb{E}{e}^{\theta A(t)}=e^{t\Lambda (\theta )}$. If $\Lambda (\theta )\le C\theta$, then $e^{\theta (A(t)-Ct)}$ is a nonnegative supermartingale. Applying the maximal inequality for a nonnegative supermartingale up to time $T$ and then letting $T\to \infty$ givesThe stationary workload of a stable queue fed by $A$ has this supremum distribution, using the past input and stationary increments and independent increments. For independent flows, the constraint is ${\sum }_j{a}_j(\theta )\le C$, where $a_j(\theta )={\Lambda }_j(\theta )/\theta$ is the long-time effective bandwidth. The result bounds a continuous-time supremum directly; a one-time Chernoff bound by itself does not do that.

Independent increments and stationary increments make a Lévy process's characteristic functions multiplicative in time. Together with continuity of Lévy characteristic functions and the value one at time zero, this forces an exponential. A continuous local complex logarithm is additive: its failure of additivity would be a continuous integer multiple of $2\pi i$, hence zero. The Cauchy functional equation and subdivision then give the exponential at every time. This construction of the characteristic exponent of a Lévy process precedes the Lévy–Khintchine formula describing the exponent's possible form.

A stochastic process has independent increments when, for every $t_0<t_1<\cdots <t_k$, the random variables $X_{t_1}-X_{t_0},\dots ,X_{t_k}-X_{t_{k-1}}$ are independent. Adjacent intervals are allowed: each increment represents the change after its left endpoint up to its right endpoint. A Poisson process, Brownian motion, and a random walk with independent steps have this property. Their values at different times need not be independent: for a fair simple random walk on the integer line, $\mathrm{Cov}(S_m,S_n)=\min (m,n)$ because both positions contain the same first $\min (m,n)$ increments. Independent increments need not be stationary increments; summing independent steps with time-dependent variances gives a counterexample.

A Lévy process starts at zero, has independent stationary increments, is stochastically continuous, and is conventionally taken to have càdlàg paths.

True. The unit-time increments $Z_k=B_k-B_{k-1}$ are independent and identically distributed random variables with standard normal distribution, hence integrable with mean zero. The strong law of large numbers gives $B_n/n\to 0$ almost surely at integer times.

To control the intervals between integers, put $M_n={\sup }_{0\le u\le 1}|B_{n+u}-B_n|$. Stationary increments and the Doob L2 maximal inequality give $\mathbb{E}{M}_n^2\le 4$. Thus the Markov inequality yieldsThe Borel-Cantelli lemma, followed by a countable intersection over positive rational $\epsilon$, implies $M_n/n\to 0$ almost surely. For $n\le t\le n+1$,Both terms tend to zero on the same probability-one event, proving $\boxed{B_t/t\to 0\text{ almost surely}}$ for continuous time as well.

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

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.

Under the path convention used here, a real Lévy process $(L_a{)}_{a\ge 0}$ starts at zero, has independent increments and stationary increments, is stochastically continuous, and has càdlàg paths almost surely. We verify all four requirements for $S$.

The immediate positive excursions proved in part (b) give $S_0=0$. For a deterministic $a\ge 0$, $S_a$ is an almost-surely finite stopping time, and continuity gives $X_{S_a}=a$. The Strong Markov property implies that ${\tilde{X}}_t=X_{S_a+t}-a$ is a fresh Brownian motion independent of the information up to $S_a$. For $b\ge 0$,No crossing of $a+b$ could have occurred before $S_a$. The increment thus has the same law as $S_b$ and is independent of the past. Apply this successively at a finite increasing sequence of deterministic levels to prove stationary independent increments.

For $h>0$ and $\epsilon >0$, part (b) and reflection giveHence $S_h\to 0$ in probability. Stationarity of the increments gives stochastic continuity at every level, from either side where defined.

For path regularity, on a single probability-one event the Brownian running maximum $M_t={\max }_{s\le t}{X}_s$ is continuous, nondecreasing and unbounded, andThis inverse is nondecreasing. If $a_j\downarrow a$, then $S_{a_j}\downarrow L\ge S_a$. For any $t>S_a$, some $u<t$ has $M_u>a$; for all sufficiently large $j$, $M_u>a_j$, so $S_{a_j}\le u<t$. Thus $L=S_a$, proving right continuity at every level simultaneously. On each bounded level interval it is bounded by the finite value at a larger integer level. Monotonicity then gives finite left limits. ThereforeThe Brownian first-passage subordinator is the level-indexed increasing process, rather than a process indexed by the original Brownian time.

For $H$, the inverse relations for a continuous running maximum areFor example, if $b<a$ then $S_b\le H_a$; while for $t<H_a$, $M_t<a$ and any $b\in (M_t,a)$ has $S_b>t$. Taking limits gives $S_{a-}=H_a$. Similarly $H_b\ge S_a$ for $b>a$, and a crossing of any level strictly above $a$ before a given $t>S_a$ bounds $H_b$ from above for $b$ sufficiently close to $a$. This proves $H_{a+}=S_a$. These are the non-strict inverse of a continuous nondecreasing function identities.

At the random level $A$ in part (b), $H_A<S_A=H_{A+}$, so $H$ fails right continuity almost surely. Thus it is not a Lévy process under the stated càdlàg-path convention. It does have the same finite-dimensional distributions and stochastic continuity as $S$, so definitions of a Lévy process that require only the increment and stochastic-continuity properties would call $H$ a Lévy process with a càdlàg modification. The reason for excluding the particular version $H$ here is its path regularity.

A real Lévy process is a real-valued stochastic process $(X_t{)}_{t\ge 0}$ with the following properties:

One convention also includes càdlàg paths in the definition. Equivalently, under the intrinsic definition above one chooses the càdlàg modification, which exists for such a stochastic process. Thus the usual working version of a Lévy process has right-continuous paths with left limits. There is no assumption of finite moments or continuous paths.

Fix $u\in \mathbb{R}$ and let $s\to t$. For any $\delta >0$, use $|e^{ia}-e^{ib}|\le \min (2,|a-b|)$ to obtainThe probability tends to zero by stochastic continuity. Taking the limit superior and then letting $\delta \downarrow 0$ provesThis proves continuity of Lévy characteristic functions from the elementary estimate, without requiring moments or replacing convergence in probability by an unjustified almost sure limit. At $t=0$, time approaches from the right. If stochastic continuity is formulated only at zero, stationary increments give the same argument at every $t$: the absolute value of $X_s-X_t$ has the law of $|X_{|s-t|}|$. For $u=0$ the characteristic function is identically one.

For fixed $u$, write $h(t)={\phi }_{X_t}(u)$. Independent increments and stationary increments giveThe preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m{)}^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to 1$.

Here is a direct proof of the exponential form of Lévy characteristic functions. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal complex logarithm gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\ge 0$ with $s+t\le a$, the multiplicative identity impliesThis difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive Cauchy functional equation locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in [0,a]$.

Define $\eta (u)=L(a)/a$. For any $t\ge 0$, choose an integer $m$ with $t/m\le a$; thenThe coefficient is unique: if two coefficients give the same exponential for every $t\ge 0$, their derivatives at zero agree. In particular $\eta (0)=0$, and $\mathrm{Re}\eta (u)\le 0$ follows from $|h(t)|\le 1$. The characteristic exponent of a Lévy process has therefore been obtained from first principles, without invoking the Lévy–Khintchine formula.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

Let $Z=X+Y$, where the two Lévy processes are independent as processes. For any disjoint ordered time intervals, the increment vectors of $X$ and $Y$ are independent of one another, and each vector has independent coordinates. Thus the pairs of corresponding increments are independent across intervals, and so are their sums. The law of each summed increment is the convolution of the two increment laws, depending only on the interval length. This proves independent increments and stationary increments for $Z$.

Also $Z_0=0$, and for every $\epsilon >0$,Thus $Z$ is stochastically continuous. The sum of two càdlàg functions is càdlàg. All defining properties hold, so $X+Y$ is a Lévy process. Independence of the entire two processes, not merely equality of some one-time laws, is what supplies independent summed increments.

First, convergence in probability at $t=0$ gives $X_0=0$ almost surely, since every $X_0^n=0$. For any finite set of times, the corresponding vectors converge in probability: the union bound controls the probability that any coordinate differs by more than a fixed tolerance. The same holds for their increment vectors, hence also for their convergence in distribution.

For $0=t_0<t_1<\cdots <t_m$, set $D_j^n=X_{t_j}^n-X_{t_{j-1}}^n$ and $D_j=X_{t_j}-X_{t_{j-1}}$. The characteristic function of a random vector factors for the independent increments of each $X^n$:Pass to the limit using convergence in distribution and bounded continuous functions. The resulting factorization of the characteristic function of a random vector, with the uniqueness theorem for characteristic functions, proves independence of the $D_j$. Similarly $X_{s+t}^n-X_s^n\stackrel{d}{=}{X}_t^n$ passes to the limit, proving stationary increments for $X$.

It remains to prove stochastic continuity. For $\epsilon >0$, the triangle inequality and the union bound imply, for every fixed $n$,The second term vanishes by stochastic continuity of $X^n$. Now let $n\to \infty$ and use the additional near-zero approximation hypothesis. We obtain $X_t\to 0$ in probability as $t\downarrow 0$. The stationary increments transfer this to every time: both $X_{s+h}-X_s$ and $X_s-X_{s-h}$ have the law of $X_h$ for $h>0$ when defined. Thus $X$ has all the intrinsic Lévy process properties, proving closure of Lévy processes under locally controlled convergence in probability.

If one requires the supplied process itself to be càdlàg, the hypotheses justify a càdlàg modification, rather than that stronger pathwise assertion. To see the distinction, take $X^n\equiv 0$, let $U$ have uniform distribution on $(0,1)$, and put $X_t=1_{\{t=U\}}$. At each fixed time $t$, $X_t=0$ almost surely, so both approximation hypotheses hold with zero error probability. Nevertheless every path has an isolated spike at $U$ and is not right-continuous there. Its identically zero modification of a stochastic process is a Lévy process with càdlàg paths. The conclusion is exact under the intrinsic definition, and exact up to modification under the convention requiring càdlàg paths.