Convolution semigroups, canonical processes, and Brownian motion

1 Convolution semigroups, projective families, and canonical processes

Let

$$E={ℝ}^d$$

and let $ℰ={ℬ}_{{ℝ}^d}$, the Borel $\sigma$-algebra of ${ℝ}^d$, and let $𝒫(E)$ be the collection of Borel probability measures on ${ℝ}^d$. With the narrow topology, $𝒫(E)$ is a Polish space. For a nonempty set $J$, we write

$${ℰ}^J=\underset{t\in J}{\otimes }ℰ,$$

the product $\sigma$-algebra.

Let $A:E\times E\to E$ be $A(x_1,x_2)=x_1+x_2$. For ${\nu }_1,{\nu }_2\in 𝒫(E)$, the convolution of ${\nu }_{\mathrm{1}}$ and ${\nu }_{\mathrm{2}}$ is the pushforward of the product measure ${\nu }_1\times {\nu }_2$ by $A$:

$${\nu }_1*{\nu }_2=A_{*}({\nu }_1\times {\nu }_2).$$

The convolution ${\nu }_1*{\nu }_2$ is an element of $𝒫(E)$.

Let

$$I={ℝ}_{\ge 0}.$$

A convolution semigroup is a family ${({\nu }_t)}_{t\in I}$ of elements of $𝒫(E)$ such that for $s,t\in I$,

$${\nu }_{s+t}={\nu }_s*{\nu }_t.$$

From this, it turns out that ${\mu }_0={\delta }_0$. A convolution semigroup is called continuous when the map $t\mapsto {\nu }_t$ is continuous $I\to 𝒫(E)$.

For $\nu \in 𝒫(E)$ and $x\in E$, and for $B\in ℰ$,

$$(\nu *{\delta }_x)(B)={\int }_E\left({\int }_E{1}_B(x_1+x_2)𝑑{\delta }_x(x_1)\right)𝑑\nu (x_2)=\nu (B-x),$$

and we define ${\nu }^x\in 𝒫(E)$ by

$${\nu }^x=\nu *{\delta }_x.$$

For $\nu \in 𝒫(E)$ and for a Borel measurable function $f:E\to [0,\mathrm{\infty }]$, write

$$\nu f={\int }_{E}f𝑑\mu .$$

For $x\in E$, using the change of variables formula and Fubini’s theorem,

${\nu }^{x}f$ $={\displaystyle {\int }_E}fd(\nu *{\delta }_x)$

$={\displaystyle {\int }_{E\times E}}f\circ Ad(\nu \times {\delta }_x)$

$={\displaystyle {\int }_E}\left({\displaystyle {\int }_E}f(x_1+x_2)𝑑{\delta }_x(x_2)\right)𝑑\nu (x_1)$

$={\displaystyle {\int }_E}f(x_1+x)𝑑\nu (x_1).$

That is, for $\nu \in 𝒫(E)$, for $f:E\to [0,\mathrm{\infty }]$ Borel measurable, and for $x\in E$,

$${\nu }^{x}f={\int }_{E}f𝑑{\nu }^x={\int }_{E}f(x+y)𝑑\nu (y).$$ (1)

For nonempty subsets $J$ and $K$ of $I$ with $J\subset K$, let

$${\pi }_{K,J}:E^K\to E^J$$

be the projection map. Let $𝒦=𝒦(I)$ be the collection of finite nonempty subsets of $I$. Let $(\mathrm{\Omega },ℱ,P,{(X_t)}_{t\in I})$ be a stochastic process with state space $E$. For $J\in 𝒦$, with elements , we define

$$X_J=X_{t_1}\otimes \mathrm{\cdots }\otimes X_{t_n},$$

which is measurable $ℱ\to {ℰ}^J$. The joint distribution $P_J$ of the family of random variables ${(X_t)}_{t\in J}$ is the distribution of $X_J$, i.e.

$$P_J=X_J{}_{*}P.$$

The family of finite-dimensional distributions of $X$ is the family ${(P_J)}_{J\in 𝒦}$. For $J,K\in 𝒦$ with $J\subset K$,

$$X_J={\pi }_{K,J}\circ X_K,$$

from which

$${({\pi }_{K,J})}_{*}{P}_K=P_J.$$ (2)

Forgetting the stochastic process $X$, a family of probability measures $P_J$ on ${ℰ}^J$, for $J\in 𝒦$, is called a projective family when (2) is true. The Kolmogorov extension theorem tells us that if ${(P_J)}_{J\in 𝒦}$ is a projective family, then there is a unique probability measure $P_I$ on ${ℰ}^I$ such that for any $J\in 𝒦$,

$${({\pi }_{I,J})}_{*}{P}_I=P_J.$$ (3)

Then for $\mathrm{\Omega }=E^I$ and $ℱ={ℰ}^I$, $(\mathrm{\Omega },ℱ,P_I)$ is a probability space, and for $t\in I$ we define $X_t:\mathrm{\Omega }\to E$ by

$$X_t(\omega )={\pi }_{I,\{t\}}(\omega )=\omega (t),$$ (4)

which is measurable $ℱ\to ℰ$, and thus the family ${(X_t)}_{t\in I}$ is a stochastic process with state space $E$. For $J\in 𝒦$ it is immediate that

$$X_J={\pi }_{I,J}.$$

For $B\in {ℰ}^J$, applying (3) gives

$$(X_J{}_{*}{P}_I)(B)=({({\pi }_{I,J})}_{*}{P}_I)(B)=P_J(B),$$

which means that $X_J{}_{*}{P}_I=P_J$, namely, ${(P_J)}_{J\in 𝒦}$ is the family of finite-dimensional distributions of the stochastic process ${(X_t)}_{t\in I}$. We call the stochastic process (4) the canonical process associated with the projective family ${\mathrm{(}{P}_J\mathrm{)}}_{J\mathrm{\in }\mathrm{K}}$.

Let ${({\nu }_t)}_{t\in I}$ be a convolution semigroup and let $\mu \in 𝒫(E)$. For $J\in 𝒦$, with elements , and for $B\in {ℰ}^J$, define

(5)

We say that ${(P_J)}_{J\in 𝒦}$ is the family of measures induced by the convolution semigroup ${\mathrm{(}{\nu }_t\mathrm{)}}_{t\mathrm{\in }I}$. It is proved that ${(P_J)}_{J\in 𝒦}$ is a projective family. Therefore, from the Kolmogorov extension theorem it follows that there is a unique probability measure $P^{\mu }$ on ${ℰ}^I$ such that

$${({\pi }_{I,J})}_{*}{P}^{\mu }=P_J.$$ (6)

For $\mathrm{\Omega }=E^I$ and $ℱ={ℰ}^I$, $(\mathrm{\Omega },ℱ,P^{\mu })$ is a probability space. For $t\in I$ define $X_t:\mathrm{\Omega }\to E$ by

$$X_t(\omega )={\pi }_{I,\{t\}}(\omega )=\omega (t).$$

${(X_t)}_{t\in I}$ is a stochastic processes whose family of finite-dimensional distributions is ${(P_J)}_{J\in 𝒦}$, i.e. for $J\in 𝒦$ with elements and for $B\in {ℰ}^J$,

Applying this with $\mu ={\delta }_x$ yields

$$({(X_{t_1}\otimes \mathrm{\cdots }\otimes X_{t_n})}_{*}{P}^{{\delta }_x})(B)={\int }_E\mathrm{\cdots }{\int }_E{1}_B(x_1,\mathrm{\dots },x_n)𝑑{\nu }_{t_n-t_{n-1}}^{x_{n-1}}(x_n)\mathrm{\cdots }𝑑{\nu }_{t_1}^x(x_1),$$

and thus, for any $\mu \in 𝒫(E)$,

That is, for $\mu \in 𝒫(E)$, for $J\in 𝒦$, and $B\in {ℰ}^J$,

$$(X_J{}_{*}{P}^{\mu })(B)={\int }_E(X_J{}_{*}{P}^{{\delta }_x})(B)𝑑\mu (x).$$ (7)

For $J\in 𝒦$, $A_t\in ℰ$ for $t\in J$, and $A={\prod }_{t\in J}{A}_t\times {\prod }_{t\in I\setminus J}E\in ℱ={ℰ}^I$, namely $A$ is a cylinder set, let $B={\pi }_{I,J}(A)={\prod }_{t\in J}{A}_t\in {ℰ}^J$,

$$X_J^{-1}(B)={\pi }_{I,J}^{-1}(B)=A,$$

so by (7),

$$P^{\mu }(A)={\int }_E{P}^{{\delta }_x}(A)𝑑\mu (x).$$ (8)

Because this is true for all cylinder sets in the product $\sigma$-algebra ${ℰ}^I$ and ${ℰ}^I$ is generated by the collection of cylinder sets, (8) is true for all $A\in ℱ$.

Let $J\in 𝒦$, with elements , and let ${\sigma }_n:E^{n+1}\to E^n$ be

$${\sigma }_n(x_0,x_1,\mathrm{\dots },x_n)=(x_0+x_1,x_0+x_1+x_2,\mathrm{\dots },x_0+x_1+x_2+\mathrm{\cdots }+x_n).$$

For $B\in {ℰ}^n$ using (1) we obtain by induction

Thus, with $P_J$ the probability measure on ${ℰ}^J$ defined in (5),

$${\int }_{E^J}{1}_B𝑑P_J=P_J(B)={\int }_E{\int }_E\mathrm{\cdots }{\int }_E{1}_B\circ {\sigma }_n𝑑{\nu }_{t_n-t_{n-1}}(x_n)\mathrm{\cdots }𝑑{\nu }_{t_1}(x_1)𝑑\mu (x_0).$$

For $f:E^n\to [0,\mathrm{\infty }]$ a Borel measurable function, there is a sequence of measurable simple functions pointwise increasing to $f$, and applying the monotone convergence theorem yields

$${\int }_{E^n}f𝑑P_J={\int }_E{\int }_E\mathrm{\cdots }{\int }_{E}f\circ {\sigma }_n𝑑{\nu }_{t_n-t_{n-1}}(x_n)\mathrm{\cdots }𝑑{\nu }_{t_1}(x_1)𝑑\mu (x_0).$$ (9)

2 Increments

Let $(\mathrm{\Omega },ℱ,P,{(X_t)}_{t\in I})$ be a stochastic process with state space $E$. $X$ is said to have stationary increments if there is a family ${({\nu }_t)}_{t\in I}$ of probability measures on $ℰ$ such that for all $s,t\in I$ with $s\le t$,

$$P_{*}(X_t-X_s)={\nu }_{t-s}.$$

In particular, for $s=t$ this implies that $P_{*}(0)={\nu }_0$, hence ${\nu }_0={\delta }_0$.

A stochastic process is said to have independent increments if for any $J\in 𝒦$, with elements , the random variables

$$X_{t_0},X_{t_1}-X_{t_0},\mathrm{\dots },X_{t_n}-X_{t_{n-1}}$$

are independent.

We now prove that the canonical process associated with the projective family of probability measures induced by a convolution semigroup and any initial distribution has stationary and independent increments.

Theorem 1.

Let ${({\nu }_t)}_{t\in I}$ be a convolution semigroup, let ${(P_J)}_{J\in 𝒦}$ be the family of measures induced by this convolution semigroup, let $\mu \in 𝒫(E)$, and let $(\mathrm{\Omega },ℱ,P^{\mu },{(X_t)}_{t\in I})$, $\mathrm{\Omega }=E^I$ and $ℱ={ℰ}^I$, be the associated canonical process. $X$ has stationary increments,

$${(X_t-X_s)}_{*}{P}^{\mu }={\nu }_{t-s},s\le t,$$ (10)

and has independent increments.

Proof.

${\nu }_0={\delta }_0$, so (10) is immediate when $s=t$. When , let

$$Y=X_s\otimes X_t=X_{\{s,t\}}={\pi }_{I,\{s,t\}},$$

which is measurable $ℱ\to ℰ\otimes ℰ$, and let $q:E\times E\to E$ be $(x_1,x_2)\mapsto x_2-x_1$, which is continuous and hence Borel measurable. Then $q\circ Y$ is measurable $ℱ\to ℰ$, and for $B\in ℰ$,

${(q\circ Y)}^{-1}(B)$ $=\{\omega \in \mathrm{\Omega }:(q\circ Y)(\omega )\in B\}$

$=\{\omega \in \mathrm{\Omega }:X_t(\omega )-X_s(\omega )\in B\}$

$={(X_t-X_s)}^{-1}(B),$

and thus

$${(q\circ Y)}_{*}{P}^{\mu }={(X_t-X_s)}_{*}{P}^{\mu }.$$ (11)

Now, according to (6),

$$Y_{*}{P}^{\mu }={({\pi }_{I,\{s,t\}})}_{*}{P}^{\mu }=P_{\{s,t\}}.$$

Therefore, using that $x_2-x_1\in B$ if and only if $x_2\in x_1+B$ and also using ${\nu }_{t-s}^{x_1}(x_1+B)={\nu }_{t-s}(B)$,

${(X_t-X_s)}_{*}{P}^{\mu }(B)$ $={(q\circ Y)}_{*}{P}^{\mu }(B)$

$=Y_{*}{P}^{\mu }(q^{-1}(B))$

$=P_{\{s,t\}}{}_{*}(q^{-1}(B))$

$={\displaystyle {\int }_E}{\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{q^{-1}(B)}(x_1,x_2)𝑑{\nu }_{t-s}^{x_1}(x_2)𝑑{\nu }_s^x(x_1)𝑑\mu (x)$

$={\displaystyle {\int }_E}{\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{x_1+B}(x_2)𝑑{\nu }_{t-s}^{x_1}(x_2)𝑑{\nu }_s^x(x_1)𝑑\mu (x)$

$={\nu }_{t-s}(B){\displaystyle {\int }_E}{\displaystyle {\int }_E}𝑑{\nu }_s^x(x_1)𝑑\mu (x)$

$={\nu }_{t-s}(B){\displaystyle {\int }_E}{\nu }_s^x(E)𝑑\mu (x)$

$={\nu }_{t-s}(B){\displaystyle {\int }_E}𝑑\mu (x)$

$={\nu }_{t-s}(B),$

which shows that

$${(X_t-X_s)}_{*}{P}^{\mu }={\nu }_{t-s},$$

and thus that $X$ has stationary increments.

Let , let $J=\{t_0,t_1,\mathrm{\dots },t_n\}\in 𝒦$, write $X_{t_{-1}}=0$, and let

$$Y_0=X_{t_0}-X_{t_{-1}},Y_1=X_{t_1}-X_{t_0},\mathrm{\dots },Y_n=X_{t_n}-X_{t_{n-1}}.$$

For the random variables $Y_0,\mathrm{\dots },Y_n$ to be independent means for their joint distribution to be equal to the product of the distributions of each, i.e. to prove that $X$ has independent increments, writing

$$Z=Y_0\otimes \mathrm{\cdots }\otimes Y_n={\tau }_n\circ (X_{t_0}\otimes \mathrm{\cdots }\otimes X_{t_n})={\tau }_n\circ X_J={\tau }_n\circ {\pi }_{I,J},$$

with ${\tau }_n:E^{n+1}\to E^n$ defined by

$${\tau }_n(x_0,x_1,\mathrm{\dots },x_n)=(x_0,x_1-x_0,\mathrm{\dots },x_n-x_{n-1}),$$

we have to prove that

$$Z_{*}{P}^{\mu }=\prod _{j=0}^{n}Y_j{}_{*}{P}^{\mu }.$$

To prove this, it suffices (because the collection of cylinder sets generates the product $\sigma$-algebra) to prove that for any $A_0,\mathrm{\dots },A_n\in ℰ$ and for $A={\prod }_{j=0}^n{A}_j\in {ℰ}^{n+1}$,

$$(Z_{*}{P}^{\mu })(A)=\left(\prod _{j=0}^{n}Y_j{}_{*}{P}^{\mu }\right)(A),$$

i.e. that

$$(Z_{*}{P}^{\mu })(A)=\prod _{j=0}^n(Y_{j_{*}}{P}^{\mu })(A_j).$$

We now prove this. Using the change of variables theorem and (6),

$(Z_{*}{P}^{\mu })(A)$ $={\displaystyle {\int }_{E^{n+1}}}{1}_{A}d(Z_{*}{P}^{\mu })$

$={\displaystyle {\int }_{\mathrm{\Omega }}}{1}_A\circ Z𝑑P^{\mu }$

$={\displaystyle {\int }_{\mathrm{\Omega }}}{1}_A\circ {\tau }_n\circ (X_{t_0}\otimes \mathrm{\cdots }\otimes X_{t_n})𝑑P^{\mu }$

$={\displaystyle {\int }_{E^J}}{1}_A\circ {\tau }_{n}d(X_J{}_{*}{P}^{\mu })$

$={\displaystyle {\int }_{E^J}}{1}_A\circ {\tau }_n𝑑P_J.$

Then applying (9) with $f=1_A\circ {\tau }_n$,

and because $t_0=0$ and ${\nu }_0={\delta }_0$,

${\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{A_0}(x_{-1}+x_0)𝑑{\nu }_{t_0}(x_0)𝑑\mu (x_{-1})$ $={\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{A_0}(x_{-1}+x_0)𝑑{\delta }_0(x_0)𝑑\mu (x_{-1})$

$={\displaystyle {\int }_E}{1}_{A_0}(x_{-1})𝑑\mu (x_{-1})$

$=\mu (A_0),$

and therefore

$$(Z_{*}{P}^{\mu })(A)=\mu (A_0)\cdot \prod _{j=1}^n{\nu }_{t_j-t_{j-1}}(A_j).$$

But we have already proved that (10), which tells us that for each $j$,

$$Y_j{}_{*}{P}^{\mu }={(X_{t_j}-X_{t_{j-1}})}_{*}{P}^{\mu }={\nu }_{t_j-t_{j-1}},$$

and thus

$$(Z_{*}{P}^{\mu })(A)=\mu (A_0)\cdot \prod _{j=1}^n(Y_j{}_{*}{P}^{\mu })(A_j).$$

But $Y_0{}_{*}{P}^{\mu }=X_0{}_{*}{P}^{\mu }$ and from (7) we have

$$(X_0{}_{*}{P}^{\mu })(A_0)={\int }_E(X_0{}_{*}{P}^{{\delta }_x})(A_0)𝑑\mu (x)={\int }_E(\pi _0{}_{*}{P}^{{\delta }_x})(A_0)𝑑\mu (x),$$

and, from (5),

$(\pi _0{}_{*}{P}^{{\delta }_x})(A_0)$ $={\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{A_0}(x_0)𝑑{\nu }_0^y(x_0)𝑑{\delta }_x(y)$

$={\displaystyle {\int }_E}{\displaystyle {\int }_E}{1}_{A_0}(x_0)𝑑{\delta }_y(x_0)𝑑{\delta }_x(y)$

$={\displaystyle {\int }_E}{1}_{A_0}(y)𝑑{\delta }_x(y)$

$=1_{A_0}(x),$

thus

$$(X_0{}_{*}{P}^{\mu })(A_0)={\int }_E{1}_{A_0}(x)𝑑\mu (x)=\mu (A_0).$$

Therefore

$$(Z_{*}{P}^{\mu })(A)=(X_0{}_{*}{P}^{\mu })(A_0)\cdot \prod _{j=1}^n(Y_j{}_{*}{P}^{\mu })(A_j)=\prod _{j=0}^n(Y_j{}_{*}{P}^{\mu })(A_j),$$

which completes the proof that $X$ has independent increments. ∎

3 The Brownian convolution semigroup and Brownian motion

For $a\in ℝ$ and $\sigma >0$, let ${\gamma }_{a,{\sigma }^2}$ be the Gaussian measure on $ℝ$, the probability measure on $ℝ$ whose density with respect to Lebesgue measure is

$$p(x,a,{\sigma }^2)=\frac{1}{\sqrt{2\pi {\sigma }^2}}\exp \left(-\frac{{(x-a)}^2}{2{\sigma }^2}\right).$$

For $\sigma =0$, let

$${\gamma }_{a,0}={\delta }_a.$$

Define for $t\in I$,

$${\nu }_t=\prod _{k=1}^d{\gamma }_{0,t},$$

which is an element of $𝒫(E)$. For $s,t\in I$, we calculate

$${\nu }_s*{\mu }_t=\left(\prod _{k=1}^d{\gamma }_{0,s}\right)*\left(\prod _{k=1}^d{\gamma }_{0,t}\right)=\prod _{k=1}^d({\gamma }_{0,s}*{\gamma }_{0,t})=\prod _{k=1}^d{\gamma }_{0,s+t}={\nu }_{s+t},$$

showing that ${({\nu }_t)}_{t\in I}$ is a convolution semigroup. It is proved using Lévy’s continuity theorem that $t\mapsto {\nu }_t$ is continuous $I\to 𝒫(E)$, showing that ${({\nu }_t)}_{t\in I}$ is a continuous convolution semigroup.

We first prove a lemma (which is made explicit in Isserlis’s theorem) about the moments of random variables with Gaussian distributions.

Lemma 2.

If $Z:\mathrm{\Omega }\to E$ is a random variable with Gaussian distribution ${\nu }_{\tau }$, $\tau >0$, then for each $n$ there is some $C_n>0$ such that

$$E({|Z|}^{2n})=C_n{\tau }^n.$$

In particular, $C_2=d$ and $C_4=d(d+2)$.

Proof.

That $Z$ has distribution ${\nu }_{\tau }$ means that

$$Z_{*}P={\nu }_{\tau }=\prod _{j=1}^d{\gamma }_{0,\tau }.$$

Write $Z=Z_1\otimes \mathrm{\cdots }\otimes Z_d$, each of which has distribution ${\gamma }_{0,\tau }$, and $Z_{*}P={\prod }_{j=1}^{d}Z_j{}_{*}P$, which means that $Z_1,\mathrm{\dots },Z_d$ independent. Let $U_j={\tau }^{-1/2}{Z}_j$ for $j=1,\mathrm{\dots },d$, and then $U_1,\mathrm{\dots },U_d$ are independent random variables each with distribution ${\gamma }_{0,1}$. Then using the multinomial formula,

$E({|Z|}^{2n})$ $=E({(Z_1^2+\mathrm{\cdots }+Z_d^2)}^n)$

$={\tau }^n\cdot E({(U_1^2+\mathrm{\cdots }+U_d^2)}^n)$

$={\tau }^n\cdot E\left({\displaystyle \sum _{k_1+\mathrm{\cdots }+k_d=n}}{\displaystyle \frac{n!}{k_1!\mathrm{\cdots }{k}_d!}}{\displaystyle \prod _{1\le i\le d}}{U}_j^{2k_i}\right)$

$={\tau }^n\cdot {\displaystyle \sum _{k_1+\mathrm{\cdots }+k_d=n}}{\displaystyle \frac{n!}{k_1!\mathrm{\cdots }{k}_d!}}E\left({\displaystyle \prod _{1\le i\le d}}{U}_j^{2k_i}\right).$

For $n=2$, since $E(U_i{U}_j)=E(U_i)E(U_j)=0$ for $i\ne j$,

$${\tau }^2\cdot \sum _{k_1+\mathrm{\cdots }+k_d=2}\frac{2}{k_1!\mathrm{\cdots }{k}_d!}E\left(\prod _{1\le i\le d}{U}_i^{2k_i}\right)={\tau }^2\cdot \sum _{j=1}^{d}E(U_j^2)={\tau }^2\cdot \sum _{j=1}^{d}1=d{\tau }^2,$$

showing that $C_2=d$. ∎

A stochastic process $(\mathrm{\Omega },ℱ,P,{(X_t)}_{t\in I})$ with state space $E$ is called a $d$-dimensional Brownian motion when:

1. For $s\le t$,

   $${(X_t-X_s)}_{*}P={\nu }_{t-s},$$

   and thus $X$ has stationary increments.

2. $X$ has independent increments.

3. For almost all $\omega \in \mathrm{\Omega }$, the path $t\mapsto X_t(\omega )$ is continuous $I\to E$.

We call $X_0{}_{*}P$ the initial distribution of the Brownian motion. When $X_0{}_{*}P={\delta }_x$ for some $x\in E$, we say that $x$ is the starting point of the Brownian motion. We now prove that for any Borel probability measure on $ℰ$, in particular ${\delta }_x$, there is a $d$-dimensional Brownian motion which has this as its initial distribution.

Theorem 3 (Brownian motion).

For any $\mu \in 𝒫(E)$, there is a $d$-dimensional Brownian motion with initial distribution $\mu$.

Proof.

Let ${(P_J)}_{J\in 𝒦}$ be the family of measures induced by the Brownian convolution semigroup

$${\nu }_t=\prod _{k=1}^d{\gamma }_{0,t},t\in I,$$

and let $(\mathrm{\Omega },ℱ,P^{\mu },{(X_t)}_{t\in I})$, $\mathrm{\Omega }=E^I$ and $ℱ={ℰ}^I$, be the associated canonical process. Theorem 10 tells us that $X$ has stationary increments,

$${(X_t-X_s)}_{*}{P}^{\mu }={\nu }_{t-s},s\le t,$$ (12)

and has independent increments. For $\tau =t-s>0$, by (12) and Lemma 2,

$$E({|X_t-X_s|}^4)=d(d+2){\tau }^2=d(d+2){|t-s|}^2.$$

Because $E({|X_t-X_t|}^4)=E(0)=0$, we have that for any $s,t\in I$,

$$E({|X_t-X_s|}^4)=d(d+2){|t-s|}^2.$$

The initial distribution of $X$ is $X_0{}_{*}{P}^{\mu }=\mu$. For $\alpha =4,\beta =1,c=d(d+2)$, the Kolmogorov continuity theorem tells us that there is a continuous modification $B$ of $X$. That is, there is a stochastic process ${(B_t)}_{t\in I}$ such that for each $\omega \in \mathrm{\Omega }$, the path $t\mapsto B_t(\omega )$ is continuous $I\to E$, namely, $B$ is a continuous stochastic process, and for each $t\in I$,

$$P(X_t=B_t)=1,$$

namely, $B$ is a modification of $X$. Because $B$ is a modification of $X$, $B$ has the same finite-dimensional distributions as $X$, from which it follows that $B$ satisfies (12) and has independent increments. For $A\in ℰ$, because $B$ is a modification of $X$,

$$(B_0{}_{*}{P}^{\mu })(A)=P^{\mu }(B_0\in A)=P^{\mu }(X_0\in A)=(X_0{}_{*}{P}^{\mu })(A),$$

thus $B_0{}_{*}{P}^{\mu }=X_0{}_{*}{P}^{\mu }=\mu$, namely, $B$ has initial distribution $\mu$. Therefore, $B$ is a Brownian motion (indeed, all the paths of $B$ are continuous, not merely almost all of them) that has initial distribution $\mu$, proving the claim. ∎

For $\mu \in 𝒫(E)$, let $(\mathrm{\Omega },ℱ,P^{\mu },{(B_t)}_{t\in I})$ be the $d$-dimensional Brownian motion with initial distribution $\mu$ constructed in Theorem 3; we are not merely speaking about some $d$-dimensional Brownian motion but about this construction, for which $\mathrm{\Omega }=E^I$, all whose paths are continuous rather than merely almost all whose paths are continuous. For a measurable space $(A,𝒜)$ and topological spaces $X$ and $Y$, a function $f:X\times A\to Y$ is called a Carathéodory function if for each $x\in X$, the map $a\mapsto f(x,a)$ is measurable $𝒜\to {ℬ}_Y$, and for each $a\in A$, the map $x\mapsto f(x,a)$ is continuous $X\to Y$. It is a fact that if $X$ is a separable metrizable space and $Y$ is a metrizable space, then any Carathéodory function $f:X\times A\to Y$ is measurable ${ℬ}_X\otimes 𝒜\to {ℬ}_Y$, namely it is jointly measurable. $B:I\times \mathrm{\Omega }\to E$ is a Carathéodory function. $I={ℝ}_{\ge 0}$, with the subspace topology inherited from $ℝ$, is a separable metrizable space, and $E={ℝ}^d$ is a metrizable space, and therefore the $d$-dimensional Brownian motion $B$ is jointly measurable.

The Kolmogorov-Chentsov theorem says that if a stochastic process ${(X_t)}_{t\in I}$ with state space $E$ satisfies, for $\alpha ,\beta ,c>0$,

$$E({|X_s-X_s|}^{\alpha })\le c{|t-s|}^{1+\beta },s,t\in I,$$

and almost every path of $X$ is continuous, then for almost every $\omega \in \mathrm{\Omega }$, for every the map $t\mapsto X_t(\omega )$ is locally $\gamma$-Hölder continuous: for each $t_0\in I$ there is some and some $C_{t_0}$ such that

For $\mu \in 𝒫(E)$, let $(\mathrm{\Omega },ℱ,P^{\mu },{(B_t)}_{t\in I})$ be the $d$-dimensional Brownian motion with initial distribution $\mu$ formed in Theorem 3. For $s\le t$, ${(B_t-B_s)}_{*}{P}^{\mu }={\nu }_{t-s}$, and thus Lemma 2 tells us that for each $n\ge 1$ there is some $C_n$ with which $E({|B_t-B_s|}^{2n})=C_n{(t-s)}^n$ for all . Then $E({|B_t-B_s|}^{2n})\le C_n{|t-s|}^n$ for all $s,t\in I$. For $n>1$ and for ${\alpha }_n=2n$ and ${\beta }_n=n-1$,

$$\frac{{\beta }_n}{{\alpha }_n}=\frac{n-1}{2n}=\frac{1}{2}-\frac{1}{2n},$$

and for $n>2$, take some . Let $N_n$ be the set of those $\omega \in \mathrm{\Omega }$ for which $t\mapsto B_t(\omega )$ is not locally ${\gamma }_n$-Hölder continuous. Then the Kolmogorov-Chentsov theorem yields $P^{\mu }(N_n)=0$. Let $N={\bigcup }_{n>2}{N}_n$, which is a $P^{\mu }$-null set. For $\omega \in \mathrm{\Omega }\setminus N$ and for any , there is some ${\gamma }_n$ satisfying , and hence the map $t\mapsto B_t(\omega )$ is locally ${\gamma }_n$-Hölder continuous, which implies that this map is locally $\gamma$-Hölder continuous. We summarize what we have just said in the following theorem.

Theorem 4.

Let $\mu \in 𝒫(E)$ and let $(\mathrm{\Omega },ℱ,P^{\mu },{(B_t)}_{t\in I})$ be the $d$-dimensional Brownian motion with initial distribution $\mu$ formed in Theorem 3. For almost all $\omega \in \mathrm{\Omega }$, for all , the map $t\mapsto B_t(\omega )$ is locally $\gamma$-Hölder continuous.

4 Lévy processes

A stochastic process ${(X_t)}_{t\in I}$ with state space $E$ is called a Lévy process if (i) $X_0=0$ almost surely, (ii) $X$ has stationary and independent increments, and (iii) for any $a>0$,

$$\underset{t\downarrow 0}{lim}P(|X_t|\ge ϵ)=0.$$

Because $X_0=0$ almost surely and $X$ has stationary increments, (iii) yields for any $t\in I$,

$$\underset{s\to t}{lim}P(|X_s-X_s|\ge ϵ)=0.$$ (13)

In any case, (13) is sufficient for (iii) to be true. Moreover, (iii) means that $X_s\to X_t$ in the topology of convergence in probability as $s\to t$, and if $X_s\to X_t$ almost surely then $X_s\to X_t$ in the topology of convergence in probability; this is proved using Egorov’s theorem. Thus, a $d$-dimensioanl Brownian motion with starting point $0$ is a Lévy process; we do not merely assert that the Brownian motion formed in Theorem 3 is a Lévy process. There is much that can be said generally about Lévy processes, and thus the fact that any $d$-dimensional Brownian motion with starting point $0$ is a Lévy process lets us work in a more general setting in which some results may be more naturally proved: if we work merely with a Lévy process we know less about the process and thus have less open moves.