Stochastic Differential Equations

An ordinary differential equation is a promise: tell me where the system is now, and I will tell you exactly where it goes next. That promise is useless for a share price. Nothing about today's close determines tomorrow's, and a model that returns a single number is answering a question nobody asked. What you actually want is a rule that produces a spread of tomorrows, with the right centre and the right width. A stochastic differential equation is exactly that rule.

The picture before any symbols

Think of a small boat crossing a choppy lake under engine power. Two things move it. The engine pushes steadily in a chosen direction — over an hour it covers a predictable distance. The chop shoves it about randomly — over an hour those shoves partly cancel, so the boat ends up displaced, but far less than the sum of the individual shoves.

That asymmetry is the entire mathematical content. Steady push accumulates in proportion to time. Random shoves accumulate in proportion to the square root of time, because they cancel each other out on average. Over a minute the chop dominates and you cannot tell where the engine is pointing. Over a day the engine wins and the chop is a blur around the track. Everything below is bookkeeping for those two rates.

Writing the rule down

The standard form is

$$dX_t \;=\; \mu(X_t, t)\,dt \;+\; \sigma(X_t, t)\,dW_t .$$

Reading it symbol by symbol: $X_t$ is the thing you are modelling at time $t$ — a price, a rate, a spread. $\mu$ is the drift, the engine: how much $X$ is expected to move per unit of time, possibly depending on where $X$ currently is. $\sigma$ is the diffusion coefficient or volatility, the chop: how large the random kicks are. $dW_t$ is an increment of Brownian motion — a fresh, unpredictable shock with mean zero and standard deviation $\sqrt{dt}$.

In plain English: over the next sliver of time, $X$ moves by a predictable amount proportional to the length of the sliver, plus a random amount proportional to the square root of that length.

The $d$'s are not derivatives; Brownian paths have no derivative anywhere. The equation above is shorthand for an honest statement about Itô integrals:

$$X_T \;=\; X_0 \;+\; \int_0^T \mu(X_t,t)\,dt \;+\; \int_0^T \sigma(X_t,t)\,dW_t .$$

In words: where you end up equals where you started, plus everything the engine did, plus everything the chop did. The second integral is the one that needs Itô's construction; the first is ordinary calculus.

Why $\sqrt{dt}$ changes how you think

Put numbers on the boat. Take a stock with drift $\mu = 0.08$ per year and volatility $\sigma = 0.20$ per year. Over one trading day, $\Delta t = 1/252 = 0.003968$:

• drift contribution: $0.08 \times 0.003968 = 0.000317$, about 0.03 percent
• noise standard deviation: $0.20 \times \sqrt{0.003968} = 0.20 \times 0.06299 = 0.0126$, about 1.26 percent

The noise is roughly 40 times the drift over a single day. That is why one day of price action tells you essentially nothing about expected return. The two are equal when $\mu t = \sigma\sqrt{t}$, that is when $\sqrt{t} = \sigma/\mu = 0.20/0.08 = 2.5$, so $t = 6.25$ years. Below six years of data, the noise in a mean-return estimate is larger than the mean itself. This single calculation explains why Sharpe ratios need decades to measure and why risk models are estimable from a few months of data.

Worked example 1: solving the equation for a stock

The standard equity model is geometric Brownian motion, where both drift and volatility scale with the price:

$$dS_t = \mu S_t\,dt + \sigma S_t\,dW_t .$$

In words: the stock moves by a percentage each instant, not a dollar amount, so a 100-dollar stock and a 10-dollar stock have the same return dynamics.

To solve it, apply Itô's lemma to $Y_t = \ln S_t$. The first derivative is $1/S$, the second is $-1/S^2$, and the lemma's extra half-second-derivative term contributes $\tfrac12(-1/S^2)(\sigma S)^2 = -\tfrac12\sigma^2$:

$$dY_t = \left(\mu - \tfrac{1}{2}\sigma^2\right)dt + \sigma\,dW_t .$$

The right-hand side no longer mentions $Y$, so it integrates directly and exponentiates back:

$$S_T = S_0 \exp\!\left[\left(\mu - \tfrac{1}{2}\sigma^2\right)T + \sigma W_T\right].$$

In words: the log price is a straight-line drift plus a Brownian wobble, so the price itself is lognormal.

Numbers, with $S_0 = 100$, $\mu = 0.08$, $\sigma = 0.20$, $T = 1$:

• log-drift: $\mu - \tfrac12\sigma^2 = 0.08 - 0.02 = 0.06$
• suppose the year's shock lands at $W_1 = +0.5$. Exponent $= 0.06 + 0.20(0.5) = 0.16$, so $S_1 = 100\,e^{0.16} = 117.35$
• expected price: $E[S_1] = 100\,e^{0.08} = 108.33$
• median price: $100\,e^{0.06} = 106.18$

The mean and the median differ by more than two dollars, and the median is the lower one. Half of all outcomes land below 106.18 even though the average is 108.33 — a handful of large upside paths drag the mean up. And the chance of simply losing money over the year is $\Phi(-0.06/0.20) = \Phi(-0.3) = 0.38$, so 38 percent of one-year paths finish below where they started despite a positive drift. That $-\tfrac12\sigma^2$ term is where most of the intuition damage happens.

Worked example 2: simulating one you cannot solve

Most SDEs have no closed form. You step them numerically with Euler–Maruyama: replace $dt$ with $\Delta t$ and $dW$ with $\sqrt{\Delta t}\,z$, where $z$ is a standard normal draw.

$$X_{k+1} = X_k + \mu(X_k)\,\Delta t + \sigma(X_k)\,\sqrt{\Delta t}\;z_{k+1}.$$

Take a short rate that is pulled toward 4 percent: $dr = 3(0.04 - r)\,dt + 0.02\,dW$, starting at $r_0 = 0.06$, stepping quarterly so $\Delta t = 0.25$ and $\sigma\sqrt{\Delta t} = 0.02 \times 0.5 = 0.01$. Suppose the four normal draws are $z = (+0.8,\,-1.2,\,+0.3,\,-0.5)$.

step	pull term $3(0.04-r)\Delta t$	shock $0.01 z$	new $r$
1	$3(-0.020)(0.25) = -0.01500$	$+0.0080$	$0.05300$
2	$3(-0.013)(0.25) = -0.00975$	$-0.0120$	$0.03125$
3	$3(+0.00875)(0.25) = +0.00656$	$+0.0030$	$0.04081$
4	$3(-0.00081)(0.25) = -0.00061$	$-0.0050$	$0.03520$

Read the middle column down: it is negative while the rate sits above 4 percent, flips positive the moment the rate falls below, and shrinks to almost nothing once the rate is close. That is a drift that depends on the state — the thing an ODE-shaped intuition keeps missing. The path goes 6.00, 5.30, 3.13, 4.08, 3.52 percent: overshooting, getting hauled back, overshooting less. This is the Ornstein–Uhlenbeck process, the workhorse for spreads and rates.

Set the explorer below to gbm and drag the drift to zero: the fan of paths still spreads out, it just stops leaning. Then switch to reverting and watch the fan stop widening altogether — the drift term is now pulling every path back and the spread reaches a ceiling. Same equation shape, completely different long-run behaviour, and the only thing that changed is whether $\mu$ depends on $X$.

What this means in practice

• Pricing. Black–Scholes is one SDE (geometric Brownian motion) plus a change of drift under the risk-neutral measure. Heston is two coupled SDEs, one for the price and one for its variance. Reading a pricing paper is largely reading which SDE the author chose.
• Choosing the diffusion term. $\sigma\,dW$ (constant, additive) lets $X$ go negative — fine for spreads, wrong for prices. $\sigma X\,dW$ keeps $X$ positive but makes vol proportional to level. $\sigma\sqrt{X}\,dW$ sits between and is why CIR rates stay non-negative. The shape of $\sigma$ is a modelling decision with visible consequences at the boundaries.
• Calibration. The drift is nearly unobservable on any realistic sample, as the 6.25-year calculation showed. The diffusion is estimable from high-frequency data almost exactly, because quadratic variation converges quickly. Trust your fitted $\sigma$; treat your fitted $\mu$ as a prior.
• Risk systems. Scenario generators, VaR engines and stress libraries are all SDE simulators. If your engine steps too coarsely near a barrier or a default boundary, discretisation error shows up as understated tail risk rather than as an obvious bug.