Markov kernels, convolution semigroups, and projective families of probability measures

1 Transition kernels

For a measurable space $(E,ℰ)$, we denote by ${ℰ}_{+}$ the set of functions $E\to [0,\mathrm{\infty }]$ that are $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$ measurable. It can be proved that if $I:{ℰ}_{+}\to [0,\mathrm{\infty }]$ is a function such that (i) $f=0$ implies that $I(f)=0$, (ii) if $f,g\in {ℰ}_{+}$ and $a,b\ge 0$ then $I(af+bg)=aI(f)+bI(g)$, and (iii) if $f_n$ is a sequence in ${ℰ}_{+}$ that increases pointwise to an element $f$ of ${ℰ}_{+}$ then $I(f_n)$ increases to $I(f)$, then there a unique measure $\mu$ on $ℰ$ such that $I(f)=\mu f$ for each $f\in {ℰ}_{+}$.

Let $(E,ℰ)$ and $(F,ℱ)$ be a measurable space. A transition kernel is a function

$$K:E\times ℱ\to [0,\mathrm{\infty }]$$

such that (i) for each $x\in E$, the function $K_x:ℱ\to [0,\mathrm{\infty }]$ defined by

$$B\mapsto K(x,B)$$

is a measure on $ℱ$, and (ii) for each $B\in ℱ$, the map

$$x\mapsto K(x,B)$$

is measurable $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$.

If $\mu$ is a measure on $ℰ$, define

$$(K_{*}\mu )(B)={\int }_{E}K(x,B)𝑑\mu (x),B\in ℱ.$$

If $B_n$ are pairwise disjoint elements of $ℱ$, then using that $B\mapsto K(x,B)$ is a measure and the monotone convergence theorem,

$(K_{*}\mu )\left({\displaystyle \bigcup _n}{B}_n\right)$ $={\displaystyle {\int }_E}K(x,{\displaystyle \bigcup _n}{B}_n)𝑑\mu (x)$

$={\displaystyle {\int }_E}{\displaystyle \sum _n}K(x,B_n)d\mu (x)$

$={\displaystyle \sum _n}{\displaystyle {\int }_E}K(x,B_n)𝑑\mu (x)$

$={\displaystyle \sum _n}(K_{*}\mu )(B_n),$

showing that $K_{*}\mu$ is a measure on $ℱ$.

If $f\in {ℱ}_{+}$, define $K^{*}f:E\to [0,\mathrm{\infty }]$ by

$$(K^{*}f)(x)={\int }_{F}f(y)𝑑K_x(y),x\in E.$$ (1)

For $\varphi ={\sum }_{j=1}^k{b}_j{1}_{B_j}$ with $b_j\ge 0$ and $B_j\in ℱ$, because $x\mapsto K(x,B_j)$ is measurable $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$ for each $j$,

$$(K^{*}\varphi )(x)={\int }_F\sum _{j=1}^k{b}_j{1}_{B_j}(y)d{K}_x(y)=\sum _{j=1}^k{b}_j{K}_x(B_j)=\sum _{j=1}{b}_{j}K(x,B_j),$$

is measurable $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$. For $f\in {ℱ}_{+}$, there is a sequence of simple functions ${\varphi }_n$ with $0\le {\varphi }_1\le {\varphi }_2\le \mathrm{\cdots }$ that converges pointwise to $f$, and then by the monotone convergence theorem, for each $x\in E$ we have

$$(K^{*}{\varphi }_n)(x)={\int }_F{\varphi }_n(y)𝑑K_x(y)\to {\int }_{F}f(y)𝑑K_x(y)=(K^{*}f)(x),$$

showing $K^{*}{\varphi }_n$ converges pointwise to $K^{*}f$, and because each $K^{*}{\varphi }_n$ is measurable $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$, $K^{*}f$ is measurable $ℰ\to {ℬ}_{[0,\mathrm{\infty }]}$. Therefore, if $f\in {ℱ}_{+}$ then $K^{*}f\in {ℰ}_{+}$. In particular, if $K$ is a transition kernel from $(E,ℰ)$ to $(F,ℱ)$,

$$(K^{*}{1}_B)(x)={\int }_F{1}_B(y)𝑑K_x(y)=K_x(B)=K(x,B),x\in E,B\in ℱ.$$ (2)

The following gives conditions under which (2) defines a transition kernel.

Lemma 1.

Suppose that $N:{ℱ}_{+}\to {ℰ}_{+}$ satisfies the following properties:

1. $N(0)=0$.

2. $N(af+bg)=aN(f)+bN(g)$ for $f,g\in {ℱ}_{+}$ and $a,b\ge 0$.

3. If $f_n$ is a sequence in ${ℱ}_{+}$ increasing to $f\in {ℱ}_{+}$, then $N(f_n)\uparrow N(f)$.

Then

$$K(x,B)=(N(1_B))(x),x\in E,B\in ℱ,$$

is a transition kernel from $(E,ℰ)$ to $(F,ℱ)$. $K$ is the unique transition kernel satisfying

$$K^{*}f=N(f),f\in {ℱ}_{+}.$$

If $K$ is a transition kernel from $(E,ℰ)$ to $(F,ℱ)$ and $L$ is a transition kernel from $(F,ℱ)$ to $(G,𝒢)$, the function $K^{*}\circ L^{*}:{𝒢}_{+}\to {ℰ}_{+}$ satisfies (i) $(K^{*}\circ L^{*})(0)=K^{*}(0)=0$, (ii) if $f,g\in {𝒢}_{+}$ and $a,b\ge 0$,

$(K^{*}\circ L^{*})(af+bg)$ $=K^{*}(a{L}^{*}(f)+b{L}^{*}(g))$

$=a{K}^{*}(L^{*}(f))+K^{*}(L^{*}(g))$

$=a(K^{*}\circ L^{*})(f)+b(K^{*}\circ L^{*})(g),$

and (iii) if $f_n\uparrow f$ in ${𝒢}_{+}$, then by the monotone convergence theorem, $L^{*}(f_n)\uparrow L^{*}(f)$, and then again applying the monotone convergence theorem, $K^{*}(L^{*}(f_n))\uparrow K^{*}(L^{*}(f))$, i.e.

$$(K^{*}\circ L^{*})(f_n)\uparrow (K^{*}\circ L^{*})(f).$$

Therefore, from Lemma 1 we get that there is a unique transition kernel from $(E,ℰ)$ to $(G,𝒢)$, denoted $KL$ and called the product of $K$ and $L$, such that

$${(KL)}^{*}f=(K^{*}\circ L^{*})(f),f\in {𝒢}_{+}.$$

For $f\in {𝒢}_{+}$ and $x\in E$,

${(KL)}^{*}(f)(x)$ $=(K^{*}(L^{*}f))(x)$

$={\displaystyle {\int }_F}(L^{*}f)(y)𝑑K_x(y)$

$={\displaystyle {\int }_F}\left({\displaystyle {\int }_G}f(z)𝑑L_y(z)\right)𝑑K_x(y).$

In particular, for $C\in 𝒢$,

$${(KL)}^{*}(1_C)(x)={\int }_F{L}_y(C)𝑑K_x(y)={\int }_{F}L(y,C)𝑑K_x(y).$$ (3)

2 Markov kernels

A Markov kernel from $(E,ℰ)$ to $(F,ℱ)$ is a transition kernel $K$ such that for each $x\in E$, $K_x$ is a probability measure on $ℱ$. The unit kernel from $(E,ℰ)$ to $(E,ℰ)$ is

$$I(x,A)={\delta }_x(A).$$ (4)

It is apparent that the unit kernel is a Markov kernel.

If $K$ is a Markov kernel from $(E,ℰ)$ to $(F,ℱ)$ and $L$ is a Markov kernel from $(F,ℱ)$ to $(G,𝒢)$, then for $x\in E$, by (3) we have

$${(KL)}^{*}(1_G)(x)={\int }_F𝑑K_x(y)=K_x(F)=K(x,F)=1,$$

and thus by (2),

$${(KL)}_x(G)=(KL)(x,G)=1,$$

showing that for each $x\in E$, ${(KL)}_x$ is a probability measure. Therefore, the product of two Markov kernels is a Markov kernel.

Let $(E,ℰ)$ be a measurable space and let

$$B_b(ℰ)$$

be the set of bounded functions $E\to ℝ$ that are measurable $ℰ\to {ℬ}_{ℝ}$. $B_b(ℰ)$ is a Banach space with the uniform norm

$${\parallel f\parallel }_u=\underset{x\in E}{sup}|f(x)|.$$

For $K$ a Markov kernel from $(E,ℰ)$ to $(F,ℱ)$ and for $f\in B_b(ℱ)$, define $K^{*}f:E\to ℝ$ by

$$(K^{*}f)(x)={\int }_{F}f(y)𝑑K_x(y),x\in E,$$

for which

$$|(K^{*}f)(x)|\le {\int }_F|f(y)|𝑑K_x(y)\le {\parallel f\parallel }_u{K}_x(F)={\parallel f\parallel }_u,$$

showing that ${\parallel K^{*}f\parallel }_u\le {\parallel f\parallel }_u$. Furthermore, there is a sequence of simple functions ${\varphi }_n\in B_b(ℱ)$ that converges to $f$ in the norm $\parallel \cdot {\parallel }_u$. For $x\in E$, by the dominated convergence theorem we get that

$$(K^{*}{\varphi }_n)(x)={\int }_F{\varphi }_n(y)𝑑K_x(y)\to {\int }_{F}f(y)𝑑K_x(y)=(K^{*}f)(x).$$

Each $K^{*}{\varphi }_n$ is measurable $ℰ\to {ℬ}_{ℝ}$, hence $K^{*}f$ is measurable $ℰ\to {ℬ}_{ℝ}$ and so belongs to $B_b(ℰ)$.

3 Markov semigroups

Let $(E,ℰ)$ be a measurable space and for each $t\ge 0$, let $P_t$ be a Markov kernel from $(E,ℰ)$ to $(E,ℰ)$. We say that the family ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ is a Markov semigroup if

$$P_{s+t}=P_s{P}_t,s,t\in {ℝ}_{\ge 0}.$$

For $x\in E$ and $A\in ℰ$ and for $s,t\ge 0$, by (2) and (3),

$$(P_s{P}_t)(x,A)=({(P_s{P}_t)}^{*}{1}_A)(x)={\int }_E{P}_t(y,A)d{(P_s)}_x(y)$$

Thus

$$P_{s+t}(x,A)={\int }_E{P}_t(y,A)d{(P_s)}_x(y),$$ (5)

called the Chapman-Kolmogorov equation.

4 Infinitely divisible distributions

Let $𝒫({ℝ}^d)$ be the collection of Borel probability measures on ${ℝ}^d$. For $\mu \in 𝒫({ℝ}^d)$, its characteristic function $\tilde{\mu }:{ℝ}^d\to ℂ$ is defined by

$$\tilde{\mu }(x)={\int }_{{ℝ}^d}{e}^{i⟨x,y⟩}𝑑\mu (y).$$

$\tilde{\mu }$ is uniformly continuous on ${ℝ}^d$ and $|\tilde{\mu }(x)|\le \tilde{\mu }(0)=1$ for all $x\in {ℝ}^d$. For ${\mu }_1,\mathrm{\dots },{\mu }_n\in 𝒫({ℝ}^d)$, let $\mu$ be their convolution:

$$\mu ={\mu }_1*\mathrm{\cdots }*{\mu }_n,$$

which for $A$ a Borel set in ${ℝ}^d$ is defined by

$$\mu (A)={\int }_{{({ℝ}^d)}^n}{1}_A(x_1+\mathrm{\cdots }+x_n)d({\mu }_1\times \mathrm{\cdots }\times {\mu }_n)(x_1,\mathrm{\dots },x_n).$$

One computes that

$$\tilde{\mu }={\tilde{\mu }}_1\mathrm{\cdots }{\tilde{\mu }}_n.$$

An element $\mu$ of $𝒫({ℝ}^d)$ is called infinitely divisible if for each $n\ge 1$, there is some ${\mu }_n\in 𝒫({ℝ}^d)$ such that

$$\mu =\underset{n}{\underbrace{{\mu }_n*\mathrm{\cdots }*{\mu }_n}}.$$ (6)

Thus,

$$\tilde{\mu }={({\tilde{\mu }}_n)}^n.$$ (7)

On the other hand, if ${\mu }_n\in 𝒫({ℝ}^d)$ is such that (7) is true, then because the characteristic function of ${\mu }_n*\mathrm{\cdots }*{\mu }_n$ is ${({\tilde{\mu }}_n)}^n$ and the characteristic function of $\mu$ is $\tilde{\mu }$ and these are equal, it follows that ${\mu }_n*\mathrm{\cdots }*{\mu }_n$ and $\mu$ are equal.

The following theorem is useful for doing calculations with the characteristic function of an infinitely divisible distribution.

Theorem 2.

Suppose that $\mu$ is an infinitely divisible distribution on ${ℝ}^d$. First,

$$\tilde{\mu }(x)\ne 0,x\in {ℝ}^d.$$

Second, there is a unqiue continuous function $\varphi :{ℝ}^d\to ℝ$ satisfying $\varphi (0)=0$ and

$$\tilde{\mu }=|\tilde{\mu }|e^{i\varphi }.$$

Third, for each $n\ge 1$, there is a unique ${\mu }_n\in 𝒫({ℝ}^d)$ for which $\mu ={\mu }_n*\mathrm{\cdots }*{\mu }_n$. The characteristic function of this unique ${\mu }_n$ is

$${\tilde{\mu }}_n={|\tilde{\mu }|}^{\frac{1}{n}}{e}^{i\frac{\varphi }{n}}.$$

A convolution semigroup is a family ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ of elements of $𝒫({ℝ}^d)$ such that for $s,t\in {ℝ}_{\ge 0}$,

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

The convolution semigroup is called continuous when $t\mapsto {\mu }_t$ is continuous ${ℝ}_{\ge 0}\to 𝒫({ℝ}^d)$, where $𝒫({ℝ}^d)$ has the narrow topology.

The following theorem connects convolution semigroups and infinitely divisible distributions.

Theorem 3.

If ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a convolution semigroup on ${ℬ}_{{ℝ}^d}$, then for each $t$, the measure ${\mu }_t$ is infinitely divisible.

If $\mu \in 𝒫({ℝ}^d)$ is infinitely divisible and $t_0>0$, then there is a unique continuous convolution semigroup ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ such that ${\mu }_{t_0}=\mu$.

It follows from the above theorem that for a convolution semigroup ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ on ${ℬ}_{{ℝ}^d}$, ${\mu }_1$ is infinitely divisible and therefore by Theorem 2, ${\tilde{\mu }}_1(x)\ne 0$ for all $x$. But ${\mu }_0*{\mu }_1={\mu }_1$, so ${\tilde{\mu }}_0{\tilde{\mu }}_1={\tilde{\mu }}_1$, and ${\tilde{\mu }}_0(x)=1$ for each $x$. But ${\tilde{\delta }}_0(x)=1$ for all $x$, so

$${\mu }_0={\delta }_0.$$ (8)

5 Translation-invariant semigroups

Let ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ be a Markov semigroup on $({ℝ}^d,{ℬ}_{{ℝ}^d})$. We say that ${(P_t)}_{t\in ℝ}$ is translation-invariant if for all $x,y\in {ℝ}^d$, $A\in {ℬ}_{{ℝ}^d}$, and $t\in {ℝ}_{\ge 0}$,

$$P_t(x,A)=P_t(x+y,A+y).$$

In this case, for $t\ge 0$ and for $A\in {ℬ}_{{ℝ}^d}$, define

$${\mu }_t(A)=P_t(0,A).$$

Each ${\mu }_t$ is a probability measure on ${ℬ}_{{ℝ}^d}$, and

$${\mu }_t(A-x)=P_t(0,A-x)=P_t(x,(A-x)+x)=P_t(x,A).$$

Using that the Chapman-Kolmogorov equation (5) and as ${(P_s)}_0(B)=P_s(0,B)={\mu }_s(B)$,

${\mu }_{s+t}(A)$ $=P_{s+t}(0,A)$

$={\displaystyle {\int }_{{ℝ}^d}}{P}_t(y,A)d{(P_s)}_0(y)$

$={\displaystyle {\int }_{{ℝ}^d}}{\mu }_t(A-y)𝑑{\mu }_s(y)$

$=({\mu }_t*{\mu }_s)(A),$

showing that ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a convolution semigroup on ${ℬ}_{{ℝ}^d}$.

On the other hand, if ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a convolution semigroup of probability measures on ${ℬ}_{{ℝ}^d}$, for $t\ge 0$, $x\in {ℝ}^d$, and $A\in {ℬ}_{{ℝ}^d}$ define

$$P_t(x,A)={\mu }_t(A-x).$$

Let $t\ge 0$. For $x\in {ℝ}^d$, the map $A\mapsto P_t(x,A)={\mu }_t(A-x)$ is a probability measure on ${ℬ}_{{ℝ}^d}$. The map $(x,y)\mapsto x+y$ is continuous ${ℝ}^d\times {ℝ}^d\to {ℝ}^d$, and for $A\in {ℬ}_{{ℝ}^d}$, the map $1_A:{ℝ}^d\to ℝ$ is measurable ${ℬ}_{{ℝ}^d}\to {ℬ}_{ℝ}$. Hence, as ${ℬ}_{{ℝ}^d\times {ℝ}^d}={ℬ}_{{ℝ}^d}\otimes {ℬ}_{{ℝ}^d}$, the map $(x,y)\mapsto 1_A(x+y)$ is measurable ${ℬ}_{{ℝ}^d}\otimes {ℬ}_{{ℝ}^d}\to {ℬ}_{ℝ}$. Thus by Fubini’s theorem,

$$x\mapsto {\int }_{{ℝ}^d}{1}_A(x+y)𝑑{\mu }_t(y)={\int }_{{ℝ}^d}{1}_{A-x}(y)𝑑{\mu }_t(y)={\mu }_t(A-x)$$

is measurable ${ℬ}_{{ℝ}^d}\to {ℬ}_{ℝ}$. Hence $P_t$ is a Markov kernel, and thus ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ is a translation-invariant Markov semigroup.

Define $S:{ℝ}^d\to {ℝ}^d$ by $S(x)=-x$. For $\mu ,\nu \in 𝒫({ℝ}^d)$,

$S_{*}(\mu *\nu )(A)$ $=(\mu *\nu )(-A)$

$={\displaystyle {\int }_{{ℝ}^d}}\mu (-A-y)𝑑\nu (y)$

$={\displaystyle {\int }_{{ℝ}^d}}\mu (-A+y)𝑑\overline{\nu }(y)$

$={\displaystyle {\int }_{{ℝ}^d}}\overline{\mu }(A-y)𝑑\overline{\nu }(y)$

$=(\overline{\mu }*\overline{\nu })(A),$

thus

$$S_{*}(\mu *\nu )=(S_{*}\mu )*(S_{*}\nu ).$$ (9)

For $\mu \in 𝒫({ℝ}^d)$, write

$$\overline{\mu }=S_{*}\mu \in 𝒫({ℝ}^d),$$

i.e.,

$$\overline{\mu }(A)=\mu (S^{-1}(A))=\mu (S(A))=\mu (-A).$$

We calculate

$$(P_t^{*}{1}_A)(x)=P_t(x,A)={\mu }_t(A-x)={\int }_{{ℝ}^d}{1}_A(x+y)𝑑{\mu }_t(y).$$

Then if $f$ is a simple function, $f={\sum }_k{a}_k{1}_{A_k}$,

$$(P_t^{*}f)(x)=\sum _k{a}_k{\int }_{{ℝ}^d}{1}_{A_k}(x+y)𝑑{\mu }_t(y)={\int }_{{ℝ}^d}f(x+y)𝑑{\mu }_t(y).$$

For $f\in B_b({ℬ}_{{ℝ}^d})$, there is a sequence of simple functions $f_n$ that converge to $f$ in the uniform norm, and then by the dominated convergence theorem we get

$$(P_t^{*}f)(x)={\int }_{{ℝ}^d}f(x+y)𝑑{\mu }_t(y).$$

But

${\displaystyle {\int }_{{ℝ}^d}}f(x+y)𝑑{\mu }_t(y)$ $={\displaystyle {\int }_{{ℝ}^d}}f(x+S(S(y)))𝑑{\mu }_t(y)$

$={\displaystyle {\int }_{{ℝ}^d}}f(x+S(y))d(S_{*}{\mu }_t)(y)$

$={\displaystyle {\int }_{{ℝ}^d}}f(x-y)𝑑{\overline{\mu }}_t(y)$

$=(f*{\overline{\mu }}_t)(x).$

Therefore for $t\ge 0$ and $f\in B_b({ℬ}_{{ℝ}^d})$,

$$P_t^{*}f=f*{\overline{\mu }}_t.$$ (10)

For $s,t\ge 0$ and $f\in B_b({ℬ}_{{ℝ}^d})$, by (10), the fact that ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a convolution semigroup, and (9), we get

$P_{s+t}^{*}f$ $=f*(S_{*}{\mu }_{s+t})$

$=f*(S_{*}({\mu }_s*{\mu }_t))$

$=f*((S_{*}{\mu }_s)*(S_{*}{\mu }_t))$

$=(f*(S_{*}{\mu }_s))*(S_{*}{\mu }_t)$

$=(P_s^{*}f)*(S_{*}{\mu }_t)$

$=P_t^{*}(P_s^{*}f).$

This shows that ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ is a Markov semigroup. Moreover, by (8) it holds that ${\mu }_0={\delta }_0$, and hence

$$P_0(x,A)={\mu }_0(A-x)={\delta }_0(A-x)={\delta }_x(A).$$

Namely, $P_0$ is the unit kernel (4).

If ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a convolution semigroup and some ${\mu }_t$ has density $q_t$ with respect to Lebesgue measure ${\lambda }_d$ on ${ℝ}^d$,

$${\mu }_t=q_t{\lambda }_d,$$

then writing ${\overline{q}}_t(x)=q_t(-x)$, for $f\in B_b({ℬ}_{{ℝ}^d})$ by (10) we have

$$(P_t^{*}f)(x)=(f*{\overline{\mu }}_t)(x)={\int }_{{ℝ}^d}f(x-y)𝑑{\overline{\mu }}_t(y)={\int }_{{ℝ}^d}f(x+y)q_t(y)𝑑{\lambda }_d(y)$$

so

$$P_t*f=f*{\overline{q}}_t.$$ (11)

6 The Brownian semigroup

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 $t\in {ℝ}_{\ge 0}$,

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

which is an element of $𝒫({ℝ}^d)$. For $s,t\in {ℝ}_{\ge 0}$, we calculate

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

Lévy’s continuity theorem states that if ${\nu }_n$ is a sequence in $𝒫({ℝ}^d)$ and there is some $\varphi :{ℝ}^d\to ℂ$ that is continuous at $0$ and to which ${\tilde{\nu }}_n$ converges pointwise, then there is some $\nu \in 𝒫({ℝ}^d)$ such that $\varphi =\tilde{\nu }$ and such that ${\nu }_n\to \nu$ narrowly. But for $t\in {ℝ}_{\ge 0}$ and $x\in {ℝ}^d$, we calculate

$${\tilde{\mu }}_t(x)={\int }_{{ℝ}^d}{e}^{i⟨x,y⟩}𝑑{\mu }_t(y)=\exp \left(-\frac{t{|x|}^2}{2}\right).$$ (12)

Let $\varphi (x)=1$ for all $x$, for which ${\tilde{\delta }}_0=\varphi$. For $t_n\in {ℝ}_{\ge 0}$ tending to $0$, let ${\nu }_n={\mu }_{t_n}$. Then by (12), ${\tilde{\nu }}_n$ converges pointwise to $\varphi$, so by Lévy’s continuity theorem, ${\nu }_n$ converges narrowly to ${\delta }_0$. Moreover, because ${ℝ}^d$ is a Polish space, $𝒫({ℝ}^d)$ is a Polish space, and in particular is metrizable. It thus follows that ${\mu }_t$ converges narrowly to ${\delta }_0$ as $t\to 0$. It then follows that $t\mapsto {\mu }_t$ is continuous ${ℝ}_{\ge 0}\to 𝒫({ℝ}^d)$. Summarizing, ${({\mu }_t)}_{t\in {ℝ}_{\ge 0}}$ is a continuous convolution semigroup.

For $t>0$, ${\mu }_t$ has density

$$g_t(x)=\prod _{j=1}^d{(2\pi t)}^{-1/2}{e}^{-\frac{x_j^2}{2t}}={(2\pi t)}^{-d/2}{e}^{-\frac{{|x|}^2}{2t}}$$

with respect to Lebesgue measure ${\lambda }_d$ on ${ℝ}^d$. For $t\ge 0$, let

$$P_t(x,A)={\mu }_t(A-x).$$

We have established that ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ is a translation-invariant Markov semigroup for which $P_0(x,A)={\delta }_x(A)$. We call ${(P_t)}_{t\in {ℝ}_{\ge 0}}$ the Brownian semigroup. For $t>0$ and $f\in B_b({ℬ}_{{ℝ}^d})$, because ${\overline{g}}_t=g_t$ we have by (11),

$$(P_{t}f)(x)=(f*g_t)(x)={(2\pi t)}^{-d/2}{\int }_{{ℝ}^d}f(x-y)e^{-\frac{{|y|}^2}{2t}}𝑑{\lambda }_d(y).$$

7 Projective families

For a nonempty set $I$, let $𝒦(I)$ denote the family of finite nonempty subsets of $I$. We speak in this section about projective families of probability measures.

The following theorem shows how to construct a projective family from a Markov semigroup on a measurable space and a probability measure on this measurable space.

Theorem 4.

Let $I={ℝ}_{\ge 0}$, let $(E,ℰ)$ be a measurable space, let ${(P_t)}_{t\in I}$ be a Markov semigroup on $ℰ$, and let $\mu$ be a probability measure on $ℰ$. For $J\in 𝒦(I)$, with elements , and for $A\in {ℰ}^J$, let

$$P_J(A)=\underset{n+1}{\underbrace{{\int }_E{\int }_E\mathrm{\cdots }{\int }_E}}{1}_A(x_1,\mathrm{\dots },x_n)d{(P_{t_n-t_{n-1}})}_{x_{n-1}}(x_n)\mathrm{\cdots }d{(P_{t_1})}_{x_0}(x_1)d\mu (x_0).$$

Then ${(P_J)}_{J\in 𝒦(I)}$ is a projective family of probability measures.

Proof.

Let $A_k$ be pairwise disjoint elements of ${ℰ}^J$, and call their union $A$. Then $1_A={\sum }_k{1}_{A_k}$, and applying the monotone convergence theorem $n+1$ times,

i.e.

$$P_J(A)=\sum _k{P}_J(A_k).$$

Furthermore, because ${(P_t)}_x$ is a probability measure for each $t$ and for each $x$ and $\mu$ is a probability measure, we calculate that

$$P_J(E^J)=1.$$

Thus, $P_J$ is a probability measure on ${ℰ}^J$.

To prove that ${(P_J)}_{J\in 𝒦(I)}$ is a projective family, it suffices to prove that when $J,K\in 𝒦(I)$, $J\subset K$, and $K\setminus J$ is a singleton, then ${({\pi }_{K,J})}_{*}{P}_K=P_J$. Moreover, because (i) the product $\sigma$-algebra ${ℰ}^J$ is generated by the collection of cylinder sets, i.e. sets of the form ${\prod }_{t\in J}{A}_t$ for $A_t\in ℰ$, and (ii) the intersection of finitely many cylinder sets is a cylinder sets, it is proved using the monotone class theorem that if two probability measures on ${ℰ}^J$ coincide on the cylinder sets, then they are equal. Let be the elements of $J$. To prove that ${({\pi }_{K,J})}_{*}{P}_K$ and $P_J$ are equal, it suffices to prove that for any $A_1,\mathrm{\dots },A_n\in ℰ$,

$${({\pi }_{K,J})}_{*}{P}_K\left(\prod _{j=1}^n{A}_j\right)=P_J\left(\prod _{j=1}^n{A}_j\right).$$

Moreover, for $A={\prod }_{j=1}^n{A}_j$,

$$1_A=1_{A_1}\otimes \mathrm{\cdots }\otimes 1_{A_n},$$

thus

Let $K\setminus J=\{t^{\prime }\}$. Either , or $t^{\prime }>t_n$, or there is some $1\le j\le n-1$ for which . Take the case . Then

$${\pi }_{K,J}^{-1}\left(\prod _{j=1}^n{A}_j\right)=\prod _{k=0}^n{B}_k,$$

where $B_0=E$ and $B_j=A_j$ for $1\le j\le n$. Then

for

$$f(x_1)={\int }_{A_2}\mathrm{\cdots }{\int }_{A_n}d{(P_{t_n-t_{n-1}})}_{x_{n-1}}(x_n)\mathrm{\cdots }d{(P_{t_2-t_1})}_{x_1}(x_2).$$

By (1) and because ${(P_t)}_{t\in I}$ is a Markov semigroup,

Thus

This shows that the claim is true in the case . ∎

Thus, if $E$ is a Polish space with Borel $\sigma$-algebra $ℰ$, let $I={ℝ}_{\ge 0}$, let ${(P_t)}_{t\in I}$ be a Markov semigroup on $ℰ$, and let $\mu$ be a probability measure on $ℰ$. The above theorem tells us that ${(P_J)}_{𝒦(I)}$ is a projective family, and then the Kolmogorov extension theorem tells us that there is a probability measure $P^{\mu }$ on ${ℰ}^I$ such that for any $J\in 𝒦(I)$, $\pi _J{}_{*}{P}^{\mu }=P_J^{\mu }$. This implies that there is a stochastic process ${(X_t)}_{t\in I}$ whose finite-dimensional distributions are equal to the probability measures $P_J$ defined in Theorem 4 using the Markov semigroup ${(P_t)}_{t\in I}$ and the probability measure $\mu$.