Given any continuous function and its infinitesimal increment, the Taylor Series helps us obtain a discrete approximation for the function at any point (defined by a series of increments) in its continuous domain. We almost immediately run into problems with stochastic differential equations because Brownian motion is not differentiable. However, Ito’s lemma provides us with an essential work around provided we disregard Taylor Series terms higher than the second order which we reasonably assume to sum up to zero. At this juncture we have the tools required to value derivatives approximated by a range of stochastic models comprised of a single Brownian motion.

Solving the Black Scholes equation, perhaps by first transforming it to the heat equation, yields a solution that is comprised of an infinite integral. If we are to compute the value of a general derivative this could therefore prove impracticable. Integrals expressed as expectations provide a way out of this constraint. This is the fundamental theory behind Monte Carlo simulation.

Fourier transforms can be shown to be expressible as expectations. We therefore have an alternative to PDE’s for determining the value of derivatives.

Partial Differential Equations

Many of the methods deployed in financial engineering are borrowed from the physical sciences. Many of the methods studied here may be more easily understood if explored using examples from the physical sciences. For PDE’s and their finite solution, I found the paper by Joaquim Peiro and Spencer Sherwin Finite Difference, Finite Element and Finite Volume Methods for Partial Differential Equations particularly useful as it helped me draw strong parallels between my experience with finite element method in engineering and the finite difference method in finance. Both are about discretizing continua in order to take advantage of computer software to solve a set of equations.

Properties of Brownian Motion

One of the most important characteristics of Brownian motion is that as increment $ h \rightarrow 0 $, we get a limiting random function $$dW_t = \sqrt{dt}Z_t$$ where $dt$ is infintessimally small such that
\begin{equation}dt^\alpha
\begin{cases}
0 &\text{ if } \alpha > 1 \\
dt^\alpha &\text{ if } 0 < \alpha < 1\\ \end{cases} \end{equation} The increments do not converge confirming that at any single point Brownian motion is not differentiable. Each increment is independent and normally distributed with mean $\mu$ and variance $\sigma^2$. However right from the onset we benefit from working with a standard Brownian motion normally distributed with mean $0$ and variance $1.$ $$W_{kh} \; – W_{(k \; – 1)h} = Z \qquad \mathcal{N}(0 \; , 1)$$

The nondifferentiability property can be shown by proving the finiteness of the quadratic variation of a Brownian process. This stems from the result in calculus that differentiability implies vanishing of the quadratic variation of the function.[ref]Mathematical Models of Financial Derivatives, Y. K. Kwok.[/ref]

As each increment is a random walk, over any number of discrete intervals $h$, the variance becomes the sum of each incremental variance $$ \text{Var} \; W_{kh} = \sum_{i=1}^k \text{Var} \; \mathcal{z} = kh$$ such that $$ W_t \sim \mathcal{N} (0 \; , t) \qquad \text{for all }t > 0.$$ In the analogy of zooming in on a curve, one would see a straight line with increasing resolution or a high zoom level in photographic terms. This is not the case with Brownian motion. The jagged edges constantly remain visible at all zoom levels. We see the phenomenon present itself again if we consider the task of calculating the exact length of a country’s coastline. The coastline can only be determined for a ‘zoom’ or resolution. A higher resolution will continue to change the result due to the increased number of jagged edges that would otherwise be collectively approximated as a straight line at lower resolutions.

Statistical observations of Brownian motion

Denote standard Brownian motion by $Z$
\begin{align}\mathbb{E}[dW_t] &= \mathbb{E} [Z_t \sqrt{dt}] = \sqrt{dt}\mathbb{E} [Z_t] = 0 \tag{1} \\ \\
\text{var} (dW_t) &= \mathbb{E}[dW_t^2] = \mathbb{E}[dt Z_t^2] = dt\mathbb{E}[Z_t^2] = dt \tag{2} \\ \\
(\text{Note:} \quad \text{var} (dW_t) &= \mathbb{E}[dW_t^2] = \text{mean} (dW_t^2) = dt)\\ \\
\text{variance of r.v } (dW_t^2) &= \text{expectation of (the r.v – mean of the r.v)}^2 = \mathbb{E}[(dW_t – dt)^2]\\
&= \mathbb{E}[(dt Z_t^2 – dt)^2]\\ \\
&= \mathbb{E}[dt^2 (Z_t^2 – 1)^2]\\ \\
&= dt^2 \mathbb{E}[(Z_t^2 – 1)^2]\\ \\
\text{var} (dW_t^2) & =0 \tag{3} \\ \\
\mathbb{E}[dW_t dt] &= \mathbb{E} [dt \sqrt{dt}Z_t] = dt^{3/2}\mathbb{E}Z_t=0 \tag{4}\\ \\
\text{var} (dt dW_t)&=\mathbb{E} (dt^2 dW_t^2)\\ \\
&= \mathbb{E} (dt^2 dt Z_t^2)\\ \\
\text{var} (dt dW_t) &= dt^3 \mathbb{E} Z_t^2 = 0 \tag{5}
\end{align}
Any variable with zero variance is deterministic and not random and is value must match its mean. In view of this,
\begin{align}
(2) \; \& \; (3) \; \Rightarrow \quad dW_t^2 = dt \\ \\
(4) \; \& \; (5) \; \Rightarrow \quad dW_t dt = 0
\end{align}

Geometric Brownian Motion Model of Asset Prices

Asset prices may be approximated in their evolution by geometric Brownian motion. $$G_t = e^{\alpha + \beta t + \sigma W_t}$$ As this represents the future value we must discount in order to obtain today’s value. So we say $$e^{-rt}G_t = e^{\alpha + ( \beta -r )t + \sigma W_t}$$ We have the requirement that the mathematical representations above must be martingales – they must be fair. (What goes up must come down). For this to happen, \begin{align*}\beta \; – r &= \; – \frac{\sigma^2}{2}\\ \beta &= r \; – \frac{\sigma^2}{2}\end{align*} The asset price (discounted to today) may be given as $$S_t = e^{\alpha + \beta t + \sigma W_t} = e^{\alpha + (r \; – \frac{\sigma^2}{2} )t + \sigma W_t}$$ or more usefully as $$S_t = S_0 e^{ (r \; – \frac{\sigma^2}{2} )t + \sigma W_t}$$where

• $r$ is the risk-free rate
• $\sigma$ is the volatility of the asset

Rather than being stuck with just the asset price $S_t$ at $t$ relative to $S_0$ at $t=0$, we can generalise for any time interval $\tau = T \; – t$ with $W_T \; – W_t = \sqrt{\tau} Z$ and $Z \sim \mathcal{N}(0 \; , 1)$ $$S_T = S_t e^{ (r \; – \frac{\sigma^2}{2} )\tau + \sigma \sqrt{\tau}Z_T} \qquad Z_T \sim \mathcal{N}(0 \; , 1)$$

We have an expression for the value of $S_T$ that is comprised of a standard Brownian motion – a random variable. To make any practical sense of this analytical solution we at least want to know the expectation (mean) of the exponential function $e^x$ when $x$ is comprised of a standard Brownian motion – a random variable.