Why Couple Two Euler Grids?
Euler-Maruyama approximations at two time resolutions are most informative when they are driven by the same Brownian path. This does not mean their numerical values should agree: in general, coarse and fine recursions differ. Rather, shared randomness isolates the effect of changing the time step. If the approximations use independently simulated Brownian paths, their difference also contains fluctuations between unrelated noise realizations.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) with respect to this filtration. Thus \(B\) is adapted, has continuous paths and independent normal increments, and \(B_t-B_s\) is independent of \(\mathcal{F}_s\) whenever \(s<t\). The filtration condition matters: adaptedness alone does not imply independence of future increments from present information.
For an SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. Its integral form is \(X_t=X_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s\). The Ito integral is defined for a progressively measurable integrand \(H_s=\sigma(s,X_s)\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); assume also that the drift integral is well-defined. The coupling construction below concerns the Brownian increments used in numerical recursions.
A strong solution is defined on a specified probability space and driven by its specified Brownian motion. A weak solution specifies a joint law for a solution and Brownian motion without fixing that realization in advance. Strong-error comparisons use values formed with the same driving noise. Pathwise uniqueness means solutions with the same initial value and Brownian motion agree almost surely at all times; uniqueness in law means their distributions agree. These are distinct properties, and the increment construction here does not prove either uniqueness property.
Nested Grids and Shared Increments
Fix \(T=NH\), with \(N\) a positive integer and \(H\) the coarse step. Choose a positive integer \(m\) and set \(h=H/m\). The coarse times are \(t_j=jH\), \(j=0,\ldots,N\), and each coarse interval is divided into \(m\) fine intervals. Define the fine Brownian increments in block \(j\) by
The coarse increment is the sum of the fine increments in its block:
For uniform grids, the coarse Euler-Maruyama recursion is
The fine recursion is applied afresh at every fine-grid state:
Both paths start from the same initial value. In simulation, generate the fine increments and sum them to form each coarse increment. Equivalently, generate independent standard normal variables \(Z_{j,k}\), set \(\Delta B_{j,k}=\sqrt{h}\,Z_{j,k}\), and use \(\sqrt{h}\sum_{k=1}^m Z_{j,k}\) as the coarse increment. Do not generate a second independent normal variable for that coarse increment.
Proof. Within a block the \(m\) fine increments are independent \(N(0,h)\) random variables. Their sum is normal with mean \(0\) and variance \(mh=H\). Increments on distinct coarse blocks are independent because Brownian increments on disjoint intervals are independent. To verify independence from \(\mathcal{F}_{t_j}\), take real numbers \(u_1,\ldots,u_m\). Brownian increments relative to the filtration give the conditional characteristic function
To obtain it, condition successively from the last increment backwards. Earlier increments are measurable at the left endpoint of the increment being conditioned on, and that Brownian increment is independent of the information at its left endpoint. The resulting conditional characteristic function is deterministic, so the vector of fine increments, and therefore its sum, is independent of \(\mathcal{F}_{t_j}\). This proves the assertions. \(\square\)
The proposition uses Brownian motion relative to the filtration, not just adaptedness. Its variance calculation also explains the scaling: the sum of \(m\) independent \(N(0,h)\) increments has variance \(mh=H\), the variance required for a Brownian increment over a coarse interval.
Worked Euler Comparisons
Worked Example: Recalculating Coefficients on a Fine Grid
Consider \(dX_t=X_t\,dt+X_t\,dB_t\), \(X_0=1\), on \([0,1]\). Use coarse step \(H=\tfrac12\), fine step \(h=\tfrac14\), and fine increments \(\tfrac14,\tfrac14,-\tfrac14,-\tfrac14\). These are illustrative simulated inputs. First check the block sums: the first is \(\tfrac14+\tfrac14=\tfrac12\), and the second is \(-\tfrac14-\tfrac14=-\tfrac12\). These, not separately generated values, are the coarse increments.
For a step of length \(q\), starting at the current state \(x\), the update is \(x+xq+x\Delta B=x(1+q+\Delta B)\). The coefficient values must be evaluated at the current state before each update. For the coarse path, the first step starts at \(1\), so both coefficients are \(1\), giving \(X^c_1=1+\tfrac12+\tfrac12=2\). The second coarse step starts at \(2\), so both coefficients are now \(2\); hence
For the fine path, the current state and the coefficients change recursively. On the first step the current state and both coefficients equal \(1\), so \(X^f_1=1+\tfrac14+\tfrac14=\tfrac32\). On the second step both coefficients are \(\tfrac32\), so
On the third step the state and coefficients are \(\tfrac94\), and the increment is \(-\tfrac14\). The drift contribution is \(\tfrac9{16}\) and the diffusion contribution is \(-\tfrac9{16}\), giving \(X^f_3=\tfrac94\). On the fourth step the state and both coefficients are again \(\tfrac94\), so the same contributions give \(X^f_4=\tfrac94\).
Check: the block sums used above are exactly \(\tfrac12\) and \(-\tfrac12\); substituting them into the two coarse updates gives terminal value \(2\). Applying all four fine updates recursively gives \(\tfrac94\), so the coupled terminal difference is \(\tfrac94-2=\tfrac14\). The fine path is not obtained by freezing the coefficient at its initial value: each displayed update uses the current fine-grid state. The coarse and fine paths need not agree because they evaluate the coefficients at different times and states.
Worked Example: An Additive SDE with Matching Endpoints
Consider \(dX_t=2\,dt+3\,dB_t\), \(X_0=1\), on \([0,1]\). Use \(H=\tfrac12\), \(h=\tfrac14\), and fine increments \(\tfrac15,-\tfrac1{10},\tfrac25,-\tfrac12\). The first block sum is \(\tfrac15-\tfrac1{10}=\tfrac1{10}\); the second is \(\tfrac25-\tfrac12=-\tfrac1{10}\).
Each coarse drift contribution is \(2H=1\), and each coarse diffusion contribution is three times its block sum. Thus
On each fine step the drift contribution is \(2h=\tfrac12\). The four total changes, including diffusion, are \(\tfrac12+3(\tfrac15)=\tfrac{11}{10}\), \(\tfrac12+3(-\tfrac1{10})=\tfrac15\), \(\tfrac12+3(\tfrac25)=\tfrac{17}{10}\), and \(\tfrac12+3(-\tfrac12)=-1\). Consequently,
Check: the four fine changes sum to \(\tfrac{11}{10}+\tfrac15+\tfrac{17}{10}-1=2\), so the fine terminal value is \(1+2=3\), equal to the coarse terminal value. For constant drift \(a\) and diffusion \(\sigma\), each Euler step adds \(ah+\sigma\Delta B\), so summing over the full interval gives \(aT+\sigma(B_T-B_0)\) on every grid. This exact agreement at the endpoint is special to constant coefficients.
Worked Example: Independent Paths Add Noise to the Difference
Take \(dX_t=dB_t\), \(X_0=0\), \(T=1\). On any grid the Euler endpoint is the sum of the Brownian increments. With shared increments, both coarse and fine sums equal \(B_1-B_0\), so their endpoint difference is zero.
If the coarse and fine simulations are independent, their endpoints \(C\) and \(F\) are independent \(N(0,1)\) random variables. Therefore
For one concrete pair of independent simulation inputs, coarse increments \(\tfrac14,\tfrac14\) give \(C=\tfrac12\). Fine increments \(\tfrac14,\tfrac14,\tfrac14,\tfrac14\) give \(F=1\), so this pair has difference \(\tfrac12\) and squared difference \(\tfrac14\). That single value need not equal the expected squared difference \(2\); the expectation averages over the independent simulations.
Check: with shared increments, the sums telescope to the same Brownian endpoint for every realization, giving squared difference \(0\). With independent paths, the calculation gives expected squared difference \(2\), even though each scheme is exact at the endpoint. The independent comparison measures variation between unrelated noise realizations, not a discretization discrepancy.
What a Strong Comparison Measures
Suppose \(X_T\) is an exact strong solution and \(X^h_T\) is its Euler approximation. A strong mean-square error compares them on the same probability space and with the same Brownian motion, through a quantity such as \(\mathbb{E}[|X^h_T-X_T|^2]\). Likewise, \(\mathbb{E}[|X^h_T-X^H_T|^2]\) is a paired comparison of fine and coarse schemes when both are built from the same Brownian path. It is not automatically the exact error of either scheme.
This differs from comparing marginal distributions. A weak comparison may concern expectations of test functions of the approximations without pairing their values path by path. Shared increments are important when the objective is a strong, pathwise comparison: both recursions respond to the same random forcing, while different time steps and coefficient evaluations may still produce different values.
For each coarse block, generate \(m\) independent \(N(0,h)\) increments, or standard normal variables multiplied by \(\sqrt h\).
Define the coarse increment as the sum of those fine increments. Do not draw a separate coarse increment.
Use the block sum in the coarse update and individual increments in the fine updates. Reevaluate drift and diffusion at the current state on every step.
Subtract values at matching times, such as the terminal time, before forming a squared error or another pathwise discrepancy.
Common Mistake: Matching Distributions Is Not Coupling
A common error is to give a coarse step a fresh \(N(0,H)\) draw while also generating the fine increments over that block. Each draw may have the correct marginal distribution, but they no longer represent the same Brownian increment. For example, with \(H=\tfrac12\) and two fine increments both equal to \(\tfrac14\), the coarse increment on that path must be \(\tfrac12\). Replacing it with a separately simulated value \(-\tfrac12\) breaks the coupling. For \(dX_t=dB_t\), the fine block change is \(\tfrac12\), whereas that assigned coarse change is \(-\tfrac12\); their difference is \(1\).
The correction is structural: generate one set of fine increments and define each coarse increment by summation. The resulting schemes retain their appropriate marginal increment distributions while their differences reflect the resolution change under common noise.
Check Your Understanding
For each question, distinguish the distribution of an increment from the way increments are paired across grids.
- If a coarse interval of length \(H\) is divided into \(m\) fine steps of length \(h\), express its coarse increment using the fine increments.
- Why does adaptedness alone not show that a future Brownian increment is independent of \(\mathcal{F}_s\)? What additional property was assumed?
- For constant drift and diffusion coefficients, explain why coupled coarse and fine Euler endpoints agree.
- For independent \(N(0,1)\) endpoints \(C\) and \(F\), compute \(\mathbb{E}[(F-C)^2]\). Why is this not a discretization error in the additive Brownian example?
- Does sharing increments force coarse and fine Euler paths to agree for nonconstant coefficients? Explain using the states at which the coefficients are evaluated.