How to solve a stochastic differential equation: the four moves quant interviews test

When an interviewer writes an SDE on the board and says "solve it", they rarely mean "find the solution as a function of the Brownian path". Sometimes they do, for geometric Brownian motion or the Ornstein–Uhlenbeck process. More often they want a number: the mean at time one, the probability of hitting a barrier first, the variance in the long run. Each of those has a shortcut that avoids solving anything, and taking the long way round is the most common way candidates run out of time.

Move 1: apply Itô's lemma to the right function

For multiplicative noise, look for a function that turns the equation into one with constant coefficients. Geometric Brownian motion, dS = μS dt + σS dW, is the standard case. The logarithm removes the S on the right:

d(ln S) = dS/S − ½ (dS)²/S² = (μ − ½σ²) dt + σ dW

Integrate both sides and S_t = S₀ exp((μ − ½σ²)t + σW_t). The −½σ² comes from the (dS)² term, which ordinary calculus would drop and Itô calculus cannot, because (dW)² = dt. It is the single most important correction in the subject, and the one interviewers listen for.

It has a consequence worth knowing in numbers. Take S₀ = 100, μ = 6% and σ = 25% over one year. The mean is 100e^0.06 ≈ 106.2, but the median is 100e^{0.06 − 0.03125} ≈ 102.9. Half of all paths end below 102.9. The difference is volatility drag, and it is why the growth rate of wealth is μ − ½σ², not μ.

Move 2: an integrating factor, for linear equations with additive noise

The Ornstein–Uhlenbeck process, dX = −θX dt + σ dW, pulls X back towards zero. It is linear in X and the noise does not depend on X, which is the signal to use the same trick as for a linear ODE. Multiply by e^θt:

d(e^θtX) = θe^θtX dt + e^θt dX = σe^θt dW

The drift cancels, so X_t = e^−θtX₀ + σ∫₀^t e^−θ(t−s) dW_s. The integral of a deterministic function against dW is normal, with variance given by the Itô isometry, so X_t is normal with mean e^−θtX₀ and variance σ²(1 − e^−2θt)/(2θ). As t grows the variance settles at σ²/(2θ).

With θ = 2 and σ = 0.4 the long-run standard deviation is 0.2, and a deviation halves in ln 2/θ ≈ 0.35 time units. That half-life is what a pairs trader estimates first, because it says how long a position on a mean-reverting spread should take to work.

Move 3: when you only need moments, don't solve at all

Take expectations of both sides. The dW term has mean zero whatever multiplies it, so the mean obeys an ordinary differential equation, and the noise disappears from it entirely. For the second moment, apply Itô to X² first and then take expectations.

The CIR model for interest rates, dr = κ(θ − r) dt + σ√r dW, has no Gaussian solution, and you do not need one. With κ = 1.5, θ = 3%, σ = 0.15 and r₀ = 5%, the mean solves m′ = κ(θ − m), so E[r₁] = 0.03 + 0.02e^−1.5 ≈ 3.45%. Itô on r² gives the stationary variance θσ²/(2κ) = 0.000225, a standard deviation of 1.5 percentage points. And because 2κθ = 0.09 exceeds σ² = 0.0225 (the Feller condition), the rate never touches zero.

The same move answers a question that looks harder than it is. For dX = (aX + b) dt + σX dW the volatility does not appear in the mean at all: m′ = am + b, whatever σ is. Candidates who start solving the SDE lose five minutes finding that out.

Move 4: for hitting probabilities, find the scale function

"Which barrier does X reach first?" is answered by a function s with s(X_t) a martingale. For dX = μ dt + σ dW that means ½σ²s″ + μs′ = 0, so s(x) = e^−2μx/σ². Optional stopping then gives the probability of reaching b before a, starting from x:

P = (s(a) − s(x)) / (s(a) − s(b))

With μ = 1, σ = 2, starting at 0, the probability of reaching +2 before −1 is (e^0.5 − 1)/(e^0.5 − e⁻¹) ≈ 0.506. The upper barrier is twice as far, and the drift almost exactly makes up for it. The expected time to leave the interval follows from a second martingale, X_t − μt: E[X_T] = μE[T], which gives about 0.52.

Let the far barrier go to infinity and the same formula says a process drifting upwards ever falls by a with probability e^−2μa/σ². The largest drawdown of a drifting Brownian motion is exponentially distributed, which is a useful thing to have in your head in a risk interview.

When none of them works: simulate, carefully

For anything else, Euler–Maruyama is the default: X + μ(X)Δt + σ(X)ΔW, step by step. Its pathwise error shrinks only like √Δt. The Milstein scheme adds ½σσ′((ΔW)² − Δt) and gets order Δt. When the transition is known exactly, as for geometric Brownian motion and the Ornstein–Uhlenbeck process, step with that instead and the discretisation error is zero.

Hitting times need more care than moments. When we checked the exit-time question in our SDE question bank by simulation, a step of 0.001 gave an expected exit time of 0.96 against an exact 0.924. The simulation was not noisy, it was biased: checking the barrier only at grid points misses crossings that happen between them, so every path appears to leave late. Halving the error means cutting the step by a factor of four. If an interviewer asks how you would price a barrier option by Monte Carlo, this is the problem they want you to raise, and the fix is a Brownian-bridge correction for the chance of a crossing between steps.

Which move, at a glance

Fifty questions that drill exactly these moves, each with a full solution and checked by simulation, are in the stochastic differential equations bank. The Brownian-motion groundwork behind them, Itô's lemma, martingales and Girsanov, is in the stochastic calculus bank.

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