\documentclass{beamer} %\usepackage{beamerthemesplit} %\usepackage{graphicx} \usepackage{amssymb} \usepackage{amsmath} %\usetheme{default} %\usetheme{Warsaw} \usetheme{Copenhagen} \newcommand{\CC}{{{\mathbb C}}} \newcommand{\EE}{{{\mathbb E}}} \newcommand{\PP}{{{\mathbb P}}} \newcommand{\QQ}{{{\mathbb Q}}} \newcommand{\RR}{{{\mathbb R}}} \newcommand{\VV}{{{\mathbb V}}} \newcommand{\ZZ}{{{\mathbb Z}}} \title{Exponential Brownian Motion \& Approximation Theory} \author{Brad Baxter\\ Birkbeck College, University of London} \begin{document} \begin{frame} \titlepage \end{frame} \begin{frame} Collaborators: R.~Brummelhuis, S.~Fretwell \end{frame} \begin{frame} Consider exponential Brownian motion \[ S(t) = e^{(r - \sigma^2/2)t + \sigma W_t}, \qquad t \ge 0, \] where $W_t$ is Brownian motion, $r \ge 0$, $\sigma \in \RR$ constants. Its time average is \[ A(T) = \frac{1}{T} \int_0^T S(t)\,dt, \qquad T > 0. \] \alert<1>{Empirical discovery}: $S(T)$ and $A(T)$ typically highly correlated -- coefficient $\approx 0.85$. \alert<1>{Problem}: Calculating correlation coefficient is tricky. %Hence we calculate %\[ %R = \frac{\EE S(T)A(T) - \EE S(T) \EE A(T)}{\sqrt{\VV S(T) \VV A(T)}}. %\] \alert<1>{Surprise}: Divided differences occur naturally in the analysis, leading to great simplification and new insights from approximation theory. \end{frame} \begin{frame} \alert<1>{L\'evy and Cieselski subdivision-style construction of Brownian motion} Let $\{Z(q) : q \in \QQ\}$ be independent normalized Gaussian random variables. Define $B(0)=1$ and \[ B(k) = B(k-1) + Z(k), \qquad k = 1, 2, \ldots. \] Then define \begin{align*} &B\biggl(\frac{k+1/2}{2^n}\biggr)\cr &\!\!= \frac{1}{2} \biggl(\! B\biggl(\frac{k}{2^n}\biggr)\! + B\biggl(\frac{k+1}{2^n}\biggr)\!\biggr)\! + 2^{-1-n/2} Z\biggl(\frac{k+1/2}{2^n}\biggr)\cr &\!\!. \end{align*} \end{frame} \begin{frame} Now it was already known that $$\EE \left(A(T)^2\right)$$ is given by \[ \frac{2e^{(2r+\sigma^2)T}}{(r+\sigma^2)(2r+\sigma^2)T} +\frac{2}{rT^2}\left(\frac{1}{2r+\sigma^2} - \frac{e^{rT}}{r+\sigma^2}\right). \] \alert<1>{Surprise:} This {\em is} a divided difference: \[ \EE \left(A(T)^2\right) = 2 \exp[0, rT, (2r+\sigma^2)T]. \] \end{frame} \begin{frame} Key fact: $\EE S(t) = e^{rt}$. Simple link with divided differences: \begin{eqnarray*} \EE A(T) &=& \frac{1}{T} \int_0^T \EE S(t) \,dt\\ &=& \frac{e^{rT} - 1}{rT}\\ &=& \exp[0, rT]. \end{eqnarray*} Coincidence? Let's try another. \end{frame} \begin{frame} We need a simple Lemma: \[ \EE S(a)S(b) = e^{a(r+\sigma^2)} e^{br}, \qquad\hbox{ for } 0 \le a \le b. \] {\bf Proof:} Straightforward Brownian motion exercise. %\begin{eqnarray*} %\EE S(a) S(b) %&=& \EE e^{(r-\sigma^2/2)a + (r-\sigma^2/2)b + 2\sigma W_a + \sigma (W_b - W_a)}\\ %&=& e^{(r-\sigma^2/2)a + (r-\sigma^2/2)b} \left(\EE e^{2\sqrt{a} Z_1}\right) % \left( e^{\sigma \sqrt{b-a} Z_2}\right)\\ %&=& e^{(r-\sigma^2/2)a + (r-\sigma^2/2)b + 2\sigma^2 a + \sigma^2 (b-a)/2 }\\ %&=& e^{a(r+\sigma^2)} e^{br}. %\end{eqnarray*} Then \begin{eqnarray*} \EE S(T) A(T) &=& T^{-1} \int_0^T \EE S(t)S(T)\,dt\\ &=& T^{-1} \int_0^T e^{(r+\sigma^2)t}e^{rT}\,dt\\ &=& \frac{e^{(2r+\sigma^2)T} - e^{rT}}{(r + \sigma^2)T}\\ &=& \exp[rT, (2r+\sigma^2)T]. \end{eqnarray*} \end{frame} \begin{frame} Similarly \begin{eqnarray*} \EE (A(T)^2) &=& T^{-2} \int_0^T \Bigl( \int_0^T \EE S(t_1) S(t_2)\, dt_2 \Bigr)\,dt_1\\ &=& 2T^{-2} \int_0^T \Bigl( \int_0^{t_1} \EE S(t_1) S(t_2)\, dt_2 \Bigr)\,dt_1\\ %\end{eqnarray*} %} %\begin{frame} %\begin{eqnarray*} &=& 2T^{-2} \int_0^T \Bigl( \int_0^{t_1} e^{r(t_1 + t_2)} e^{\sigma^2 t_2}\, dt_2 \Bigr)\,dt_1\\ &=& 2T^{-2} \int_0^T e^{rt_1}\Bigl(\frac{e^{(r+\sigma^2)t_1} - 1}{r+\sigma^2} \Bigr)\,dt_1\\ &=& \frac{2}{(r+\sigma^2)T} \left[ \exp[0,(2r+\sigma^2)T] - \exp[0,rT] \right]\\ &=& 2 \exp[0, rT, (2r+\sigma^2)T], \end{eqnarray*} \end{frame} \begin{frame} Now we {\em expect} to see divided differences: \begin{eqnarray*} \lefteqn{\EE S(T) A(T) - \EE S(T) \EE A(T)}\\ &=& \exp[rT,(2r+\sigma^2)T] - e^{rT}(e^{rT} - 1)/(rT)\\ &=& \exp[rT,(2r+\sigma^2)T] - \exp[rT,2rT]\\ &=& \sigma^2 T \exp[rT, 2rT, (2r+\sigma^2)T], \end{eqnarray*} and for the variance \begin{eqnarray*} \lefteqn{\VV S(T)}\\ &=& \EE(S(T)^2) - (\EE S(T))^2\\ &=& e^{(2r+\sigma^2)T} - e^{2rT}\\ &=& \sigma^2 T \exp[2rT, (2r+\sigma^2)T]. \end{eqnarray*} \end{frame} \begin{frame} Finally, the correlation coefficient $R$ is given by \[ \frac{ \exp[rT, 2rT, (2r+\sigma^2)T] } { \sqrt{ 2\exp[2rT, (2r+\sigma^2)T] \exp[0, rT, 2rT, (2r+\sigma^2)T]} }. \] Two obvious questions arise: \begin{itemize} \item Why do these iterated integrals lead to divided differences? \item So what? \end{itemize} \end{frame} \begin{frame} {Hermite--Genocchi} Let $f \in C^{(n)}(\RR)$ and let $a_0, a_1, \ldots, a_n$ be real numbers. Then \begin{eqnarray*} \lefteqn{f[a_0, a_1, \ldots, a_n] } \\ &=& \int_{S_n} f^{(n)}(t_0 a_0 + t_1 a_1 + \cdots + t_n a_n)\,dt_1 \cdots dt_n,\\ &=& \int_0^1 dt_1 \cdots \int_0^{1 - \sum_{k=1}^{n-1}t_k} dt_n \ f^{(n)}(\sum_{k=0}^n t_k a_k) \end{eqnarray*} integrating over the simplex \[ S_n = \{ t = (t_1, t_2, \ldots, t_n) \in \RR_+^n \colon \sum_{k=1}^n t_k \le 1\} \] and \[ t_0 = 1 - \sum_{k=1}^n t_k. \] \end{frame} \begin{frame} For the exponential function, \[ \exp[a_0, \ldots, a_n] = \int_{S_n} \exp(\sum_{k=0}^n t_k a_k)\,dt_1 \cdots dt_n. \] For any nonsingular matrix \[ V = \left( v_1 \quad \cdots \quad v_n\right) \in \RR^{n \times n}, \] let \[ K(V) = \hbox{conv} \{0, v_1, \ldots, v_n\}. \] Then \[ \frac{1}{|\det V|} \int_{K(V)} \exp(a^T y)\,dy \] is equal to \[ \exp[0, (V^T a)_1, \ldots, (V^T a)_n]. \] [$(V^T a)_j$ is $j$th component of $V^T a$.] \end{frame} \begin{frame} If \[ V = \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ \vdots & & \ddots & \\ 1 & 1 & \cdots & 1 \end{pmatrix}, \] then \begin{align*} &\int_0^1 dx_n \int_0^{x_n} dx_{n-1} \cdots \int_0^{x_2} dx_1 \exp \left(\sum_{k=1}^n a_k x_k\right)\\ &= \exp[0, a_n, a_n + a_{n-1}, \ldots, a_n + a_{n-1} + \cdots + a_1]. \end{align*} \end{frame} \begin{frame} Now we can compute higher moments of $A(T)$. We obtain \[ \EE\left(A(T)^m\right) = m! \exp[ b_0 T, b_1 T, \ldots, b_m T], \] where \[ b_k = r k + \sigma^2 k(k-1)/2, \quad k \ge 0. \] \end{frame} \begin{frame} So what? Divided differences allow us to use the rich analytic toolbox of approximation theory: \begin{itemize} \item If $r = \sigma^2$, then the correlation coefficient $R = \sqrt{3}/2 = 0.866\ldots$ \item \alert<1>{Theorem}[B and Fretwell] For any $r \ge 0$ and $\sigma$, the correlation coefficient satisfies $R \ge \frac{1}{\sqrt{2}} = 0.7071\ldots$ \end{itemize} Thus the time-average is a remarkably good predictor for asset's price in the geometric Brownian motion universe. \end{frame} \begin{frame} In fact the correlation coefficient inequality is a special case of the following \alert<1>{Theorem}\ Let $h \ge 0$ and define \[ E_n(x) = \exp[0, -h, -2h, \ldots, -nh, x], \qquad x \in \RR, \ n \ge 0. \] Then $(E_n(x))$ is a log-concave sequence, i.e. \[ E_{n+1}(x) E_{n-1}(x) \le E_n(x)^2, \qquad \hbox{for } n \ge 1. \] \alert<1>{Log-concave sequences:} Enormous literature. See, e.g., Wilf, {\em Generatingfunctionology}. \end{frame} \begin{frame} \alert<1>{Special Case:} Define \[ R_m(\alpha) = e^\alpha - \sum_{k=0}^m \frac{\alpha^k}{k!}, \label{cc1} \] for non-negative integer $m$ and $\alpha \in \RR$. Thus $R_m(\alpha)$ is the Taylor remainder (after $m+1$ terms) for the exponential function. Further \[ R_m(\alpha) = \alpha^{m+1}\exp[\underbrace{0, 0, \ldots, 0}_{m+1}, \alpha]. \label{cc2} \] Furthermore, \[ R_m^\prime(\alpha) = R_{m-1}(\alpha), \qquad\hbox{ for } m \ge 1, \alpha \in \RR. \label{cc3} \] \end{frame} \begin{frame} \alert<1>{Lemma} The exponential function Taylor remainders satisfy \[ \frac{R_{m+1}(\alpha)}{R_m(\alpha)} = 1 - \frac{1}{(m+1)! \exp[\underbrace{0, 0, \ldots, 0}_{m+1}, \alpha]}. \label{cc4} \] \alert<1>{Proof} \begin{align*} 1 - \frac{R_{m+1}(\alpha)}{R_m(\alpha)} &= \frac{R_m(\alpha) - R_{m+1}(\alpha)}{R_m(\alpha)}\\ &= \frac{p_{m+1}(\alpha) - p_m(\alpha)}{R_m(\alpha)}\\ &= \frac{\alpha^{m+1}}{(m+1)! R_m(\alpha)}\\ &= \frac{1}{(m+1)! \exp[\underbrace{0, 0, \ldots, 0}_{m+1}, \alpha]}. \end{align*} \end{frame} \begin{frame} However $\alpha \mapsto \exp[\underbrace{0, 0, \ldots, 0}_{m+1},\alpha]$ is an increasing function, with derivative \[ \exp[\underbrace{0, 0, \ldots, 0}_{m+1},\alpha, \alpha]. \] \alert<1>{Corollary} $R_{m+1}(\alpha)/R_m(\alpha)$ is an increasing function. \alert<1>{Proof } \[ \frac{d}{d\alpha} \frac{R_{m+1}(\alpha)}{R_m(\alpha)} = \frac{\exp[\overbrace{0, 0, \ldots, 0}^{m+1},\alpha, \alpha]} {(m+1)! \exp[\underbrace{0, 0, \ldots, 0}_{m+1},\alpha]^2}. \] Hence \[ R_m(\alpha)^2 \ge R_{m+1}(\alpha) R_{m-1}(\alpha), \qquad\hbox{ for } m \ge 1 \hbox{ and } \alpha \in \RR. \] because \[ 0 \le \frac{d}{d\alpha} \frac{R_{m+1}(\alpha)}{R_m(\alpha)} = \frac{R_m(\alpha)^2 - R_{m+1}(\alpha) R_{m-1}(\alpha)}{R_m(\alpha)^2}. \] \end{frame} \begin{frame} Now, when $h=0$, \[ R_m(\alpha) = E_m(\alpha) \alpha^{m+1}, \] so $(E_m(\alpha))$ is also log-concave, i.e. \[ \exp[\underbrace{0, 0, \ldots, 0}_{m+1},\alpha]\exp[\underbrace{0, 0, \ldots, 0}_{m-1},\alpha] \le \exp[\underbrace{0, 0, \ldots, 0}_{m},\alpha]^2. \] \end{frame} \begin{frame} Is it only true for exponentials? Maple experiments show that \[ f[0, h, 2h, \ldots, nh], \qquad h > 0, \] is a log-concave sequence for many (all?) completely monotonic functions, i.e. $(-1)^n f^{(n)}(x) \ge 0$, $x \ge 0$. \alert<1>{Bernstein--Widder Theorem:} $f: [0,\infty) \to \RR$ is completely monotonic if and only if \[ f(x) = \int_0^\infty e^{-xs}\,d\mu(s), \qquad x \ge 0, \] for some positive Borel measure $\mu$ on $[0, \infty)$. \end{frame} \begin{frame} Let $X$ be a L\'{e}vy-Stable process. Then the natural logarithm of its characteristic function is given by n\[ \ln \EE [e^{iX\theta}] = \begin{cases} -\kappa^{\alpha}\lvert\theta\rvert^{\alpha}(1 - i\beta (\textrm{sign } \theta)\tan\frac{\alpha\pi}{2})+im\theta & \mbox{if $\alpha \neq 1$}\\ -\kappa\lvert\theta\rvert(1+i\beta\frac{2}{\pi}(\textrm{sign }\theta)\ln \lvert \theta \rvert ) + im\theta & \mbox{if $\alpha = 1$} \end{cases} \] where $\alpha \in (0,2]$, $\kappa > 0$, and $\beta \in [-1,1]$; we write $X \sim S_\alpha(\kappa, \beta, m)$ \end{frame} \begin{frame} Then \[ S(T) = S(t)\exp\left((r+\mu)(T-t)-\sigma X_{T-t}\right), \] where $X_{T-t} \sim S_{\alpha} \left( (T-t)^{1/\alpha},1,0 \right)$. For risk-neutrality, $\mu = \sigma^{\alpha}\sec(\alpha \pi/2)$. The correlation coefficient satisfies \[ R = \frac{\exp[rT,2rT,(2r+\mu(2-2^{\alpha}))T]}{\sqrt{2\exp[2rT,(2r+\mu(2-2^{\alpha}))T]\exp[0,rT,2rT,(2r+\mu(2-2^{\alpha}))T]}} \] \alert<1>{Theorem} $R \ge 1/\sqrt{2}$. \end{frame} \end{document}