Wiener measure and Donsker’s theorem

1 Relatively compact sets of Borel probability measures on C[0,1]

Let $E=C[0,1]$, let ${ℬ}_E$ be the Borel $\sigma$-algebra of $E$, and let ${𝒫}_E$ be the collection of Borel probability measures on $E$. We assign $𝒫$ the narrow topology, the coarsest topology on ${𝒫}_E$ such that for each $F\in C_b(E)$ the map $\mu \mapsto {\int }_{E}F𝑑\mu$ is continuous.

For $f\in E$ and $\delta >0$ we define

$${\omega }_f(\delta )=\underset{s,t\in [0,1],|s-t|\le \delta }{sup}|f(s)-f(t)|.$$

For $f\in E$, ${\omega }_f(\delta )\downarrow 0$ as $\delta \downarrow 0$, and for $\delta >0$, $f\mapsto {\omega }_f(\delta )$ is continuous. We shall use the following characterization of a relatively compact subset $A$ of $E$, which is proved using the Arzelà-Ascoli theorem.

Lemma 1.

Let $A$ be a subset of $E$. $\overline{A}$ is compact if and only if

and

$$\underset{f\in A}{sup}{\omega }_f(\delta )\downarrow 0,\delta \downarrow 0.$$

We shall use Prokhorov’s theorem: for $X$ a Polish space and for $\mathrm{\Gamma }\subset {𝒫}_X$, $\overline{\mathrm{\Gamma }}$ is compact if and only if for each $ϵ>0$ there is a compact subset $K_{ϵ}$ of $X$ such that $\mu (K_{ϵ})\ge 1-ϵ$ for all $\mu \in \mathrm{\Gamma }$. Namely, a subset of ${𝒫}_X$ is relatively compact if and only if it is tight. We use Prokhorov’s theorem to prove a characterization of relatively compact subsets of ${𝒫}_E$, which we then use to prove the characterization in Theorem 3.

Lemma 2.

Let $\mathrm{\Gamma }$ be a subset of ${𝒫}_E$. $\overline{\mathrm{\Gamma }}$ is compact if and only if for each $ϵ>0$ there is some and a function $\delta \mapsto {\omega }_{ϵ}(\delta )$ satisfying ${\omega }_{ϵ}(\delta )\downarrow 0$ as $\delta \downarrow 0$ and such that for all $\mu \in \mathrm{\Gamma }$,

$$\mu (A_{ϵ})\ge 1-\frac{ϵ}{2},\mu (B_{ϵ})\ge 1-\frac{ϵ}{2},$$

where

$$A_{ϵ}=\{f\in E:|f(0)|\le M_{ϵ}\},B_{ϵ}=\{f\in E:{\omega }_f(\delta )\le {\omega }_{ϵ}(\delta )\text{ for all }\delta >0\}.$$

Proof.

Suppose that $\mathrm{\Gamma }$ satisfies the above conditions. Because $f\mapsto |f(0)|$ is continuous, $A_{ϵ}$ is closed. For $\delta >0$, suppose that $f_n$ is a sequence in $B_{ϵ}$ tending to some $f\in E$. Because $g\mapsto {\omega }_g(\delta )$ is continuous, ${\omega }_{f_n}(\delta )\to {\omega }_f(\delta )$, and because ${\omega }_{f_n}(\delta )\le {\omega }_{ϵ}(\delta )$ for each $n$, we get ${\omega }_f(\delta )\le {\omega }_{ϵ}(\delta )$ and hence $f\in B_{ϵ}$, showing that $B_{ϵ}$ is closed. Therefore $K_{ϵ}=A_{ϵ}\cap B_{ϵ}$ is closed, i.e. $K_{ϵ}=\overline{K_{ϵ}}$. The set $K_{ϵ}$ satisfies

$$\underset{f\in K_{ϵ}}{sup}|f(0)|\le M_{ϵ}$$

and

$$\underset{\delta \downarrow 0}{lim\; sup}\underset{f\in K_{ϵ}}{sup}{\omega }_f(\delta )\le \underset{\delta \downarrow 0}{lim\; sup}{\omega }_{ϵ}(\delta )=0,$$

thus by Lemma 1, $K_{ϵ}$ is compact. For $\mu \in \mathrm{\Gamma }$,

$$\mu (K_{ϵ})\ge 1-\frac{ϵ}{2},$$

and because $K_{ϵ}$ is compact, this means that $\mathrm{\Gamma }$ is tight, so by Prokhorov’s theorem, $\mathrm{\Gamma }$ is relatively compact.

Now suppose that $\mathrm{\Gamma }$ is relatively compact and let $ϵ>0$. By Prokhorov’s theorem, there is a compact set $K_{ϵ}$ in $E$ such that $\mu (K_{ϵ})\ge 1-\frac{ϵ}{2}$ for all $\mu \in \mathrm{\Gamma }$. Define

$$M_{ϵ}=\underset{f\in K_{ϵ}}{sup}|f(0)|,{\omega }_{ϵ}(\delta )=\underset{f\in K_{ϵ}}{sup}{\omega }_f(\delta ),\delta >0.$$

Because $K_{ϵ}$ is compact, by Lemma 1 we get that and ${\omega }_{ϵ}(\delta )\downarrow 0$ as $\delta \downarrow 0$. For $\mu \in \mathrm{\Gamma }$,

$$\mu (A_{ϵ})\ge \mu (K_{ϵ})\ge 1-\frac{ϵ}{2},\mu (B_{ϵ})\ge \mu (K_{ϵ})\ge 1-\frac{ϵ}{2},$$

showing that $\mathrm{\Gamma }$ satisfies the conditions of the theorem. ∎

We now prove the characterization of relatively compact subsets of ${𝒫}_E$ that we shall use in our proof of Donsker’s theorem.

Theorem 3 (Relatively compact sets in $𝒫$).

Let $\mathrm{\Gamma }$ be a subset of ${𝒫}_E$. $\overline{\mathrm{\Gamma }}$ is compact if and only if the following conditions are satisfied:

1. For each $ϵ>0$ there is some such that

   $$\mu (f:|f(0)|\le M_{ϵ})\ge 1-\frac{ϵ}{2},\mu \in \mathrm{\Gamma }.$$

2. For each $ϵ>0$ and $\delta >0$ there is some $\eta =\eta (ϵ,\delta )>0$ such that

   $$\mu (f:{\omega }_f(\eta )\le \delta )\ge 1-\frac{ϵ}{2},\mu \in \mathrm{\Gamma }.$$

Proof.

Suppose that $\overline{\mathrm{\Gamma }}$ is compact and let $ϵ>0$. By Lemma 2, there is some and a function $\eta \mapsto {\omega }_{ϵ}(\eta )$ satisfying ${\omega }_{ϵ}(\eta )\downarrow 0$ as $\eta \downarrow 0$ and

$$\mu (A_{ϵ})\ge 1-\frac{ϵ}{2},\mu (B_{ϵ})\ge 1-\frac{ϵ}{2},\mu \in \mathrm{\Gamma }.$$

For $\delta >0$, there is some $\eta =\eta (ϵ,\delta )$ with ${\omega }_{ϵ}(\eta )\le \delta$. Then for $\mu \in \mathrm{\Gamma }$,

$$\mu (f:{\omega }_f(\eta )\le \delta )\ge \mu (f:{\omega }_f(\eta )\le {\omega }_{ϵ}(\eta ))\ge \mu (B_{ϵ})\ge 1-\frac{ϵ}{2}.$$

Now suppose that the conditions of the theorem hold. For each $ϵ>0$ and $n\ge 1$ there is some ${\eta }_{ϵ,n}>0$ such that

$$\mu (F_{ϵ,n})\ge 1-\frac{ϵ}{2^{n+1}},\mu \in \mathrm{\Gamma },$$

where

$$F_{ϵ,n}=\{f:{\omega }_f({\eta }_{ϵ,n})\le \frac{1}{n}\}.$$

Let

$$K_{ϵ}=\{f:|f(0)|\le M_{ϵ}\}\cap \bigcap _{n=1}^{\mathrm{\infty }}{F}_{ϵ,n},$$

for which

$$\mu (K_{ϵ})\ge \mu (f:|f(0)|\le M_{ϵ})\ge 1-\frac{ϵ}{2},\mu \in \mathrm{\Gamma }.$$

For $f\in K_{ϵ}$, then for each $n\ge 1$ we have $f\in F_{ϵ,n}$, which means that ${\omega }_f({\eta }_{ϵ,n})\le \frac{1}{n}$, and therefore

$$\underset{f\in K_{ϵ}}{sup}{\omega }_f({\eta }_{ϵ,n})\le \frac{1}{n}.$$

Thus for $n\ge 1$, if then

$$\underset{f\in K_{ϵ}}{sup}{\omega }_f(\eta )\le \frac{1}{n},$$

which shows ${sup}_{f\in K_{ϵ}}{\omega }_f(\eta )\downarrow 0$ as $\eta \downarrow 0$. Then because

$$\underset{f\in K_{ϵ}}{sup}|f(0)|\le M_{ϵ},$$

applying Lemma 1 we get that $\overline{K_{ϵ}}$ is compact. The map $f\mapsto {\omega }_f({\eta }_{ϵ,n})$ is continuous, so the set $F_{ϵ,n}$ is closed, and therefore the set $K_{ϵ}$ is closed. Because $K_{ϵ}$ is compact and $\mu (K_{ϵ})\ge 1-\frac{ϵ}{2}$ for all $\mu \in \mathrm{\Gamma }$, it follows from by Prokhorov’s theorem that $\mathrm{\Gamma }$ is relatively compact. ∎

2 Wiener measure

For $t_1,\mathrm{\dots },t_d\in [0,1]$, , define ${\pi }_{t_1,\mathrm{\dots },t_d}:E\to {ℝ}^d$ by

$${\pi }_{t_1,\mathrm{\dots },t_d}(f)=(f(t_1),\mathrm{\dots },f(t_d)),f\in E,$$

which is continuous. We state the following results, which we will use later.

Theorem 4 (The Borel $\sigma$-algebra of $E$).

${ℬ}_E$ is equal to the $\sigma$-algebra generated by $\{{\pi }_t:t\in [0,1]\}$.

Two elements $\mu$ and $\nu$ of ${𝒫}_E$ are equal if and only if for any $d$ and any , the pushforward measures

$${\mu }_{t_1,\mathrm{\dots },t_d}={({\pi }_{t_1,\mathrm{\dots },t_d})}_{*}\mu ,{\nu }_{t_1,\mathrm{\dots },t_d}={({\pi }_{t_1,\mathrm{\dots },t_d})}_{*}\nu$$

are equal.

Let ${({\xi }_t)}_{t\in [0,1]}$ be a stochastic process with state space $ℝ$ and sample space $(\mathrm{\Omega },ℱ,P)$. For , let ${\xi }_{t_1,\mathrm{\dots },t_d}={\xi }_{t_1}\otimes \mathrm{\cdots }\otimes {\xi }_{t_d}$ and let $P_{t_1,\mathrm{\dots },t_d}={({\xi }_{t_1,\mathrm{\dots },t_d})}_{*}P$: for $B\in {ℬ}_{ℝ}^d$,

$$P_{t_1,\mathrm{\dots },t_d}(B)=({({\xi }_{t_1,\mathrm{\dots },t_d})}_{*}P)(B)=P({\xi }_{t_1,\mathrm{\dots },t_d}^{-1}(B))=P(({\xi }_{t_1},\mathrm{\dots },{\xi }_{t_d})\in B).$$

$P_{t_1,\mathrm{\dots },t_d}$ is a Borel probability measure on ${ℝ}^d$ and is called a finite-dimensional distribution of the stochastic process.

The Kolmogorov continuity theorem tells us that if there are $\alpha ,\beta ,K>0$ such that for all $s,t\in [0,1]$,

$$E{|{\xi }_t-{\xi }_s|}^{\alpha }\le K{|t-s|}^{1+\beta },$$

then there is a unique $\mu \in {𝒫}_E$ such that for all $k$ and for all ,

$${\mu }_{t_1,\mathrm{\dots },t_d}=P_{t_1,\mathrm{\dots },t_d}.$$

We now define and prove the existence of Wiener measure.

Theorem 5 (Wiener measure).

There is a unique Borel probability measure $W$ on $E$ satisfying:

1. $W(f\in E:f(0)=0)=1$.

2. For the random variables

   $${\pi }_{t_1}-{\pi }_{t_0},{\pi }_{t_2}-{\pi }_{t_1},{\pi }_{t_3}-{\pi }_{t_2},{\pi }_{t_d}-{\pi }_{t_{d-1}}$$

   are independent $(E,{ℬ}_E,W)\to (ℝ,{ℬ}_{ℝ})$.

3. If , the random variable ${\pi }_t-{\pi }_s:(E,{ℬ}_E,W)\to (ℝ,{ℬ}_{ℝ})$ is normal with mean $0$ and variance $t-s$.

Proof.

There is a stochastic process ${({\xi }_t)}_{t\in [0,1]}$ with state space $ℝ$ and some sample space $(\mathrm{\Omega },ℱ,P)$, such that (i) $P({\xi }_0=0)=1$, (ii) ${({\xi }_t)}_{t\in [0,1]}$ has independent increments, and (iii) for , ${\xi }_t-{\xi }_s$ is a normal random variable with mean $0$ and variance $t-s$. (Namely, Brownian motion with starting point $\mathrm{0}$.) Because ${\xi }_t-{\xi }_s$ has mean $0$ and variance $t-s$, we calculate (cf. Isserlis’s theorem)

$$E{|{\xi }_t-{\xi }_s|}^4=3{|t-s|}^2.$$

Thus using the Kolmogorov continuity theorem with $\alpha =4$, $\beta =1$, $K=3$, there is a unique $W\in {𝒫}_E$ such that for all ,

$$W_{t_1,\mathrm{\dots },t_d}=P_{t_1,\mathrm{\dots },t_d},$$

i.e. for $B\in {ℬ}_{ℝ}^d$,

$$W({\pi }_{t_1}\otimes \mathrm{\cdots }\otimes {\pi }_{t_d}\in B)=P({\xi }_{t_1}\otimes \mathrm{\cdots }\otimes {\xi }_{t_d}\in B).$$

For and $B\in {ℬ}_{ℝ}^d$, with $T:{ℝ}^d\to {ℝ}^d$ defined by $T(x_1,\mathrm{\dots },x_d)=(x_1,x_2-x_1,\mathrm{\dots },x_d-x_{d-1})$,

Hence, because ${\xi }_{t_1},{\xi }_{t_2}-{\xi }_{t_1},\mathrm{\dots },{\xi }_{t_d}-{\xi }_{t_{d-1}}$ are independent,

which means that the random variables ${\pi }_{t_1},{\pi }_{t_2}-{\pi }_{t_1},\mathrm{\dots },{\pi }_{t_d}-{\pi }_{t_{d-1}}$ are independent.

If and $B_1,B_2\in {ℬ}_{ℝ}$, and for $T:{ℝ}^2\to {ℝ}^2$ defined by $T(x,y)=(x,y-x)$,

$W(({\pi }_s,{\pi }_t-{\pi }_s)\in (B_1,B_2))$ $=W(T\circ ({\pi }_s,{\pi }_t)\in (B_1,B_2))$

$=P(({\xi }_s,{\xi }_t)\in T^{-1}(B_1,B_2))$

$=P(({\xi }_s,{\xi }_t-{\xi }_s)\in (B_1,B_2)),$

which implies that ${({\pi }_t-{\pi }_s)}_{*}W={({\xi }_t-{\xi }_s)}_{*}P$, and because ${\xi }_t-{\xi }_s$ is a normal random variable with mean $0$ and variance $t-s$, so is ${\pi }_t-{\pi }_s$.

Finally,

$$W(f:f(0)=0)=W({\pi }_0=0)=P({\xi }_0=0)=1.$$

∎

$(E,{ℬ}_E,W)$ is a probability space, and the stochastic process ${({\pi }_t)}_{t\in [0,1]}$ is a Brownian motion.

3 Interpolation and continuous stochastic processes

Let ${({\xi }_t)}_{t\in [0,1]}$ be a continuous stochastic process with state space $ℝ$ and sample space $(\mathrm{\Omega },ℱ,P)$. To say that the stochastic process is continuous means that for each $\omega \in \mathrm{\Omega }$ the map $t\mapsto {\xi }_t(\omega )$ is continuous $[0,1]\to ℝ$. Define $\xi :\mathrm{\Omega }\to E$ by

$$\xi (\omega )=(t\mapsto {\xi }_t(\omega )),\omega \in \mathrm{\Omega }.$$

For $t\in [0,1]$ and $B$ a Borel set in $ℝ$,

$${\xi }^{-1}{\pi }_t^{-1}B=\{\omega \in \mathrm{\Omega }:{\xi }_t(\omega )\in B\}={\xi }_t^{-1}B,$$

and because ${\xi }_t:(\mathrm{\Omega },ℱ)\to (ℝ,{ℬ}_{ℝ})$ is measurable this belongs to $ℱ$. But by Theorem 4, ${ℬ}_E$ is generated by the collection $\{{\pi }_t^{-1}B:t\in [0,1],B\in {ℬ}_{ℝ}\}$. Now, for $f:X\to Y$ and for a nonempty collection $ℱ$ of subsets of $Y$,

$$\sigma (f^{-1}(ℱ))=f^{-1}(\sigma (ℱ)).$$

Therefore ${\xi }^{-1}({ℬ}_E)\subset ℱ$, which means that $\xi :(\mathrm{\Omega },ℱ)\to (E,{ℬ}_E)$ is measurable. This means that a continuous stochastic proess with index set $[0,1]$ induces a random variable with state space $E$. Then the pushforward measure of $P$ by $\xi$ is a Borel probability measure on $E$. We shall end up constructing a sequence of pushforward measures from a sequence of continuous stochastic processes, that converge in ${𝒫}_E$ to Wiener measure $W$.

Let ${(X_n)}_{n\ge 1}$ be a sequence of independent identically distributed random variables on a sample space $(\mathrm{\Omega },ℱ,P)$ with $E(X_n)=0$ and $V(X_n)=1$, and let $S_0=0$ and

$$S_k=\sum _{i=1}^k{X}_i.$$

Then $E(S_k)=0$ and $V(S_k)=k$. For $t\ge 0$ let

$$Y_t=S_{[t]}+(t-[t])X_{[t]+1}.$$

Thus, for $k\ge 0$ and $k\le t\le k+1$,

$Y_t$ $=S_k+(t-k)X_{k+1}$

$=S_k+(t-k)(S_{k+1}-S_k)$

$=(1-t+k)S_k+(t-k)S_{k+1}.$

For each $\omega \in \mathrm{\Omega }$, the map $t\mapsto Y_t(\omega )$ is piecewise linear, equal to $S_k(\omega )$ when $t=k$, and in particular it is continuous. For $n\ge 1$, define

$$X_t^{(n)}=n^{-1/2}{Y}_{nt}=n^{-1/2}{S}_{[nt]}+n^{-1/2}(nt-[nt])X_{[nt]+1},t\in [0,1].$$ (1)

For $0\le k\le n$,

$$X_{k/n}^{(n)}=n^{-1/2}{S}_k.$$

For each $n\ge 1$, ${(X_t^{(n)})}_{t\in [0,1]}$ is a continuous stochastic process on the sample space $(\mathrm{\Omega },ℱ,P)$, and we denote by $P_n\in {𝒫}_E$ the pushforward measure of $P$ by $X^{(n)}$.

4 Donsker’s theorem

Lemma 6.

If $Z_n$ and $U_n$ are random variables with state space ${ℝ}^d$ such that $Z_n\to Z$ in distribution and $U_n\to 0$ in distribution, then $Z_n+U_n\to 0$ in distribution.

If $Z_n$ are random variables with state space $ℝ$ that converge in distribution to some random variable $Z$ and $c_n$ are real numbers that converge to some real number $c$, then $c_n{Z}_n\to cZ$ in distribution.

For $\sigma \ge 0$, let ${\nu }_{{\sigma }^2}$ be the Gaussian measure on $ℝ$ with mean $0$ and variance ${\sigma }^2$. The characteristic function of ${\nu }_{{\sigma }^2}$ is, for $\sigma >0$,

$${\tilde{\nu }}_{{\sigma }^2}(\xi )={\int }_{ℝ}{e}^{i\xi x}𝑑{\nu }_{{\sigma }^2}(x)={\int }_{ℝ}{e}^{i\xi x}\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{x^2}{2{\sigma }^2}}𝑑x=e^{-\frac{1}{2}{\sigma }^2{\xi }^2},$$

and ${\tilde{\nu }}_0(\xi )=1$. One checks that $c_{*}{\nu }_1={\nu }_{c^2}$ for $c\ge 0$.

In following theorem and in what follows, $X^{(n)}$ is the piecewise linear stochastic process defined in (1). We prove that a sequence of finite-dimensional distributions converge to a Gaussian measure.

Theorem 7.

For , the random vectors

$$(X_{t_1}^{(n)}-X_{t_0}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}-X_{t_{d-1}}^{(n)}),(\mathrm{\Omega },ℱ,P)\to ({ℝ}^d,{ℬ}_{ℝ}^d),$$

converge in distribution to ${\nu }_{t_1-t_0}\otimes \mathrm{\cdots }\otimes {\nu }_{t_d-t_{d-1}}$ as $n\to \mathrm{\infty }$.

Proof.

For and $n\ge 1$ let

$$r_{j,n}=\frac{[n{t}_j]}{n},U_{j,n}=X_{t_j}^{(n)}-X_{r_{j,n}}^{(n)},$$

and for and $n\ge 1$ let

$$s_{j,n}=\frac{\lceil n{t}_j\rceil }{n},V_{j,n}=X_{s_{j,n}}^{(n)}-X_{t_j}^{(n)},$$

with which

$(X_{t_1}^{(n)}-X_{t_0}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}-X_{t_{d-1}}^{(n)})$ $=(X_{r_{1,n}}^{(n)}-X_{s_{0,n}}^{(n)},\mathrm{\dots },X_{r_{d,n}}^{(n)}-X_{s_{d-1,n}}^{(n)})$

$+(U_{1,n},\mathrm{\dots },U_{d,n})+(V_{0,n},\mathrm{\dots },V_{d-1,n}).$

Because $E(X_t^{(n)})=0$,

$$E(U_{j,n})=0,E(V_{j,n})=0.$$

Furthermore,

and because this tends to $0$ as $n\to \mathrm{\infty }$. Likewise, $V(V_{j,n})\to 0$ as $n\to \mathrm{\infty }$.

For $1\le j\le d$,

$X_{r_{j,n}}^{(n)}-X_{s_{j-1,n}}^{(n)}$ $=n^{-1/2}{S}_{[n{r}_{j,n}]}+n^{-1/2}(n{r}_{j,n}-[n{r}_{j,n}])X_{[n{r}_{j,n}]+1}$

$-n^{-1/2}{S}_{[n{s}_{j-1,n}]}-n^{-1/2}(n{s}_{j-1,n}-[n{s}_{j-1,n}])X_{[n{s}_{j-1,n}]+1}$

$=n^{-1/2}{S}_{[n{t}_j]}-n^{-1/2}{S}_{\lceil n{t}_{j-1}\rceil }$

$=n^{-1/2}{\displaystyle \frac{{([n{t}_j]-\lceil n{t}_{j-1}\rceil -1)}^{1/2}}{{([n{t}_j]-\lceil n{t}_{j-1}\rceil -1)}^{1/2}}}{\displaystyle \sum _{i=\lceil n{t}_{j-1}\rceil +1}^{[n{t}_j]}}{X}_i.$

By the central limit theorem,

$${([n{t}_j]-\lceil n{t}_{j-1}\rceil -1)}^{1/2}\sum _{i=\lceil n{t}_{j-1}\rceil +1}^{[n{t}_j]}{X}_i\to {\nu }_1$$

in distribution as $n\to \mathrm{\infty }$. But

$$n^{-1/2}{([n{t}_j]-\lceil n{t}_{j-1}\rceil -1)}^{1/2}\to {(t_j-t_{j-1})}^{1/2}$$

as $n\to \mathrm{\infty }$, and ${(t_j-t_{j-1})}_{*}^{1/2}{\nu }_1={\nu }_{t_j-t_{j-1}}$, so by Lemma 6,

$$X_{r_{j,n}}^{(n)}-X_{s_{j-1,n}}^{(n)}\to {\nu }_{t_j-t_{j-1}}$$

in distribution as $n\to \mathrm{\infty }$.

For sufficiently large $n$, depending on $t_0,\mathrm{\dots },t_d$,

Check that $(U_{1,n},\mathrm{\dots },U_{d,n})\to 0$ in probability and that $(V_{0,n},\mathrm{\dots },V_{d-1,n})\to 0$ in probability, and hence these random vectors converge to $0$ in distribution as $n\to \mathrm{\infty }$. The random variables $X_{r_{1,n}}^{(n)}-X_{s_{0,n}}^{(n)},\mathrm{\dots },X_{r_{d,n}}^{(n)}-X_{s_{d-1,n}}^{(n)}$ are independent, and therefore their joint distribution is equal to the product of their distributions. Now, if ${\mu }_n={\mu }_n^1\otimes \mathrm{\cdots }\otimes {\mu }_n^d$ and ${\mu }_n^j\to {\mu }^j$ as $n\to \mathrm{\infty }$, $1\le j\le d$, then for $\xi \in {ℝ}^d$,

${\tilde{\mu }}_n(\xi )$ $={\tilde{\mu }}_n^1({\xi }_1)\mathrm{\cdots }{\tilde{\mu }}_n^d({\xi }_d)$

$\to {\tilde{\mu }}^1({\xi }_1)\mathrm{\cdots }{\tilde{\mu }}^d({\xi }_d)$

$={({\mu }^1\otimes \mathrm{\cdots }\otimes {\mu }^d)}^{\stackrel{~}{}}(\xi )$

as $n\to \mathrm{\infty }$, and therefore by Lévy’s continuity theorem, ${\mu }_n\to {\mu }^1\otimes \mathrm{\cdots }\otimes {\mu }^d$ as $n\to \mathrm{\infty }$. This means that the joint distribution of $X_{r_{1,n}}^{(n)}-X_{s_{0,n}}^{(n)},\mathrm{\dots },X_{r_{d,n}}^{(n)}-X_{s_{d-1,n}}^{(n)}$ converges to

$${\nu }_{t_1-t_0}\otimes \mathrm{\cdots }\otimes {\nu }_{t_d-t_{d-1}}$$

as $n\to \mathrm{\infty }$. Because $(U_{1,n},\mathrm{\dots },U_{d,n})\to 0$ in distribution as $n\to \mathrm{\infty }$ and $(V_{0,n},\mathrm{\dots },V_{d-1,n})\to 0$ in distribution as $n\to \mathrm{\infty }$, applying Lemma 6 we get that

$$(X_{t_1}^{(n)}-X_{t_0}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}-X_{t_{d-1}}^{(n)})\to {\nu }_{t_1-t_0}\otimes \mathrm{\cdots }\otimes {\nu }_{t_d-t_{d-1}}$$

in distribution as $n\to \mathrm{\infty }$, completing the proof. ∎

Let $t_0=0$ and let . As $X_0^{(n)}=0$, the above lemma tells us that

$$(X_{t_1}^{(n)},X_{t_2}^{(n)}-X_{t_1}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}-X_{t_{d-1}}^{(n)})\to {\nu }_{t_1}\otimes {\nu }_{t_2-t_1}\otimes \mathrm{\cdots }\otimes {\nu }_{t_d-t_{d-1}}$$

in distribution as $n\to \mathrm{\infty }$. Define $g:{ℝ}^d\to {ℝ}^d$ by

$$g(x_1,x_2,\mathrm{\dots },x_d)=(x_1,x_1+x_2,\mathrm{\dots },x_1+x_2+\mathrm{\cdots }+x_d).$$

The function $g$ is continuous and satisfies

$$g\circ (X_{t_1}^{(n)}-X_{t_0}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}-X_{t_{d-1}}^{(n)})=(X_{t_1}^{(n)},X_{t_2}^{(n)},\mathrm{\dots },X_{t_d}^{(n)}).$$

Then by the continuous mapping theorem,

$$(X_{t_1}^{(n)},X_{t_2}^{(n)},\mathrm{\dots },X_{t_d}^{(n)})\to g_{*}({\nu }_{t_1}\otimes {\nu }_{t_2-t_1}\otimes \mathrm{\cdots }\otimes {\nu }_{t_d-t_{d-1}})$$ (2)

in distribution as $n\to \mathrm{\infty }$.

We prove a result that we use to prove the next lemma, and that lemma is used in the proof of Donsker’s theorem.

Lemma 8.

For $ϵ>0$,

$$\underset{\delta \downarrow 0}{lim}\underset{n\to \mathrm{\infty }}{lim\; sup}\frac{1}{\delta }P(\underset{1\le j\le [n\delta ]+1}{\max }|S_j|>ϵn^{1/2})=0.$$

Proof.

For each $\delta >0$, by the central limit theorem,

$${([n\delta ]+1)}^{-1/2}{S}_{[n\delta ]+1}\to Z$$

in distribution as $n\to \mathrm{\infty }$, where $Z_{*}P={\nu }_1$. Because $\frac{{([n\delta ]+1)}^{1/2}}{{(n\delta )}^{1/2}}\to 1$ as $n\to \mathrm{\infty }$, by Lemma 6 we then get that

$${(n\delta )}^{-1/2}{S}_{[n\delta ]+1}\to Z$$

in distribution as $n\to \mathrm{\infty }$. Now let $\lambda >0$, and there is a sequence ${\varphi }_k$ in $C_b(ℝ)$ such that ${\varphi }_k\downarrow 1_{(-\mathrm{\infty },-\lambda ]\cup [\lambda ,\mathrm{\infty })}={\chi }_{\lambda }$ pointwise as $k\to \mathrm{\infty }$. For each $k$, writing $X=S_{[n\delta ]+1}$, using the change of variables formula,

$P(|X|\ge \lambda {(n\delta )}^{1/2})$ $={\displaystyle {\int }_{\mathrm{\Omega }}}{\chi }_{\lambda {(n\delta )}^{1/2}}(X(\omega ))𝑑P(\omega )$

$={\displaystyle {\int }_{\mathrm{\Omega }}}{\chi }_{\lambda }({(n\delta )}^{-1/2}X(\omega ))𝑑P(\omega )$

$\le {\displaystyle {\int }_{\mathrm{\Omega }}}{\varphi }_k({(n\delta )}^{-1/2}X(\omega ))𝑑P(\omega )$

$=E({\varphi }_k({(n\delta )}^{-1/2}X)).$

Therefore, by the continuous mapping theorem,

$\underset{n\to \mathrm{\infty }}{lim\; sup}P(|S_{[n\delta ]+1}|\ge \lambda {(n\delta )}^{1/2})$ $\le \underset{n\to \mathrm{\infty }}{lim}E({\varphi }_k({(n\delta )}^{-1/2}{S}_{[n\delta ]+1}))$

$=E({\varphi }_k\circ Z).$

Because ${\varphi }_k\downarrow {\chi }_{\lambda }$ pointwise as $k\to \mathrm{\infty }$, using the monotone convergence theorem and then using Chebyshev’s inequality,

$$E({\varphi }_k\circ Z)\to E({\chi }_{\lambda }\circ Z)=P(|Z|\ge \lambda )\le {\lambda }^{-3}E|Z{|}^3.$$

We have established that for each $\lambda >0$,

$$\underset{n\to \mathrm{\infty }}{lim\; sup}P(|S_{[n\delta ]+1}|\ge \lambda {(n\delta )}^{1/2})\le {\lambda }^{-3}E|Z{|}^3.$$ (3)

Define

$$\tau =\min \{j\ge 1:|S_j|>n^{1/2}ϵ\}.$$

For , it is a fact that

If $\tau (\omega )=j$ and then

$$|S_j(\omega )-S_{[n\delta ]+1}(\omega )|\ge |S_j(\omega )|-|S_{[n\delta ]+1}(\omega )|>n^{1/2}ϵ-n^{1/2}(ϵ-{(2\delta )}^{1/2})={(2n\delta )}^{1/2}.$$

But by Chebyshev’s inequality and the fact that the random variables $X_1,X_2,\mathrm{\dots }$ are independent with mean $0$ and variance $1$,

$$P(|S_j-S_{[n\delta ]+1}|>{(2n\delta )}^{1/2})\le \frac{1}{2n\delta }E({(S_j-S_{[n\delta ]+1})}^2)=\frac{1}{2n\delta }([n\delta ]-j)\le \frac{1}{2},$$

so

Therefore,

so

$$P(\underset{0\le j\le [n\delta ]+1}{\max }|S_j|>n^{1/2}ϵ)\le 2P(|S_{[n\delta ]+1}|\ge n^{1/2}(ϵ-{(2\delta )}^{1/2})).$$

Now using (3) with $\lambda =(ϵ-{(2\delta )}^{1/2}){\delta }^{-1/2}$,

$$\underset{n\to \mathrm{\infty }}{lim\; sup}P(|S_{[n\delta ]+1}|\ge (ϵ-{(2\delta )}^{1/2}){\delta }^{-1/2}{(n\delta )}^{1/2})\le {(ϵ-{(2\delta )}^{1/2})}^{-3}{\delta }^{3/2}E|Z{|}^3,$$

hence

$$\underset{n\to \mathrm{\infty }}{lim\; sup}P(\underset{0\le j\le [n\delta ]+1}{\max }|S_j|>n^{1/2}ϵ)\le 2{(ϵ-{(2\delta )}^{1/2})}^{-3}{\delta }^{3/2}E|Z{|}^3.$$

Dividing both sides by $\delta$ and then taking $\delta \downarrow 0$ we obtain the claim. ∎

We prove one more result that we use to prove Donsker’s theorem.

Lemma 9.

For $T>0$ and $ϵ>0$,

$$\underset{\delta \downarrow 0}{lim}\underset{n\to \mathrm{\infty }}{lim\; sup}P(\underset{0\le k\le [nT]+1}{\max }\underset{1\le j\le [n\delta ]+1}{\max }|S_{j+k}-S_k|>n^{1/2}ϵ)=0.$$

Proof.

For , let $m=\lceil T/\delta \rceil$, so . Then

so for all $n\ge n_{\delta }$ it is the case that . Suppose that $\omega \in \mathrm{\Omega }$ is such that there are $1\le j\le [n\delta ]+1$ and $0\le k\le [nT]+1$ satisfying

$$|S_{j+k}(\omega )-S_k(\omega )|>n^{1/2}ϵ,$$

and then let $p=[k/([n\delta ]+1)]$, which satisfies $0\le p\le m-1$ and

Because $1\le j\le [n\delta ]+1$, either

or

We separate the first case into the cases

$$|S_k(\omega )-S_{([n\delta ]+1)p}(\omega )|>\frac{1}{2}{n}^{1/2}ϵ$$

and

$$|S_{j+k}(\omega )-S_{([n\delta ]+1)p}(\omega )|>\frac{1}{2}{n}^{1/2}ϵ,$$

and we separate the second case into the cases

$$|S_k-S_{([n\delta ]+1)p}(\omega )|>\frac{1}{3}{n}^{1/2}ϵ,$$

and

$$|S_{([n\delta ]+1)p}(\omega )-S_{([n\delta ]+1)(p+1)}(\omega )|>\frac{1}{3}{n}^{1/2}ϵ,$$

and

$$|S_{([n\delta ]+1)(p+1)}(\omega )-S_{([n+\delta ]+1)(p+2)}(\omega )|>\frac{1}{3}{n}^{1/2}ϵ.$$

It follows that

For $0\le p\le m-1$,

so

Lemma 8 tells us

$$\underset{\delta \downarrow 0}{lim}\underset{n\to \mathrm{\infty }}{lim\; sup}\frac{1}{\delta }P(\underset{1\le j\le [n\delta ]+1}{\max }|S_j|>\frac{1}{3}{n}^{1/2}ϵ)=0,$$

and because $m\le \frac{T}{\delta }+1=\frac{T+\delta }{\delta }$,

$$\underset{\delta \downarrow 0}{lim}\underset{n\to \mathrm{\infty }}{lim\; sup}P\{\underset{1\le j\le [n\delta ]+1}{\max }\underset{0\le k\le [nT]+1}{\max }|S_{j+k}-S_k|>n^{1/2}ϵ\}=0,$$

proving the claim. ∎

In the following, $P_n\in {𝒫}_E$ denotes the pushforward measure of $P$ by $X^{(n)}$, for $X^{(n)}$ defined in (1). We now prove Donsker’s theorem.

Theorem 10 (Donsker’s theorem).

$P_n\to W$.

Proof.

We shall use Theorem 3 to prove that $\mathrm{\Gamma }=\{P_n:n\ge 1\}$ is relatively compact in ${𝒫}_E$. For $n\ge 1$,

$$P_n(f\in E:|f(0)|=0)=P(\omega \in \mathrm{\Omega }:|X_0^{(n)}(\omega )|=0)=1,$$

thus the first condition of Theorem 3 is satisfied with $M_{ϵ}=0$. For the second condition of Theorem 3 to be satisfied it suffices that for each $ϵ>0$,

$$\underset{\delta \downarrow 0}{lim}\underset{n\to \mathrm{\infty }}{lim\; sup}P(\underset{0\le s,t\le 1,|s-t|\le \delta }{sup}|X^{(n)}(s)-X^{(n)}(t)|>ϵ)=0.$$

Now,

$$P(\underset{0\le s,t\le 1,|s-t|\le \delta }{sup}|X_s^{(n)}-X_t^{(n)}|>ϵ)=P(\underset{0\le s,t\le n,|s-t|\le n\delta }{sup}|Y_s-Y_t|>n^{1/2}ϵ).$$

Also,

$\underset{0\le s,t\le n,|s-t|\le n\delta }{sup}|Y_s-Y_t|$ $\le \underset{0\le s,t\le n,|s-t|\le n\delta }{sup}|Y-s-Y_t|$

$\le \underset{1\le j\le [n\delta ]+1}{\max }\underset{0\le k\le n+1}{\max }|S_{j+k}-S_k|,$

so applying Lemma 9,

from which we get that $\mathrm{\Gamma }$ is tight in ${𝒫}_E$. ∎