2 The expected value as an integral over a sequence space

In this section we show that the expected value in Theorem 1.1 can be expressed as an integral over a sequence space. We remark that we will sometimes find it convenient to use interchangeably the language of measure and integration or alternatively that of probability and expectation.

Recall that the Lévy–Ciesielski expansion (1.1) expresses the Brownian path $B(t)$ in terms of an infinite sequence ${\mathbf {X}}(\omega )=(X_0, (X_{n,i})_{n\in \mathbb {N}, i=1,\ldots , 2^{n-1}})$ of independent standard normal random variables. In the following we will denote a particular realisation of this sequence ${\mathbf {X}}$ by

$$ \begin{align*} {\mathbf{x}} = (x_0,(x_{n,i})_{n\in\mathbb{N}, i=1,\ldots,2^{n-1}}) = (x_0,x_1,x_2,x_3,\ldots) \in\mathbb{R}^\infty, \end{align*} $$

where we will switch freely between the double-index labelling $(x_0,x_{1,1}, x_{2,1}, x_{2,2},\ldots )$ and a single-index labelling $(x_0,x_1,x_2,x_3,\ldots )$ as appropriate, with the indexing convention that $x_{n,i}$ becomes $x_{2^{n-1}-1+i}$ for $n\ge 1$ and $1\le i\le 2^{n-1}$ .

It is clear from (1.1) that, for $t\in [0,1]$ and a fixed $\omega \in \Omega $ ,

(2.1) $$ \begin{align} | B_{N}(t)| & \le | X_0| + \sum_{n=1}^N \Big(\max_{1\le i\le 2^{n-1}} | X_{n,i}|\Big) \sum_{i=1}^{2^{n-1}}\eta_{n,i}(t) \le | X_0| + \sum_{n=1}^N \max_{1\le i\le 2^{n-1}} | X_{n,i}| 2^{-(n+1)/2}, \end{align} $$

where in the last step we used the fact that for a given $n\geq 1$ the disjoint nature of the Faber–Schauder functions ensures that at most one value of i contributes to the sum over i, and also that the $\eta _{n,i}$ for $i=1,\ldots ,2^{n-1}$ have the same maximum value $2^{-(n+1)/2}$ .

Motivated by the bound (2.1), and following [Reference Griebel, Kuo and Sloan3], we define a norm of the sequence ${\mathbf {x}} = (x_0,(x_{n,i})_{n\in \mathbb {N}, i=1,\ldots ,2^{n-1}})$ by

$$ \begin{align*} \|{\mathbf{x}}\|_{\mathcal{X}} := |x_0|+ \sum_{n=1}^\infty \max_{1\leq i \leq 2^{n-1}} |x_{n,i}|\, 2^{-(n+1)/2}, \end{align*} $$

and we define a corresponding normed space by $\mathcal {X} := \{ {\mathbf {x}}\in \mathbb {R}^\infty : \|{\mathbf {x}}\|_{\mathcal {X}} < \infty \}$ . It is easily seen that $\mathcal {X}$ is a Banach space.

Each choice of ${\mathbf {x}}\in \mathcal {X}$ corresponds to a particular $\omega \in \Omega $ (but not vice versa, since there are sample points $\omega \in \Omega $ corresponding to sequences ${\mathbf {x}}$ for which the norm $\|{\mathbf {x}}\|_{\mathcal {X}}$ is not finite). Hence, to each ${\mathbf {x}}\in \mathcal {X}$ there corresponds a particular Brownian path via (1.1), or expressed in terms of ${\mathbf {x}}$ ,

(2.2) $$ \begin{align} B({\mathbf{x}})(t) = x_0\, \eta_0(t) +\sum_{n=1}^\infty \sum_{i=1}^{2^{n-1}} x_{n,i}\, \eta_{n,i}(t),\quad t\in[0,1]. \end{align} $$

That the resulting path is continuous on $[0,1]$ follows from the observation that the path is the pointwise limit of the truncated series

(2.3) $$ \begin{align} B_N({\mathbf{x}})(t) \,=\,x_0\, \eta_0(t) +\sum_{n=1}^N \sum_{i=1}^{2^{n-1}} x_{n,i}\, \eta_{n,i}(t),\quad t\in[0,1], \end{align} $$

which is uniformly convergent since

$$ \begin{align*} \|B_N\|_\infty \,\le\, |x_0|+ \sum_{n=1}^\infty \max_{1\leq i \leq 2^{n-1}} |x_{n,i}| 2^{-(n+1)/2} = \|{\mathbf{x}}\|_{\mathcal{X}} \,<\, \infty \quad\mbox{for}\ {\mathbf{x}}\in\mathcal{X}, \end{align*} $$

so that (2.2) does indeed define a continuous function for ${\mathbf {x}}\in \mathcal {X}$ .

We define $\mathcal {A}_{\mathbb {R}^\infty }$ to be the $\sigma $ -algebra generated by products of Borel sets of $\mathbb {R}$ (see [Reference Bogachev1, page 372]). On the Banach space $\mathcal {X}$ , we now define a product Gaussian measure $\rho (d{\mathbf {x}}) := \otimes _{j=0}^\infty \phi (x_j) d x_j$ (see [Reference Bogachev1, page 392 and Example 2.35]), where $\phi $ is the standard normal probability density $\phi (x) := \exp (- x^2/2)/\sqrt {2\pi }$ .

We next show that the space $\mathcal {X}$ has full Gaussian measure, that is,

$$ \begin{align*} \mathbb{P}\bigg(|X_0|+ \sum_{n=1}^\infty \max_{1\leq i\leq 2^{n-1}} |X_{n,i}| 2^{-(n+1)/2}<\infty\bigg) =1. \end{align*} $$

This fact is the basis of the classical proof that the Lévy–Ciesielski construction almost surely converges uniformly to the Brownian path. For a brief explanation, we define

$$ \begin{align*} H_n(\omega) := \begin{cases} |X_0(\omega)| & \mbox{for } n=0,\\ \max_{1\leq i\leq 2^{n-1}}|X_{n,i}(\omega)| 2^{-(n+1)/2} & \mbox{for } n \geq 1. \end{cases} \end{align*} $$

As a consequence of the Borel–Cantelli lemma, one can construct a sequence $(\beta _n)_{n \geq 1}$ of positive numbers such that

$$ \begin{align*} \sum_{n=1}^\infty \beta_n <\infty \quad\mbox{and}\quad \mathbb{P}(H_n(\cdot)> \beta_n \mbox{ infinitely often}) =0. \end{align*} $$

We now define $\tilde \Omega $ to be the subset of $\Omega $ consisting of the sample points $\omega $ for which $H_n(\omega )> \beta _n$ for only finitely many values of n. Then $\tilde \Omega $ is of full Gaussian measure and for each $\omega \in \tilde \Omega $ there exists $N(\omega ) \in \mathbb {N}$ such that $H_n(\omega ) \leq \beta _n$ for $n> N(\omega )$ , giving

$$ \begin{align*} \sum_{n=1}^{\infty} H_n(\omega) \le \sum_{n=1}^{N(\omega)} H_n(\omega) + \sum_{n=N(\omega)+1}^\infty \beta_n < \infty \quad \mbox{for}\; \omega \in \tilde \Omega. \end{align*} $$

Thus, $\mathbb {P}(\sum _{n=0}^\infty H_n <\infty )=1$ as claimed and $\mathcal {X}$ is of full Gaussian measure.

We now study integration on the measure space $(\mathcal {X},\mathcal {A}_{\mathbb {R}^\infty },\rho )$ and we denote the integral, or the expected value, of a measurable function f by ${\mathbb {E}[f] := \int _{\mathcal {X}} f({\mathbf {x}}) \rho (d{\mathbf {x}})}$ .

3 Expected uniform error of standard Brownian motion

We devote this section to proving Theorem 1.1. We have, from (1.1) and (1.2),

$$ \begin{align*} |B(t) - B_N(t)| = \bigg|\sum_{n=N+1}^\infty \sum_{i=1}^{2^{n-1}} X_{n,i} \eta_{n,i}(t) \bigg| \le \sum_{n=N+1}^\infty \bigg(\max_{1\le i\le 2^{n-1}} |X_{n,i}|\bigg) \sum_{i=1}^{2^{n-1}} \eta_{n,i}(t). \end{align*} $$

Using the same disjoint support argument as in (2.1), we conclude that

$$ \begin{align*} \|B - B_N\|_\infty \le \sum_{n=N+1}^\infty \max_{1\le i\le 2^{n-1}} |X_{n,i}| 2^{-(n+1)/2} = \sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{M_\ell}{2\sqrt{\ell}}, \end{align*} $$

where we introduced new random variables

$$ \begin{align*} M_\ell:= M_{2^{n-1}} := \max_{1\le i \le 2^{n-1}}|X_{n,i}| \quad\mbox{for }\ell = 2^{n-1}\mbox{ and }n\ge 1. \end{align*} $$

Thus,

(3.1) $$ \begin{align} \mathbb{E}[ \|B - B_N\|_\infty ] \le \sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{\mathbb{E}[M_\ell ]}{2\sqrt{\ell}}, \end{align} $$

and since $M_\ell $ and $M_{\ell '}$ are independent random variables for $\ell \ne \ell '$ ,

(3.2) $$ \begin{align} \mathbb{E}[ \|B - B_N\|_\infty^2 ] &\le \sum_{\ell,\ell' = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{\mathbb{E} [M_\ell M_{\ell'}]}{2\sqrt{\ell}\cdot 2\sqrt{\ell'}} \nonumber\\ &= \sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{\mathbb{E}[M_\ell^2 ]}{2\sqrt{\ell}\cdot 2\sqrt{\ell}} + \sum_{\stackrel{\scriptstyle{\ell,\ell' = 2^{N}, 2^{N+1},2^{N+2},\ldots}}{\scriptstyle{\ell\ne\ell'}}} \frac{\mathbb{E}[M_\ell]\, \mathbb{E}[M_{\ell'}]} {2\sqrt{\ell}\cdot 2\sqrt{\ell'}}. \end{align} $$

Now we are in the territory of extreme value statistics. It is known that the distribution function of the maximum of the absolute value of $\ell $ independent and identical Gaussian random variables converges (after appropriate centring and scaling, as below) to the Gumbel distribution. A first step is to obtain an explicit expression for the distribution function of $M_\ell $ . Because $X_{n,1}, X_{n,2}, \ldots ,X_{n,\ell }$ are $\mathcal {N}(0,1)$ random variables, for $x\in \mathbb {R}^+$ and $i=1,\ldots ,\ell $ , we have

$$ \begin{align*} \mathbb{P}(X_{n,i}\le x) = \int_{-\infty}^x \phi(t)\,d t =: \Phi(x), \end{align*} $$

where $\phi $ is the standard normal density. Similarly,

$$ \begin{align*} \mathbb{P}(|X_{n,i}|\le x) = \int_{-x}^x \phi(t)\,d t = \Phi(x) - \Phi(-x) = 2\Phi(x) - 1. \end{align*} $$

Therefore (since $X_{n,1}, X_{n,2}, \ldots ,X_{n,\ell }$ are independent random variables),

$$ \begin{align*} \mathbb{P}(M_\ell \le x) &= \mathbb{P}(|X_{n,1}|\le x\ \mbox{and}\ |X_{n,2}|\le x \ \mbox{and } \cdots \mbox{ and}\ |X_{n,\ell}|\le x) = (2\Phi(x) - 1)^\ell. \end{align*} $$

Thus, the distribution function of $M_\ell $ is

(3.3) $$ \begin{align} \Psi_\ell(x) := (2\Phi(x) - 1)^\ell,\quad x\in \mathbb{R^+}. \end{align} $$

We now define a new random variable $Y_\ell $ for $\ell \ge 1$ , which is a recentred and rescaled version of $M_\ell $ :

(3.4) $$ \begin{align} Y_\ell := \frac{M_\ell-a_\ell}{b_\ell},\quad \mbox{or equivalently,}\quad M_\ell = a_\ell + b_\ell Y_\ell, \quad a_\ell>0, b_\ell >0. \end{align} $$

It is known (see below) to be appropriate to take $a_\ell $ and $b_\ell $ to satisfy

(3.5) $$ \begin{align} a_\ell =\sqrt{2\ln \ell}+o(1) \quad\mbox{and}\quad b_\ell=\frac{1}{a_\ell}. \end{align} $$

More precisely, for later convenience we will define $a_\ell $ to be the unique solution of

(3.6) $$ \begin{align} \frac{1}{\ell} = \sqrt{\frac{2}{\pi}}\frac{e^{-a_\ell^2/2}}{a_\ell} =: g(a_\ell). \end{align} $$

We now show that (3.6) implies (3.5).

Proof. The fact that any solution of (3.6) is positive is immediate. Now observe that g in (3.6) is monotonically decreasing from $+\infty $ to $0$ on $\mathbb {R}^+$ . It follows immediately that there is a unique solution $a_\ell \in (0,\infty )$ for (3.6). Moreover,

$$ \begin{align*} a_\ell> 1 \Leftrightarrow \frac{1}{\ell} < \sqrt {\frac 2 \pi }\frac {e^{-1^2/2}} {1} = \sqrt {\frac 2 {\pi e} } =0.484\ldots, \end{align*} $$

which holds if and only if $\ell \ge 3$ . Now observe that (3.6) is equivalent to

(3.7) $$ \begin{align} a_\ell = \sqrt{ 2\left(\ln \ell - \ln \left(\sqrt {\frac \pi 2} a_\ell\right)\right)}. \end{align} $$

For $\ell \ge 3$ we have $a_\ell> 1$ and hence $\ln (\sqrt {\pi /2}\,a_\ell )>\ln (\sqrt {\pi /2})>0$ , so from (3.7) we have $a_\ell < \sqrt { 2 \ln \ell }$ . In turn it follows that

$$ \begin{align*} a_\ell> \sqrt{2 \ln \ell - 2 \ln\bigg(\sqrt {\frac \pi 2 }\sqrt{2 \ln \ell}\bigg)}. \end{align*} $$

Thus, for $\ell \ge 3$ we have $1\le a_\ell =\sqrt {2\ln \ell } +o(1) \le \sqrt {2\ln {\ell }}$ , completing the proof.

It is well known that the distribution function of $Y_\ell $ converges in distribution to a random variable with the Gumbel distribution $\exp (-e^{-y})$ . For later convenience we state this as a lemma and give a short proof.

Proof. The proof is based on the asymptotic version of Mill’s ratio [Reference Small7],

$$ \begin{align*} 1-\Phi(x)\sim \frac{\phi(x)}{x},\quad x\to +\infty, \end{align*} $$

where, as in the Introduction, $\sim $ means that the quotient of the two sides converges to $1$ . From this it follows that for $y\in \mathbb {R}$ ,

$$ \begin{align*} \mathbb{P}(Y_\ell\le y) &=\mathbb{P}(M_\ell\le a_\ell+b_\ell y)\,=\,(2\Phi(a_\ell+b_\ell y)-1)^\ell =(1-2[1-\Phi(a_\ell+b_\ell y)])^\ell\\ &\sim\bigg(1-\sqrt{\frac{2}{\pi}} \frac{\exp(-\tfrac{1}{2}(a_\ell + b_\ell y)^2)}{a_\ell+b_\ell y}\bigg)^\ell \sim\bigg(1-\sqrt{\frac{2}{\pi}} \frac{\exp(-\tfrac{1}{2}a_\ell^2 - a_\ell b_\ell y)}{a_\ell}\bigg)^\ell \\ &=(1- \exp(-y)/\ell)^\ell \sim\exp(-e^{-y}) \quad \mbox{as} \ \ell\to\infty, \end{align*} $$

where in the second step we dropped a higher-order term, and in the second last step we used (3.6) and $a_\ell b_\ell = 1$ , thus proving the lemma.

A deeper result, which we need, is that $Y_\ell $ converges in expectation to the limit Y. This is proved in the following lemma.

Lemma 3.3. The random variable $Y_\ell $ defined in (3.4), with $a_\ell $ defined by (3.6) and $b_\ell = 1/a_\ell $ , converges in expectation to a random variable Y with Gumbel distribution $\exp (-e^{-y})$ , thus

$$ \begin{align*} \lim_{\ell\to\infty}\mathbb{E}[Y_\ell] = \mathbb{E}[Y] = \int_{-\infty}^\infty y \exp(-y-e^{-y})\,d y = \gamma, \end{align*} $$

where $\gamma $ is Euler’s constant.

Proof. For a sequence of real-valued random variables $Y_1,Y_2,\ldots $ converging in distribution to a random variable Y, it is well known that a sufficient condition for convergence in expectation is uniform integrability of the $Y_\ell $ . In turn a sufficient condition for uniform integrability is that for sufficiently large $\ell $ ,

$$ \begin{align*} &\mathbb{P}(Y_\ell \ge y) \le Q(y) \mbox{ for } y> 0 \quad\mbox{and}\quad \mathbb{P}(Y_\ell \le y) \le R(y) \mbox{ for } y < 0, \end{align*} $$

where $Q(y)$ is integrable on $\mathbb {R}^+$ and $R(y)$ is integrable on $\mathbb {R}^-$ .

First assume $y>0$ . From (3.3),

$$ \begin{align*} \mathbb{P}(Y_\ell\ge y) &= \mathbb{P}(M_\ell\ge a_\ell + b_\ell y) = 1- \mathbb{P}(M_\ell\le a_\ell + b_\ell y) \\ &= 1- (2\Phi(a_\ell + b_\ell y)-1)^\ell = 1- (1 - 2[1-\Phi(a_\ell + b_\ell y)])^\ell \\ &\le 1-\bigg(1-2\frac{\phi(a_\ell + b_\ell y)}{a_\ell + b_\ell y}\bigg)^\ell =1-\bigg(1-\sqrt{\frac{2}{\pi}} \frac{\exp(-\tfrac{1}{2}(a_\ell + b_\ell y)^2)}{a_\ell + b_\ell y}\bigg)^\ell\\ &\le 1-\bigg(1-\sqrt{\frac{2}{\pi}}\frac{\exp(-\tfrac{1}{2}a_\ell^2 - a_\ell b_\ell y)} {a_\ell}\bigg)^\ell, \end{align*} $$

where we used the upper bound form of Mills’s ratio [Reference Small7],

$$ \begin{align*} 1-\Phi(x)< \frac{\phi(x)}{x},\quad x\in\mathbb{R}^+, \end{align*} $$

and dropped harmless terms in both the denominator and the exponent in the numerator. Using now (3.6) and also $a_\ell b_\ell = 1$ ,

(3.8) $$ \begin{align} \mathbb{P}(Y_\ell\ge y) &\le 1-\bigg(1-\frac{\exp(-y)}{\ell}\bigg)^\ell \le \exp(-y) =: Q(y), \end{align} $$

where we used the fact that the function $(1-c/x)^x$ is increasing on $[1,\infty )$ for $c\in [0,1]$ , and hence takes its minimum at $x=1$ . It follows that

$$ \begin{align*} \int_0^\infty \mathbb{P}(Y_\ell\ge y)\,d y \le \int_0^\infty \exp(-y)\,d y = 1. \end{align*} $$

Now we consider $y<0$ . Note first that $M_\ell =a_\ell +b_\ell Y_\ell $ takes only nonnegative values, thus we may restrict y to $y\ge -a_\ell /b_\ell $ . We have

$$ \begin{align*} \mathbb{P}(Y_\ell\le y) &=\mathbb{P}(M_\ell\le a_\ell + b_\ell y) = (2\Phi(a_\ell + b_\ell y)-1)^\ell. \end{align*} $$

Now for $t>0$ the standard normal distribution $\Phi $ has negative second derivative,

$$ \begin{align*} \Phi"(t)=\phi'(t) <0\quad\mbox{for } t>0, \end{align*} $$

and first derivative $\Phi '(t)=\phi (t)$ , from which it follows that

$$ \begin{align*} \Phi(a_\ell + b_\ell y)&\le \Phi(a_\ell) + b_\ell y\,\phi(a_\ell) \quad \mbox{for } y\ge -a_\ell/b_\ell. \end{align*} $$

Thus, on using $b_\ell =1/a_\ell $ , we obtain

$$ \begin{align*} \mathbb{P}(Y_\ell\le y) &\le(2\Phi(a_\ell)+2a_\ell^{-1}y\phi(a_\ell)-1)^\ell\\ &\le(1-2a_\ell^{-1}\phi(a_\ell)(1-y-a_\ell^{-2}))^\ell =\bigg(1-\frac{1}{\ell}(1-y-a_\ell^{-2})\bigg)^\ell, \end{align*} $$

where in the second step we used the lower bound form of Mills’s ratio (see [Reference Small7, page 44]),

$$ \begin{align*} 1-\Phi(t)\ge \frac{\phi(t)}{t}(1-t^{-2})\quad\mbox{for } t>0, \end{align*} $$

and in the last step we used (3.6). If we now take $\ell \ge L$ then

(3.9) $$ \begin{align} \mathbb{P}(Y_\ell\le y) &\le\bigg(1-\frac{1}{\ell}(1-y-a_L^{-2})\bigg)^\ell \nonumber\\ &\le \exp(-(1-y-a_L^{-2})) =\exp(-(1-a_L^{-2})) \exp(y) =: R(y), \end{align} $$

since the convergence in the last limit is monotone increasing. The function $R(y)$ so defined is integrable on $\mathbb {R}^-$ , completing the proof that $Y_\ell $ converges in expectation.

It then follows from Lemma 3.2 that the limit of $\mathbb {E}[Y_\ell ]$ is precisely $\mathbb {E}[Y] = \gamma $ .

Since Lemma 3.3 establishes the convergence of $\mathbb {E}[Y_\ell ]$ as $\ell \to \infty $ , it can be inferred that there exists a positive constant c such that

(3.10) $$ \begin{align} \mathbb{E}[Y_\ell]\le c \quad\mbox{and hence}\quad \mathbb{E}[M_\ell] \le a_\ell + b_\ell c \le a_\ell + c, \end{align} $$

where we used $b_\ell = a_\ell ^{-1} \le 1$ for $\ell \ge 3$ . We then conclude from (3.1) that

(3.11) $$ \begin{align} \mathbb{E}[ \|B - B_N\|_\infty ] &\le \sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{a_\ell+c}{2\sqrt{\ell}}. \end{align} $$

It only remains to estimate the sum in (3.11). Using Lemma 3.1 with $N\ge 2$ (and hence $\ell \ge 3$ ), we have $a_\ell < \sqrt {2\ln \ell }$ , and on setting $\ell = 2^{N+j}$ ,

$$ \begin{align*} \sum_{\ell=2^N, 2^{N+1}, 2^{N+2},\ldots}\frac{a_\ell}{\sqrt{\ell}} &\le \sum_{j=0}^\infty \frac {\sqrt{2(\ln 2) (N+j)}}{\sqrt{2^{N+j}}} = 2^{-(N-1)/2} \sqrt{\ln 2}\, \sum_{j=0}^\infty \frac {\sqrt{ N+j}}{2^{j/2}} .\nonumber \\ &\le 2^{-(N-1)/2} \sqrt{\ln 2}\,\sqrt{N}\,(2+\sqrt 2 )\,(1 + \mathcal{O}(N^{-1/2})), \nonumber \end{align*} $$

where in the final step we used $\sqrt {N+j} \leq \sqrt {N}+\sqrt {j}$ and $\sum _{j=0}^\infty 1/2^{j/2}=2+\sqrt {2}$ , while noting that $\sum _{j=0}^\infty \sqrt {j}/2^{j/2}$ is finite and independent of N. Moreover, by a similar argument we conclude that $\sum _{\ell =2^N, 2^{N+1}, 2^{N+2},\ldots } c/\sqrt {\ell } = \mathcal {O}(2^{-N/2})$ . Thus, altogether we obtain from (3.11) that

$$ \begin{align*} \mathbb{E}[ \|B - B_N\|_\infty ] & \le \tfrac{1}{2} \cdot 2^{-(N-1)/2}\sqrt{\ln2}\,\sqrt{N}\,(2+\sqrt{2})(1+\mathcal{O}(N^{-1/2})) \\ &= (2+\sqrt 2) \frac {\sqrt{N\ln{2}}}{\sqrt {2\cdot 2^{N}}}(1+\mathcal{O}(N^{-1/2})) = (2+\sqrt 2) \frac {\sqrt{\ln{d}}}{\sqrt {2 d}}\bigg (1+\mathcal{O}\bigg(\frac{1}{\sqrt{\ln{d}}}\bigg)\bigg), \end{align*} $$

which proves the first bound in Theorem 1.1.

To prove the second bound in Theorem 1.1, we need first to bound $\mathbb {E}[M_\ell ^2]$ . With $Y_\ell $ and Y defined as above, for $v>0$ ,

(3.12) $$ \begin{align} \mathbb{P}(Y_\ell^2 \ge v) &= \mathbb{P}(Y_\ell \ge \sqrt{v}) + \mathbb{P}(Y_\ell \le - \sqrt{v}) \nonumber\\&\to \mathbb{P}(Y \ge \sqrt{v}) + \mathbb{P}(Y \le - \sqrt{v}) = \mathbb{P}(Y^2 \ge v) \quad\mbox{as}\ \ell\to\infty, \end{align} $$

while for $v<0$ we have $\mathbb {P}(Y_\ell ^2 \ge v) = \mathbb {P}(Y^2 \ge v)= 1$ . Thus, by Lemma 3.2, $Y_\ell ^2$ converges in distribution to $Y^2$ , where Y is the Gumbel distribution. To prove convergence in expectation, we use (3.12) with (3.8) and (3.9) to give, for $\ell \ge L$ and $v>0$ ,

$$ \begin{align*} \mathbb{P}(Y_\ell^2 \ge v) \le \exp(-\sqrt{v}) + \exp(- (1-a_L^{-2})) \exp(-\sqrt{v}), \end{align*} $$

which is integrable on $\mathbb {R}^+$ , proving $\mathbb {E}[Y_\ell ^2] \to \mathbb {E}[Y^2] < \infty $ . In turn it follows that there exists $c'>0$ such that $\mathbb {E}[Y_\ell ^2] \le c'$ , and together with (3.10),

$$ \begin{align*} \mathbb{E}[M_\ell^2] \le a_\ell^2 + 2\,a_\ell\, b_\ell\, c + b_\ell^2\, c' \le a_\ell^2 + 2\,a_\ell\,c + c' \le (a_\ell + c")^2, \end{align*} $$

where we used $b_\ell = a_{\ell }^{-1} \le 1$ for $\ell \ge 3$ and introduced $c" := \max (c,\sqrt {c'})$ . Now from (3.2),

$$ \begin{align*} \mathbb{E}[ \|B - B_N\|_\infty^2 ] &\le \sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{(a_\ell + c")^2}{2\sqrt{\ell}\cdot 2\sqrt{\ell}} + \sum_{\stackrel{\scriptstyle{\ell,\ell' = 2^{N}, 2^{N+1},2^{N+2},\ldots}}{\scriptstyle{\ell\ne\ell'}}} \frac{(a_\ell + c")(a_{\ell'} + c")}{2\sqrt{\ell}\cdot 2\sqrt{\ell'}} \\ &= \bigg(\sum_{\ell = 2^{N}, 2^{N+1},2^{N+2},\ldots} \frac{a_\ell + c"}{2\sqrt{\ell}}\bigg)^2, \end{align*} $$

which is the square of the right-hand side of (3.11), with c replaced by $c"$ . The second bound in Theorem 1.1 then follows.

4 Expected uniform error of geometric Brownian motion

We can now give a proof of Corollary 1.2. From (1.5) and (1.6), $S(t) - S_N(t) = S(0) e^{(r-\sigma ^2/2)t}\, (\exp (\sigma B(t)) -\exp (\sigma B_N(t)))$ , and thus

(4.1) $$ \begin{align} \|S - S_N\|_\infty &\le S(0)\,e^{|r-\sigma^2/2|} \|\exp(\sigma B) -\exp(\sigma B_N)\|_\infty. \end{align} $$

In turn it follows that

$$ \begin{align*} \|S - S_N\|_\infty &\le S(0)\,e^{|r-\sigma^2/2|}\, \|\exp(\sigma B_N) (\exp(\sigma (B-B_N)) - 1)\|_\infty \nonumber\\ &\le S(0)\,e^{|r-\sigma^2/2|}\, \exp(\sigma \|B_N\|_\infty) (\exp(\sigma \|B-B_N\|_\infty) - 1). \end{align*} $$

Using $|{\exp} (x)-1| \le |x| \exp (|x|)$ for $x\in \mathbb {R}$ and $\|B_N\|_\infty \le \|B\|_\infty $ for $N\in \mathbb {N}$ , we have

$$ \begin{align*} \|S - S_N\|_\infty &\le S(0)\,e^{|r-\sigma^2/2|} \exp(\sigma \|B_N\|_\infty)\sigma \|B-B_N\|_\infty\exp(\sigma \|B-B_N\|_\infty) \\ &\le S(0) e^{|r-\sigma^2/2|} \sigma \exp(3\sigma \|B\|_\infty) \|B-B_N\|_\infty, \end{align*} $$

where we used $\|B-B_N\|_\infty \le \|B\|_\infty + \|B_N\|_\infty \le 2\|B\|_\infty $ . By the Cauchy–Schwarz inequality,

$$ \begin{align*} \mathbb{E} [ \|S - S_N\|_\infty ] &\le S(0)\,e^{|r-\sigma^2/2|} \sigma \sqrt{\mathbb{E} [\exp(6\sigma \|B\|_\infty)]}\, \sqrt{\mathbb{E} [\|B-B_N\|_\infty^2 ]}. \end{align*} $$

It is well known that $\mathbb {E} [\exp (\alpha \|B\|_\infty )] < \infty $ for every $\alpha> 0$ . Hence, the result now follows from the second bound of Theorem 1.1.

5 Application to option pricing

Now we consider a continuous version of a path-dependent call option with strike price K in a Black–Scholes model with risk-free interest rate $r>0$ and constant volatility $\sigma>0$ . Recall that the asset price $S(t)$ at time t is given explicitly by (1.5). The discounted payoff for the case of a continuous arithmetic Asian option with terminal time $T=1$ is therefore

(5.1) $$ \begin{align} P :&= e^{-rT}\max\bigg(\frac 1 T \int_0^T S(t) \,d t -K,0\bigg) \nonumber\\ &= e^{-r}\max\bigg( S(0)\int_0^1 \exp\bigg(\bigg(r-\frac{\sigma^2}{2}\bigg) t+\sigma B(t) \bigg)\,d t - K, 0 \bigg). \end{align} $$

The pricing problem is then to compute the expected value $\mathbb {E}(P)$ .

We use the Lévy–Ciesielski expansion (2.2) and (2.3) for $B(t)$ and $B_N(t)$ , and define

(5.2) $$ \begin{align} P_N := e^{-r}\max\bigg( S(0)\int_0^1 \exp\bigg(\bigg(r-\frac{\sigma^2}{2}\bigg) t+\sigma B_N(t) \bigg)\,d t - K, 0 \bigg). \end{align} $$

We are interested in estimating how fast $\mathbb {E} [|P - P_N|]$ converges to $0$ as $N\to \infty $ .

Corollary 5.1. For P and $P_N$ defined by (5.1) and (5.2),

$$ \begin{align*} \mathbb{E} [|P - P_N|] =\mathcal{O}\bigg(\sqrt{\frac{\ln d}{d}}\bigg), \end{align*} $$

where $d=2^N$ and the implied constant is independent of d.

Proof. It can be easily verified that

$$ \begin{align*} |\!\max(\alpha-K,0) - \max(\beta-K,0)| \le |\alpha-\beta|. \end{align*} $$

Thus,

$$ \begin{align*} |P- P_N| &\le \bigg| e^{-r} S(0)\int_0^1 e^{(r-\sigma^2/2)t} (\exp(\sigma B(t)) -\exp(\sigma B_N(t))) \,d t \bigg| \\&\le e^{-r} S(0) e^{|r-\sigma^2/2|} \int_0^1 |{\exp}(\sigma B(t)) -\exp(\sigma B_N(t))| \,d t \\&\le e^{-r} S(0)\,e^{|r-\sigma^2/2|} \|{\exp}(\sigma B) - \exp(\sigma B_N)\|_\infty, \end{align*} $$

where the last upper bound differs from the upper bound (4.1) on $\|S-S_N\|_\infty $ only by a factor of $e^{-r}$ . Hence, the result follows from Corollary 1.2.