8.5 Brownian motion, geometric Brownian motion and Itô's lemma

Motivation

The tree's log price is a scaled random walk. Its continuous limit is Brownian motion, and the calculus of functions of Brownian motion differs from ordinary calculus in one term. That term is the $-\tfrac12\sigma^2$ in every lognormal formula and the $\tfrac12\sigma^2S^2\Gamma$ in the Black–Scholes PDE. You met its shadow in M02 as volatility drag.

Brownian motion and quadratic variation#

Setup

A standard Brownian motion $W=(W_t)_{t\ge0}$ on a probability space $(\Omega,\mathcal F,\mathbb P)$ is a process with (i) $W_0=0$ ; (ii) independent increments; (iii) $W_t-W_s\sim N(0,t-s)$ for $s<t$ ; (iv) continuous paths. Its natural filtration $\mathcal F_t$ is the information revealed up to $t$ . Brownian paths are continuous but nowhere differentiable, and of unbounded total variation on every interval.

Proposition 8.5.1quadratic variation

Let $0=t_0<t_1<\dots<t_n=T$ be a partition with mesh $\max_i(t_{i+1}-t_i)\to0$ . Then $Q_n=\sum_i(W_{t_{i+1}}-W_{t_i})^2\to T$ in $L^2$ .

Proof

Write $\Delta_i=t_{i+1}-t_i$ and $D_i=(W_{t_{i+1}}-W_{t_i})^2-\Delta_i$ . The $D_i$ are independent with mean zero, and since the fourth moment of $N(0,\Delta_i)$ is $3\Delta_i^2$ , $\operatorname{Var}(D_i)=3\Delta_i^2-\Delta_i^2=2\Delta_i^2$ . So $\mathbb E[(Q_n-T)^2]=\sum_i2\Delta_i^2\le2T\max_i\Delta_i\to0$ .∎

For the uniform grid, $\operatorname{sd}(Q_n)=T\sqrt{2/n}$ . The randomness of each squared increment averages out, and the sum of squares is the deterministic number $T$ . Shorthand: $(dW_t)^2=dt$ . Similar computations give $dt\,dW_t=0$ and $(dt)^2=0$ in the same mean-square sense, because those sums are $O(\max\Delta_i^{1/2})$ and $O(\max\Delta_i)$ . m08_mc.quadratic_variation checks the convergence.

Itô processes and Itô's lemma#

An Itô process has the form $dX_t=a_t\,dt+b_t\,dW_t$ , meaning $X_t=X_0+\int_0^ta_s\,ds+\int_0^tb_s\,dW_s$ , where the second integral is the Itô integral (limit of left-endpoint Riemann sums $\sum b_{t_i}(W_{t_{i+1}}-W_{t_i})$ ). Left endpoints make the integrand non-anticipating: you choose your position before you see the next move. The Itô integral of a suitable non-anticipating integrand is a martingale with $\mathbb E[(\int_0^tb\,dW)^2]=\mathbb E\int_0^tb^2ds$ (the Itô isometry). For a full construction see Shreve II, Ch. 4.

Proposition 8.5.2Itô's lemma, heuristic derivation

Let $dX_t=a\,dt+b\,dW_t$ and $f(t,x)$ be $C^{1,2}$ . Then

df(t,X_t)=\Big(f_t+af_x+\tfrac12b^2f_{xx}\Big)dt+bf_x\,dW_t. \tag{8.14}

Derivation

Taylor-expand to second order:

df=f_t\,dt+f_x\,dX+\tfrac12f_{xx}(dX)^2+f_{tx}\,dt\,dX+\tfrac12f_{tt}(dt)^2+\dots

Now $(dX)^2=a^2(dt)^2+2ab\,dt\,dW+b^2(dW)^2=b^2\,dt$ by the multiplication rules above, and $dt\,dX=0$ , $(dt)^2=0$ . Substituting $dX=a\,dt+b\,dW$ gives (8.14). The extra term $\tfrac12b^2f_{xx}\,dt$ survives because $(dW)^2$ is of order $dt$ , not $dt^2$ : a Brownian increment over $dt$ is of size $\sqrt{dt}$ . A rigorous proof sums the second-order Taylor expansion over a partition and uses Proposition 8.5.1 to show $\sum f_{xx}(\Delta X)^2\to\int f_{xx}b^2dt$ (Shreve II, Thm 4.4.1, 4.4.6).∎

The Black–Scholes stock model#

Geometric Brownian motion The Black–Scholes model of a stock is

dS_t=\mu S_t\,dt+\sigma S_t\,dW_t, \tag{8.15}

with constant drift $\mu$ and volatility $\sigma>0$ : the instantaneous return $dS/S$ has mean $\mu\,dt$ and variance $\sigma^2dt$ .

Proposition 8.5.3solution and log-normality

The unique solution of (8.15) is

S_t=S_0\exp\!\Big(\big(\mu-\tfrac12\sigma^2\big)t+\sigma W_t\Big). \tag{8.16}

Hence $\ln S_T\sim N\big(\ln S_0+(\mu-\frac12\sigma^2)T,\ \sigma^2T\big)$ , $\mathbb E[S_T]=S_0e^{\mu T}$ , $\operatorname{Var}[S_T]=S_0^2e^{2\mu T}(e^{\sigma^2T}-1)$ , and the median of $S_T$ is $S_0e^{(\mu-\sigma^2/2)T}$ .

Proof

Apply (8.14) to $f(x)=\ln x$ with $a=\mu S$ , $b=\sigma S$ : $f_x=1/S$ , $f_{xx}=-1/S^2$ , so

d\ln S_t=\Big(\mu-\tfrac12\sigma^2\Big)dt+\sigma\,dW_t. \tag{8.17}

The right side has constant coefficients; integrate from $0$ to $t$ to get (8.16). Conversely, Itô applied to (8.16) as a function of $(t,W_t)$ recovers (8.15). Uniqueness follows from the Lipschitz conditions of the standard existence theorem (Shreve II, 6.2). For the moments, if $Y\sim N(m,v)$ then $\mathbb E[e^Y]=e^{m+v/2}$ (complete the square), so $\mathbb E[S_T]=S_0e^{(\mu-\sigma^2/2)T+\sigma^2T/2}=S_0e^{\mu T}$ ; $\mathbb E[S_T^2]=S_0^2e^{2(\mu-\sigma^2/2)T+2\sigma^2T}$ , which gives the variance. The median of $e^Y$ is $e^m$ because $\exp$ is increasing.∎

Worked example 8.6the drag in rupees

Suppose Nifty follows (8.15) with $\mu=12\%$ and $\sigma=16\%$ . The expected level in a year is $e^{0.12}=1.1275$ times today's, but the median is $e^{0.12-0.0128}=1.1131$ times. The expected log return is $10.72\%$ , not 12%. Over ten years the median outcome is $e^{1.072}=2.92\times$ against a mean of $e^{1.2}=3.32\times$ . The gap is the volatility drag of M02, now derived rather than observed. m08_mc.gbm_log_return_moments simulates (8.15) by an Euler scheme on simple returns and recovers the mean of $\ln S_T$ as $\mu-\sigma^2/2$ , not $\mu$ .

Practitioner notes

GBM assumes constant $\sigma$ , continuous paths and normal log returns. Real Nifty returns have fat tails and volatility clustering (M02). Black–Scholes survives as a quoting convention: traders quote options by the $\sigma$ that makes the formula match the price (implied volatility, 8.10), and the smile records how the model is wrong.
"Daily vol $=$ annual vol$/\sqrt{252}$" (or $\sqrt{365}$ , depending on the time convention) is the $\sqrt{t}$ scaling of $W_t$ . A 12.5% annual vol on Nifty at 24,500 means a one-standard-deviation trading-day move of about $24{,}500\times0.125/\sqrt{252}\approx193$ points.