# Convolution semigroups, canonical processes, and Brownian motion

Jordan Bell
June 16, 2015

## 1 Convolution semigroups, projective families, and canonical processes

Let

 $E=\mathbb{R}^{d}$

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

 $\mathscr{E}^{J}=\bigotimes_{t\in J}\mathscr{E},$

the product $\sigma$-algebra.

Let $A:E\times E\to E$ be $A(x_{1},x_{2})=x_{1}+x_{2}$. For $\nu_{1},\nu_{2}\in\mathscr{P}(E)$, the convolution of $\nu_{1}$ and $\nu_{2}$ 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 $\mathscr{P}(E)$.

Let

 $I=\mathbb{R}_{\geq 0}.$

A convolution semigroup is a family $(\nu_{t})_{t\in I}$ of elements of $\mathscr{P}(E)$ 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 $I\to\mathscr{P}(E)$.

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

 $(\nu*\delta_{x})(B)=\int_{E}\left(\int_{E}1_{B}(x_{1}+x_{2})d\delta_{x}(x_{1})% \right)d\nu(x_{2})=\nu(B-x),$

and we define $\nu^{x}\in\mathscr{P}(E)$ by

 $\nu^{x}=\nu*\delta_{x}.$

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

 $\nu f=\int_{E}fd\mu.$

For $x\in E$, using the change of variables formula11 1 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 484, Theorem 13.46. and Fubini’s theorem,

 $\displaystyle\nu^{x}f$ $\displaystyle=\int_{E}fd(\nu*\delta_{x})$ $\displaystyle=\int_{E\times E}f\circ Ad(\nu\times\delta_{x})$ $\displaystyle=\int_{E}\left(\int_{E}f(x_{1}+x_{2})d\delta_{x}(x_{2})\right)d% \nu(x_{1})$ $\displaystyle=\int_{E}f(x_{1}+x)d\nu(x_{1}).$

That is, for $\nu\in\mathscr{P}(E)$, for $f:E\to[0,\infty]$ 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 $\mathscr{K}=\mathscr{K}(I)$ be the collection of finite nonempty subsets of $I$. Let $(\Omega,\mathscr{F},P,(X_{t})_{t\in I})$ be a stochastic process with state space $E$. For $J\in\mathscr{K}$, with elements $t_{1}<\ldots, we define

 $X_{J}=X_{t_{1}}\otimes\cdots\otimes X_{t_{n}},$

which is measurable $\mathscr{F}\to\mathscr{E}^{J}$. 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 $(P_{J})_{J\in\mathscr{K}}$. For $J,K\in\mathscr{K}$ 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 $\mathscr{E}^{J}$, for $J\in\mathscr{K}$, is called a projective family when (2) is true. The Kolmogorov extension theorem tells us that if $(P_{J})_{J\in\mathscr{K}}$ is a projective family, then there is a unique probability measure $P_{I}$ on $\mathscr{E}^{I}$ such that for any $J\in\mathscr{K}$,

 $(\pi_{I,J})_{*}P_{I}=P_{J}.$ (3)

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

 $X_{t}(\omega)=\pi_{I,\{t\}}(\omega)=\omega(t),$ (4)

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

 $X_{J}=\pi_{I,J}.$

For $B\in\mathscr{E}^{J}$, 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, $(P_{J})_{J\in\mathscr{K}}$ 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 $(P_{J})_{J\in\mathscr{K}}$.

Let $(\nu_{t})_{t\in I}$ be a convolution semigroup and let $\mu\in\mathscr{P}(E)$. For $J\in\mathscr{K}$, with elements $t_{1}<\ldots, and for $B\in\mathscr{E}^{J}$, define

 $\begin{split}&\displaystyle P_{J}(B)\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}(x_{1},\ldots,x% _{n})d\nu_{t_{n}-t_{n-1}}^{x_{n-1}}(x_{n})\cdots d\nu_{t_{1}}^{x_{0}}(x_{1})d% \mu(x_{0}).\end{split}$ (5)

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

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

For $\Omega=E^{I}$ and $\mathscr{F}=\mathscr{E}^{I}$, $(\Omega,\mathscr{F},P^{\mu})$ is a probability space. For $t\in I$ define $X_{t}:\Omega\to E$ 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 $(P_{J})_{J\in\mathscr{K}}$, i.e. for $J\in\mathscr{K}$ with elements $t_{1}<\cdots and for $B\in\mathscr{E}^{J}$,

 $\begin{split}&\displaystyle((X_{t_{1}}\otimes\cdots\otimes X_{t_{n}})_{*}P^{% \mu})(B)\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}(x_{1},\ldots,x% _{n})d\nu_{t_{n}-t_{n-1}}^{x_{n-1}}(x_{n})\cdots d\nu_{t_{1}}^{x_{0}}(x_{1})d% \mu(x_{0}).\end{split}$

Applying this with $\mu=\delta_{x}$ yields

 $((X_{t_{1}}\otimes\cdots\otimes X_{t_{n}})_{*}P^{\delta_{x}})(B)=\int_{E}% \cdots\int_{E}1_{B}(x_{1},\ldots,x_{n})d\nu_{t_{n}-t_{n-1}}^{x_{n-1}}(x_{n})% \cdots d\nu_{t_{1}}^{x}(x_{1}),$

and thus, for any $\mu\in\mathscr{P}(E)$,

 $\begin{split}&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}(x_{1},\ldots,x_% {n})d\nu_{t_{n}-t_{n-1}}^{x_{n-1}}(x_{n})\cdots d\nu_{t_{1}}^{x}(x_{1})d\mu(x)% \\ \displaystyle=&\displaystyle\int_{E}((X_{t_{1}}\otimes\cdots\otimes X_{t_{n}})% _{*}P^{\delta_{x}})(B)d\mu(x).\end{split}$

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

 $({X_{J}}_{*}P^{\mu})(B)=\int_{E}({X_{J}}_{*}P^{\delta_{x}})(B)d\mu(x).$ (7)

For $J\in\mathscr{K}$, $A_{t}\in\mathscr{E}$ for $t\in J$, and $A=\prod_{t\in J}A_{t}\times\prod_{t\in I\setminus J}E\in\mathscr{F}=\mathscr{E% }^{I}$, namely $A$ is a cylinder set, let $B=\pi_{I,J}(A)=\prod_{t\in J}A_{t}\in\mathscr{E}^{J}$,

 $X_{J}^{-1}(B)=\pi_{I,J}^{-1}(B)=A,$

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 $\mathscr{E}^{I}$ and $\mathscr{E}^{I}$ is generated by the collection of cylinder sets, (8) is true for all $A\in\mathscr{F}$.

Let $J\in\mathscr{K}$, with elements $t_{1}<\cdots, and let $\sigma_{n}:E^{n+1}\to E^{n}$ be

 $\sigma_{n}(x_{0},x_{1},\ldots,x_{n})=(x_{0}+x_{1},x_{0}+x_{1}+x_{2},\ldots,x_{% 0}+x_{1}+x_{2}+\cdots+x_{n}).$

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

 $\begin{split}&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}(x_{1},\ldots,x_% {n-1},x_{n})d\nu_{t_{n}-t_{n-1}}^{x_{n-1}}(x_{n})\cdots d\nu_{t_{1}}^{x_{0}}(x% _{1})d\mu(x_{0})\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}(x_{1},\ldots,x% _{n-1},x_{n}+x_{n-1})d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{1}}^{x_{0}}(x_% {1})d\mu(x_{0})\\ \displaystyle=&\displaystyle\cdots\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{B}\circ\sigma_{n}% d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{1}}(x_{1})d\mu(x_{0}).\end{split}$

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

 $\int_{E^{J}}1_{B}dP_{J}=P_{J}(B)=\int_{E}\int_{E}\cdots\int_{E}1_{B}\circ% \sigma_{n}d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{1}}(x_{1})d\mu(x_{0}).$

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

 $\int_{E^{n}}fdP_{J}=\int_{E}\int_{E}\cdots\int_{E}f\circ\sigma_{n}d\nu_{t_{n}-% t_{n-1}}(x_{n})\cdots d\nu_{t_{1}}(x_{1})d\mu(x_{0}).$ (9)

## 2 Increments

Let $(\Omega,\mathscr{F},P,(X_{t})_{t\in I})$ 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 $\mathscr{E}$ such that for all $s,t\in I$ with $s\leq t$,

 $P_{*}(X_{t}-X_{s})=\nu_{t-s}.$

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 $J\in\mathscr{K}$, with elements $0=t_{0}, the random variables

 $X_{t_{0}},\quad X_{t_{1}}-X_{t_{0}},\quad\ldots,X_{t_{n}}-X_{t_{n-1}}$

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.22 2 Heinz Bauer, Probability Theory, p. 321, Theorem 37.2.

###### Theorem 1.

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

 $(X_{t}-X_{s})_{*}P^{\mu}=\nu_{t-s},\qquad s\leq t,$ (10)

and has independent increments.

###### Proof.

$\nu_{0}=\delta_{0}$, so (10) is immediate when $s=t$. When $s, let

 $Y=X_{s}\otimes X_{t}=X_{\{s,t\}}=\pi_{I,\{s,t\}},$

which is measurable $\mathscr{F}\to\mathscr{E}\otimes\mathscr{E}$, 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 $\mathscr{F}\to\mathscr{E}$, and for $B\in\mathscr{E}$,

 $\displaystyle(q\circ Y)^{-1}(B)$ $\displaystyle=\{\omega\in\Omega:(q\circ Y)(\omega)\in B\}$ $\displaystyle=\{\omega\in\Omega:X_{t}(\omega)-X_{s}(\omega)\in B\}$ $\displaystyle=(X_{t}-X_{s})^{-1}(B),$

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 $\nu_{t-s}^{x_{1}}(x_{1}+B)=\nu_{t-s}(B)$,

 $\displaystyle(X_{t}-X_{s})_{*}P^{\mu}(B)$ $\displaystyle=(q\circ Y)_{*}P^{\mu}(B)$ $\displaystyle=Y_{*}P^{\mu}(q^{-1}(B))$ $\displaystyle={P_{\{s,t\}}}_{*}(q^{-1}(B))$ $\displaystyle=\int_{E}\int_{E}\int_{E}1_{q^{-1}(B)}(x_{1},x_{2})d\nu_{t-s}^{x_% {1}}(x_{2})d\nu_{s}^{x}(x_{1})d\mu(x)$ $\displaystyle=\int_{E}\int_{E}\int_{E}1_{x_{1}+B}(x_{2})d\nu_{t-s}^{x_{1}}(x_{% 2})d\nu_{s}^{x}(x_{1})d\mu(x)$ $\displaystyle=\nu_{t-s}(B)\int_{E}\int_{E}d\nu_{s}^{x}(x_{1})d\mu(x)$ $\displaystyle=\nu_{t-s}(B)\int_{E}\nu_{s}^{x}(E)d\mu(x)$ $\displaystyle=\nu_{t-s}(B)\int_{E}d\mu(x)$ $\displaystyle=\nu_{t-s}(B),$

which shows that

 $(X_{t}-X_{s})_{*}P^{\mu}=\nu_{t-s},$

and thus that $X$ has stationary increments.

Let $0=t_{0}, let $J=\{t_{0},t_{1},\ldots,t_{n}\}\in\mathscr{K}$, write $X_{t_{-1}}=0$, and let

 $Y_{0}=X_{t_{0}}-X_{t_{-1}},\quad Y_{1}=X_{t_{1}}-X_{t_{0}},\quad\ldots,\quad Y% _{n}=X_{t_{n}}-X_{t_{n-1}}.$

For the random variables $Y_{0},\ldots,Y_{n}$ 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

 $Z=Y_{0}\otimes\cdots\otimes Y_{n}=\tau_{n}\circ(X_{t_{0}}\otimes\cdots\otimes X% _{t_{n}})=\tau_{n}\circ X_{J}=\tau_{n}\circ\pi_{I,J},$

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

 $\tau_{n}(x_{0},x_{1},\ldots,x_{n})=(x_{0},x_{1}-x_{0},\ldots,x_{n}-x_{n-1}),$

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 $A_{0},\ldots,A_{n}\in\mathscr{E}$ and for $A=\prod_{j=0}^{n}A_{j}\in\mathscr{E}^{n+1}$,

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

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),

 $\displaystyle(Z_{*}P^{\mu})(A)$ $\displaystyle=\int_{E^{n+1}}1_{A}d(Z_{*}P^{\mu})$ $\displaystyle=\int_{\Omega}1_{A}\circ ZdP^{\mu}$ $\displaystyle=\int_{\Omega}1_{A}\circ\tau_{n}\circ(X_{t_{0}}\otimes\cdots% \otimes X_{t_{n}})dP^{\mu}$ $\displaystyle=\int_{E^{J}}1_{A}\circ\tau_{n}d({X_{J}}_{*}P^{\mu})$ $\displaystyle=\int_{E^{J}}1_{A}\circ\tau_{n}dP_{J}.$

Then applying (9) with $f=1_{A}\circ\tau_{n}$,

 $\begin{split}&\displaystyle\int_{E^{J}}1_{A}\circ\tau_{n}dP_{J}\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{A}\circ\tau_{n}% \circ\sigma_{n+1}d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{0}}(x_{0})d\mu(x_{% -1})\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{A}(x_{-1}+x_{0},x% _{1},\ldots,x_{n})d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{0}}(x_{0})d\mu(x_% {-1})\\ \displaystyle=&\displaystyle\int_{E}\int_{E}\cdots\int_{E}1_{A_{0}}(x_{-1}+x_{% 0})1_{A_{1}}(x_{1})\cdots 1_{A_{n}}(x_{n})\\ &\displaystyle d\nu_{t_{n}-t_{n-1}}(x_{n})\cdots d\nu_{t_{0}}(x_{0})d\mu(x_{-1% })\\ \displaystyle=&\displaystyle\prod_{j=1}^{n}\nu_{t_{j}-t_{j-1}}(A_{j})\int_{E}% \int_{E}1_{A_{0}}(x_{-1}+x_{0})d\nu_{t_{0}}(x_{0})d\mu(x_{-1}),\end{split}$

and because $t_{0}=0$ and $\nu_{0}=\delta_{0}$,

 $\displaystyle\int_{E}\int_{E}1_{A_{0}}(x_{-1}+x_{0})d\nu_{t_{0}}(x_{0})d\mu(x_% {-1})$ $\displaystyle=\int_{E}\int_{E}1_{A_{0}}(x_{-1}+x_{0})d\delta_{0}(x_{0})d\mu(x_% {-1})$ $\displaystyle=\int_{E}1_{A_{0}}(x_{-1})d\mu(x_{-1})$ $\displaystyle=\mu(A_{0}),$

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),

 $\displaystyle({\pi_{0}}_{*}P^{\delta_{x}})(A_{0})$ $\displaystyle=\int_{E}\int_{E}1_{A_{0}}(x_{0})d\nu_{0}^{y}(x_{0})d\delta_{x}(y)$ $\displaystyle=\int_{E}\int_{E}1_{A_{0}}(x_{0})d\delta_{y}(x_{0})d\delta_{x}(y)$ $\displaystyle=\int_{E}1_{A_{0}}(y)d\delta_{x}(y)$ $\displaystyle=1_{A_{0}}(x),$

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 $a\in\mathbb{R}$ and $\sigma>0$, let $\gamma_{a,\sigma^{2}}$ be the Gaussian measure on $\mathbb{R}$, the probability measure on $\mathbb{R}$ 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 $\mathscr{P}(E)$. 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 $I\to\mathscr{P}(E)$, 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.33 3 Heinz Bauer, Probability Theory, p. 341, Lemma 40.2.

###### Lemma 2.

If $Z:\Omega\to E$ 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 $Z=Z_{1}\otimes\cdots\otimes Z_{d}$, each of which has distribution $\gamma_{0,\tau}$, and $Z_{*}P=\prod_{j=1}^{d}{Z_{j}}_{*}P$, which means that $Z_{1},\ldots,Z_{d}$ independent. Let $U_{j}=\tau^{-1/2}Z_{j}$ for $j=1,\ldots,d$, and then $U_{1},\ldots,U_{d}$ are independent random variables each with distribution $\gamma_{0,1}$. Then using the multinomial formula,

 $\displaystyle E(|Z|^{2n})$ $\displaystyle=E((Z_{1}^{2}+\cdots+Z_{d}^{2})^{n})$ $\displaystyle=\tau^{n}\cdot E((U_{1}^{2}+\cdots+U_{d}^{2})^{n})$ $\displaystyle=\tau^{n}\cdot E\left(\sum_{k_{1}+\cdots+k_{d}=n}\frac{n!}{k_{1}!% \cdots k_{d}!}\prod_{1\leq i\leq d}U_{j}^{2k_{i}}\right)$ $\displaystyle=\tau^{n}\cdot\sum_{k_{1}+\cdots+k_{d}=n}\frac{n!}{k_{1}!\cdots k% _{d}!}E\left(\prod_{1\leq i\leq d}U_{j}^{2k_{i}}\right).$

For $n=2$, since $E(U_{i}U_{j})=E(U_{i})E(U_{j})=0$ for $i\neq j$,

 $\tau^{2}\cdot\sum_{k_{1}+\cdots+k_{d}=2}\frac{2}{k_{1}!\cdots k_{d}!}E\left(% \prod_{1\leq i\leq d}U_{i}^{2k_{i}}\right)=\tau^{2}\cdot\sum_{j=1}^{d}E(U_{j}^% {2})=\tau^{2}\cdot\sum_{j=1}^{d}1=d\tau^{2},$

showing that $C_{2}=d$. ∎

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

1. 1.

For $s\leq t$,

 $(X_{t}-X_{s})_{*}P=\nu_{t-s},$

and thus $X$ has stationary increments.

2. 2.

$X$ has independent increments.

3. 3.

For almost all $\omega\in\Omega$, 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 $\mathscr{E}$, in particular $\delta_{x}$, there is a $d$-dimensional Brownian motion which has this as its initial distribution.44 4 Heinz Bauer, Probability Theory, p. 342, Theorem 40.3.

###### Theorem 3 (Brownian motion).

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

###### Proof.

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

 $\nu_{t}=\prod_{k=1}^{d}\gamma_{0,t},\qquad t\in I,$

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

 $(X_{t}-X_{s})_{*}P^{\mu}=\nu_{t-s},\qquad s\leq t,$ (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 $\omega\in\Omega$, 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 $A\in\mathscr{E}$, 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 $\mu\in\mathscr{P}(E)$, let $(\Omega,\mathscr{F},P^{\mu},(B_{t})_{t\in I})$ 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 $\Omega=E^{I}$, all whose paths are continuous rather than merely almost all whose paths are continuous. For a measurable space $(A,\mathscr{A})$ 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 $\mathscr{A}\to\mathscr{B}_{Y}$, and for each $a\in A$, the map $x\mapsto f(x,a)$ is continuous $X\to Y$. It is a fact55 5 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 153, Lemma 4.51. 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 $\mathscr{B}_{X}\otimes\mathscr{A}\to\mathscr{B}_{Y}$, namely it is jointly measurable. $B:I\times\Omega\to E$ is a Carathéodory function. $I=\mathbb{R}_{\geq 0}$, with the subspace topology inherited from $\mathbb{R}$, is a separable metrizable space, and $E=\mathbb{R}^{d}$ 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 $\alpha,\beta,c>0$,

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

and almost every path of $X$ is continuous, then for almost every $\omega\in\Omega$, for every $0<\gamma<\frac{\beta}{\alpha}$ the map $t\mapsto X_{t}(\omega)$ is locally $\gamma$-Hölder continuous: for each $t_{0}\in I$ there is some $0<\epsilon_{t_{0}}<1$ and some $C_{t_{0}}$ such that

 $|X_{t}(\omega)-X_{s}(\omega)|\leq C_{t_{0}}|t-s|^{\gamma},\qquad|s-t_{0}|<% \epsilon_{t_{0}},|t-t_{0}|<\epsilon_{t_{0}}.$

For $\mu\in\mathscr{P}(E)$, let $(\Omega,\mathscr{F},P^{\mu},(B_{t})_{t\in I})$ be the $d$-dimensional Brownian motion with initial distribution $\mu$ formed in Theorem 3. For $s\leq t$, $(B_{t}-B_{s})_{*}P^{\mu}=\nu_{t-s}$, and thus Lemma 2 tells us that for each $n\geq 1$ there is some $C_{n}$ with which $E(|B_{t}-B_{s}|^{2n})=C_{n}(t-s)^{n}$ for all $s. Then $E(|B_{t}-B_{s}|^{2n})\leq C_{n}|t-s|^{n}$ 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 $\frac{\beta_{n-1}}{\alpha_{n-1}}<\gamma_{n}<\frac{\beta_{n}}{\alpha_{n}}$. Let $N_{n}$ be the set of those $\omega\in\Omega$ 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 $\omega\in\Omega\setminus N$ and for any $0<\gamma<\frac{1}{2}$, there is some $\gamma_{n}$ satisfying $\gamma\leq\gamma_{n}<\frac{1}{2}$, 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 $\mu\in\mathscr{P}(E)$ and let $(\Omega,\mathscr{F},P^{\mu},(B_{t})_{t\in I})$ be the $d$-dimensional Brownian motion with initial distribution $\mu$ formed in Theorem 3. For almost all $\omega\in\Omega$, for all $0<\gamma<\frac{1}{2}$, 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 process66 6 See David Applebaum, Lévy Processes and Stochastic Calculus, p. 39, §1.3. if (i) $X_{0}=0$ almost surely, (ii) $X$ has stationary and independent increments, and (iii) for any $a>0$,

 $\lim_{t\downarrow 0}P(|X_{t}|\geq\epsilon)=0.$

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

 $\lim_{s\to t}P(|X_{s}-X_{s}|\geq\epsilon)=0.$ (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.