Random walks and Brownian motion

Idea: A path, not a number

A stochastic process is a family $X_t$ over time; each outcome is a whole path.

• End point: where the price closes
• Path: whether it touches the stop on the way

Formula: The simple random walk

$E[S_n]=0,\mathrm{Var}(S_n)=n$

Fair steps of +1 or −1: no drift, and the spread grows like √n, not n.

Example: Ending at +2

A fair walk takes $4$ steps of $+1$ or $-1$. What is $P(S_4=2)$?

1. Up steps needed: $\frac{4+2}{2}=3$
2. Chance: $\frac{\left(\genfrac{}{}{0ex}{}{4}{3}\right)}{2^4}=\frac{1}{4}$

Formula: Reflection principle

$P(\max S\ge a)=P(S_n\ge a)+P(S_n>a)$

Flip the path after its first touch: each touch that ends below pairs with a path that ends above.

Quick check: Touching the stop

A fair walk takes $6$ steps from $0$. What is the chance that it touches $+2$ at some point?

Idea: From ticks to Brownian motion

Shrink the ticks and the walk tends to Brownian motion $W_t$.

• $W_0=0$, continuous paths, nowhere smooth
• Independent increments: $W_t-W_s\sim N(0,t-s)$
• $\mathrm{Cov}(W_s,W_t)=\min (s,t)$

Example: Variance of a sum

Find $\mathrm{Var}(W_2+W_5)$ for Brownian motion.

1. Rewrite: $W_2+W_5=2W_2+(W_5-W_2)$
2. Independent pieces: $4\times 2+3=11$

Common mistake: Treating times as independent

Adding $2+5=7$ as if $W_2$ and $W_5$ were independent. They share the path up to time $2$.

Instead Rewrite any position in independent increments first.

Idea: Say it out loud

Say whether you answer the end point or the path.

• Risk scales with √time: $1\%$ a day is about $4\%$ over $16$ days
• A stop is hit more often than a close: about $0.32$ against $0.16$
• Real prices jump, have fat tails and clustered volatility