I got a Brownian motion $B(t)$ that starts in $0$ and want to calculate the expectated value of the supremum on the interval $[0,1]$ of the absolute value of it, i.e.
$$E \left (\sup \limits_{t \in [0,1]} |B(t) | \right ).$$
I found some information on this question in this thread.
Here user3371583 posted in the comment chain that the expected value in the case of the interval $[0,1]$ should be $\sqrt{\frac{\pi}{8}}$ and he explained how he got to this result in the same post, but I cant seem to get it done.
What I got so far is this:
\begin{align*}
E \left (\sup \limits_{t \in [0,1]} |B(t) | \right ) &= \int_0^\infty P(\sup \limits_{t \in [0,1]} |B(t) | \geq y ) dy \\
&= \int_0^\infty \sum_{k=-\infty}^\infty (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}} \right ) dy.
\end{align*}
Here Erfc is the error function:
$$\text{Erfc}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \ dt.$$
Now the post mentions splitting the series up into a series over negative $k$ and one over positive (nonnegative I guess) $k$, so that I get
\begin{align*}
= &\int_0^\infty \sum_{k=-\infty}^{-1} (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}} \right ) dy\\
+ &\int_0^\infty \sum_{k=0}^\infty (-1)^k \text{sign}((2k+1) y) \text{Erfc} \left ( \frac{\vert (2k+1) y \vert }{\sqrt{2}} \right ) dy
\end{align*}
and now the next step seems to be using a change of variable to see that both terms are actually equal, but I cannot verify this. The first term is
\begin{align*}
&\int_0^\infty \sum_{k=1}^{\infty} (-1)^{-k} \text{sign}((2(-k)+1) y) \text{Erfc} \left ( \frac{\vert ((2(-k)+1) y \vert }{\sqrt{2}} \right ) dy \\
= &\int_0^\infty \sum_{k=1}^{\infty} (-1)^{k} \text{sign}((2k+1) y) (-1)\text{Erfc} \left ( \frac{\vert ((2(-k)+1) y \vert }{\sqrt{2}} \right ) dy
\end{align*}
and this is where I'm stuck already. I know that the Errorfunction is an odd function, but I cant seem to bring it into a form where I can combine both sums into one.
Also, I don't understand how to use integration by parts to evaluate both of the integrals after combining the series.
Can anyone help me out with this? Thanks!
Best Answer
First of all, note that Erfc does not denote the Error function but its "complement", i.e.
$$\DeclareMathOperator{\erfc}{Erfc}\DeclareMathOperator{\erf}{Erf} \erfc(x) := \frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp (-y^2) \, dy = 1- \erf(x)$$
where
$$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp (-y^2) \, dy$$
is the Error function.
Since
$$|2(-k)+1| = |2k-1|$$
we have
\begin{align*} \DeclareMathOperator{\sign}{sgn} \DeclareMathOperator{\erf}{Erfc} & - \int_{(0,\infty)} \sum_{k \geq 1} (-1)^k \underbrace{\sign((2k+1)y)}_{=1} \erf \left( \frac{|(2(-k)+1)y|}{\sqrt{2}} \right) \, dy \\ &= \int_{(0,\infty)} \sum_{k \geq 1} (-1)^{k+1} \erf \left( \frac{|(2k-1)y|}{\sqrt{2}} \right) \, dy \\ &\stackrel{k \to j+1}{=} \int_{(0,\infty)} \sum_{j \geq 0} (-1)^{j} \underbrace{(-1)^2}_{=1} \erf \left( \frac{|(2j+1)y|}{\sqrt{2}} \right) \, dy. \end{align*}
Hence, by the computations from your question,
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = 2 \int_{(0,\infty)} \sum_{j \geq 0} (-1)^j \erf \left( \frac{|(2j+1)y|}{\sqrt{2}} \right) \, dy. \end{align*}
Interchanging sum and integration and applying the integration by parts formula we get
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) &= 2 \sum_{j \geq 0} (-1)^j \int_{(0,\infty)} \erf \left( \frac{(2j+1)y}{\sqrt{2}} \right) \, dy \\ &= \frac{4}{\sqrt{2\pi}} \sum_{j \geq 0} (2j+1) (-1)^j \int_{(0,\infty)} y \exp \left(- \frac{(2j+1)^2}{2} y^2 \right) \, dy \\ &= \frac{4}{\sqrt{2\pi}} \sum_{j \geq 0} (-1)^j \frac{1}{2j+1} \end{align*} As $$\arctan(x) = \sum_{j \geq 0} (-1)^j \frac{x^{2j+1}}{2j+1}$$ we conclude that
$$\mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = \frac{4}{\sqrt{2\pi}} \arctan(1) = \frac{4}{\sqrt{2\pi}} \frac{\pi}{4} = \sqrt{\frac{\pi}{2}} $$
Remark I: Feel free to check the constants appearing in my computations. In the thread, which you linked, it is claimed that
$$\mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) = \sqrt{\frac{\pi}{8}};$$
this is, however, wrong. It follows from the reflection principle that $\sup_{t \in [0,1]} B_t \sim |B_1|$, and therefore
\begin{align*} \mathbb{E} \left( \sup_{t \in [0,1]} |B_t| \right) \geq \mathbb{E} \left( \sup_{t \in [0,1]} B_t \right) &= \mathbb{E}(|B_1|) \\ &= \sqrt{\frac{2}{\pi}} \approx 0.797 \\ &> \sqrt{\frac{\pi}{8}} \approx 0.626 \end{align*}
Remark II: Let me provide a reference for the formula for $\mathbb{P}(\sup_{s \leq 1} |B_s| \geq r)$ which you used in your computations (just in case that somebody wants to look it up). In the Handbook of Brownian Motion by Borodin & Salminen it is stated on p. 339 (Part II, Chapter 3, Formula 1.1.4) that $$\mathbb{P}^x \left( \sup_{s \leq t} |B_s| \geq r \right) = \tilde{\text{cc}}_t(x,r)$$ where $$\tilde{\text{cc}}_t(x,r) := \sum_{k \in \mathbb{Z}} (-1)^k \sign(x+(2k+1)r) \erfc \left( \frac{|x+(2k+1)r|}{\sqrt{2t}} \right),$$ see p. 651 (Appendix II, Section 13).