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\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}
