Actuarial Notes

Euler-Maruyama Discretization

Study guide: Exam INV-201

Euler-Maruyama discretization approximates the solution of a stochastic differential equation by stepping it forward on a time grid, replacing dt with a step \Delta t and the Brownian increment dW with \sqrt{\Delta t} times a standard normal draw. Repeated over many simulated paths under the risk-neutral measure, it turns an SDE into a Monte Carlo price for an option that has no closed form. When the model's transition density, the distribution of X_{t+\Delta t} given X_t, is known, the path can instead be simulated exactly and the discretization error disappears.

dX_t = a(t, X_t)\,dt + b(t, X_t)\,dW_t

X_{k+1} = X_k + a(t_k, X_k)\,\Delta t + b(t_k, X_k)\sqrt{\Delta t}\,Z_{k+1}

\hat V_0 = e^{-rT}\,\frac{1}{N}\sum_{n=1}^{N} h\!\left(X_T^{(n)}\right)

\text{RMSE}^2 = \text{bias}^2 + \frac{s_h^2}{N}

  • Symbols: a is the drift and b the diffusion coefficient of the SDE, t_k = k\,\Delta t for k = 0, 1, \dots, m with m\,\Delta t = T, and Z_1, Z_2, \dots are independent standard normal draws. In the price, h is the option's payoff, X_T^{(n)} is the end of simulated path n out of N, and s_h is the sample standard deviation of the discounted payoffs.
  • Two errors: The discretization bias comes from the step size and shrinks only as \Delta t shrinks. The sampling error comes from the finite number of paths, and its standard error s_h/\sqrt{N} halves only when N is quadrupled, as the central limit theorem says. More paths do nothing for the bias, and finer steps do nothing for the sampling error.
  • Convergence: Euler-Maruyama has strong order \tfrac12, so the error on an individual path falls like \sqrt{\Delta t}, and weak order 1, so the error in an expectation such as an option price falls like \Delta t. A European price needs only weak convergence. Path-dependent payoffs that monitor a barrier or a running maximum are more sensitive to the step, because the path can cross the level between grid points.
  • Exact simulation of GBM: Under Black-Scholes, \ln S has normal increments, so a lognormal step is exact for any \Delta t: S_{t+\Delta t} = S_t\exp\big((r - \tfrac12\sigma^2)\Delta t + \sigma\sqrt{\Delta t}\,Z\big), and the lognormal distribution is the transition density. A European option then needs a single step to T.
  • Exact simulation of Vasicek: For dr = a(b - r)\,dt + \sigma\,dz, the transition density is normal, with mean r_t e^{-a\Delta t} + b(1 - e^{-a\Delta t}) and variance \sigma^2(1 - e^{-2a\Delta t})/(2a).
  • Exact simulation of CIR: For dr = a(b - r)\,dt + \sigma\sqrt{r}\,dz, r_{t+\Delta t}/c has a noncentral chi-squared distribution with c = \sigma^2(1 - e^{-a\Delta t})/(4a), d = 4ab/\sigma^2 degrees of freedom and noncentrality \lambda = r_t e^{-a\Delta t}/c. An Euler step of CIR can make the rate negative, after which \sqrt{r} is undefined. Exact sampling never can, and it stays positive when the Feller condition 2ab \geq \sigma^2 holds.
  • When there is no transition density: Stochastic volatility, local volatility and most models in an economic scenario generator have no closed-form transition density, so Euler-Maruyama, often on the log of the price, is the default. The step size and the number of paths are then chosen together, so that neither the bias nor the standard error dominates the other.

Three Euler Steps Against the Exact Draw for GBM An equity index follows geometric Brownian motion with S_0 = 100, r = 0.04 and \sigma = 0.20 under the risk-neutral measure. Simulate one path to T = 1 with three Euler-Maruyama steps of \Delta t = \tfrac13, using the normal draws Z_1 = 0.50, Z_2 = -1.20 and Z_3 = 0.80. Then use the same three draws to take the exact lognormal path, and compare the discounted payoffs of a 1-year at-the-money call.

Answer Each Euler step multiplies by 1 + r\Delta t + \sigma\sqrt{\Delta t}\,Z_k, where r\Delta t = 0.013333 and \sigma\sqrt{\Delta t} = 0.115470.

\begin{align*}
S_1 &= 100(1 + 0.013333 + 0.115470 \times 0.50) \\
&= 107.1068 \\
S_2 &= 107.1068(1 + 0.013333 - 0.115470 \times 1.20) \\
&= 93.6938 \\
S_3 &= 93.6938(1 + 0.013333 + 0.115470 \times 0.80) \\
&= 103.5981
\end{align*}

The exact path adds up the log increments. The draws sum to 0.10.

\begin{align*}
\ln\frac{S_T}{S_0} &= (0.04 - 0.02)(1) + 0.115470(0.10) \\
&= 0.031547 \\
S_T &= 100\,e^{0.031547} \\
&= 103.2050
\end{align*}

The call pays 3.5981 on the Euler path and 3.2050 on the exact one, worth e^{-0.04}(3.5981) = 3.4570 and 3.0793 today. One path proves nothing about bias, because each path is noisy. Averaged over many paths, the three-step scheme gives E[S_T] = 100(1.013333)^3 = 104.0536 against the true 100e^{0.04} = 104.0811. With twelve steps the shortfall falls from 0.0275 to 0.0069, about a quarter, as weak order 1 predicts.

Exact and Euler Steps of the Vasicek Model A short rate follows the Vasicek model dr = 0.15(0.05 - r)\,dt + 0.01\,dz under the risk-neutral measure, with r_0 = 0.02. Using the draw Z = -1.5, simulate r_1 with a single Euler step of one year and with the exact transition density. Explain the difference.

Answer The Euler step uses the drift and the volatility at r_0 for the whole year.

\begin{align*}
r_1^{\text{EM}} &= 0.02 + 0.15(0.05 - 0.02)(1) + 0.01(-1.5) \\
&= 0.0095
\end{align*}

The exact step uses e^{-0.15} = 0.860708 and the transition variance.

\begin{align*}
E[r_1] &= 0.02(0.860708) + 0.05(0.139292) \\
&= 0.024179 \\
\text{SD}[r_1] &= 0.01\sqrt{\frac{1 - e^{-0.30}}{0.30}} \\
&= 0.009295 \\
r_1^{\text{exact}} &= 0.024179 - 1.5(0.009295) \\
&= 0.010237
\end{align*}

The Euler mean, 0.0245, overshoots because it keeps pulling toward b at the full initial speed, and its standard deviation, 0.01, ignores the mean reversion that damps shocks within the year. The exact draw is right for any step, so a model with a known transition density should be stepped with it.

How it appears on the exam

  • INV-201: The objective asks candidates to "understand and apply numerical discretization methods to price options including Euler-Maruyama discretization and transition density methods". Expect to write the Euler scheme for a given SDE, take a few steps with given normal draws, and price an option from the simulated end values.
  • Exact versus approximate: Questions ask which models can be simulated exactly and how. GBM steps are lognormal, Vasicek steps are normal, and CIR steps are scaled noncentral chi-squared. The candidate then explains why an Euler step of CIR can go negative.
  • Error analysis: Written parts ask for the sources of error in a Monte Carlo price, how the bias and the standard error each respond to the step size and the number of paths, and why a barrier or lookback payoff needs finer steps than a European one.

Exam traps

  • The square root of the step: The diffusion term is b\sqrt{\Delta t}\,Z, not b\,\Delta t\,Z. Using \Delta t shrinks the volatility of a monthly step by a factor of \sqrt{12} and badly underprices options.
  • Which drift: Use the risk-neutral drift r, or r - q with a dividend yield, to price, and the real-world drift \mu only to project. An option price simulated with \mu is not an arbitrage-free price.
  • Mixing the two GBM schemes: The Euler step multiplies by 1 + r\Delta t + \sigma\sqrt{\Delta t}Z, while the exact step multiplies by e^{(r - \sigma^2/2)\Delta t + \sigma\sqrt{\Delta t}Z}. Putting -\tfrac12\sigma^2 in the Euler factor, or leaving it out of the exponent, biases the drift.
  • Same draws, same path: When a question supplies draws for several steps, use them in the order given and do not reuse a draw across steps. For two correlated factors, build the second draw as \rho Z_1 + \sqrt{1 - \rho^2}\,Z_2.
  • Discounting: Discount the average payoff once at the risk-free rate over the whole term. With a stochastic short rate, each path's payoff is discounted along its own path, by \exp(-\sum_k r_k\Delta t), before averaging.