What Strong Order One-Half Measures
Euler–Maruyama approximates an SDE by freezing its drift and diffusion coefficients at the left endpoint of each time step. Its strong error compares the numerical value with the exact solution driven by the same Brownian path. Under global Lipschitz and linear-growth assumptions, the root-mean-square error at a fixed time is bounded by a constant times the square root of the step size. The Brownian increment is of size \(\sqrt{h}\), which is central to this rate.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For a fixed horizon \(T>0\), consider the scalar SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with deterministic initial value \(X_0=x_0\). Assume that, for constants \(L\ge0\) and \(K\ge0\), all \(x,y\in\mathbb{R}\) satisfy
The first two are global Lipschitz conditions; the third is a linear-growth condition in squared form. We use the standard existence and uniqueness theorem for SDEs with globally Lipschitz coefficients: these assumptions give a strong solution with finite second moments on finite time intervals. Here a strong solution is adapted to the given filtration and Brownian motion. Pathwise uniqueness means that two solutions driven by that same Brownian motion and with the same initial value agree almost surely. These notions concern solutions of the SDE; the numerical error below is a comparison using that same Brownian motion, not a weak comparison of distributions.
This definition takes the maximum of the individual root-mean-square errors; it does not assert a bound on \(\mathbb{E}[\max_k|X_{t_k}-Y_k|^2]\). To analyze all grid points together, define the continuous Euler interpolation. For \(s\in[0,T]\), let \(\eta(s)=t_k\) when \(t_k\le s<t_{k+1}\), and set \(\eta(T)=t_{N-1}\). Then
Every stochastic integral in this proof is an Ito integral. Its integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on \([0,T]\), meaning that \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The moment estimates below verify this integrability for the Euler coefficients; the SDE solution’s integrands are square-integrable by the stated moment assumption.
Brownian Increments Give the Local Estimate
Within one step, the interpolation satisfies \(Y_s-Y_{\eta(s)}=\mu(Y_{\eta(s)})(s-\eta(s))+\sigma(Y_{\eta(s)})(B_s-B_{\eta(s)})\). The increment \(B_s-B_{\eta(s)}\) is independent of \(\mathcal{F}_{\eta(s)}\), has mean zero, and has second moment \(s-\eta(s)\). The coefficient values at \(\eta(s)\) are measurable with respect to that past. Thus the drift–Brownian cross term has expectation zero. This is why the Brownian contribution is proportional to the length of the step, not its square.
Proof. First bound the Euler values at the grid points. Conditional on \(\mathcal{F}_{t_k}\), the Brownian increment has mean zero and variance \(h\), so \(\mathbb{E}[|Y_{k+1}|^2\mid\mathcal{F}_{t_k}]=|Y_k+h\mu(Y_k)|^2+h|\sigma(Y_k)|^2\). Using \(2Y_k\mu(Y_k)\le |Y_k|^2+|\mu(Y_k)|^2\), and \(h\le1\), gives \[ \mathbb{E}|Y_{k+1}|^2 \le \mathbb{E}|Y_k|^2+h\,\mathbb{E}\big[|Y_k|^2+3K(1+|Y_k|^2)\big]. \] Consequently \(\mathbb{E}(1+|Y_{k+1}|^2)\le(1+(1+3K)h)\mathbb{E}(1+|Y_k|^2)\). Iterating and using \(1+u\le e^u\) yields \(\mathbb{E}(1+|Y_k|^2)\le e^{(1+3K)T}(1+|x_0|^2)=M_0\). Now write \(\delta=s-\eta(s)\), so \(0\le\delta\le h\). Independence and the zero conditional mean give \[ \mathbb{E}|Y_s-Y_{\eta(s)}|^2 =\delta^2\mathbb{E}|\mu(Y_{\eta(s)})|^2+\delta\mathbb{E}|\sigma(Y_{\eta(s)})|^2 \le K M_0(\delta^2+\delta)\le 2K M_0h\le R h. \] The final inequality uses \(h\le1\). This also verifies the Euler coefficient integrability needed above. \(\square\)
Worked Example: A One-Step Euler Increment
For the SDE \(dX_t=-X_t\,dt+dB_t\), take \(x_0=0\) and examine the first Euler interval of length \(h\). On that interval \(Y_0=0\), so the Euler value is \(Y_s=B_s\). The exact solution follows from Ito’s formula applied to \(f(t,x)=e^t x\), whose derivatives are \(f_t=e^t x\), \(f_x=e^t\), and \(f_{xx}=0\). The correction term \(\tfrac12 f_{xx}\,dt\) is zero, giving \(d(e^tX_t)=e^t\,dB_t\), hence \(X_h=e^{-h}\int_0^h e^s\,dB_s\).
Therefore \(X_h-Y_h=\int_0^h(e^{-(h-s)}-1)\,dB_s\). The Ito isometry gives
The integrand is nonnegative, so the expression is nonnegative. For \(h=\tfrac12\), direct substitution gives \(-1+2e^{-1/2}-\tfrac12e^{-1}\). As \(h\) becomes small, the integral is of order \(h^3\), which is a local one-step error; accumulating such errors over a fixed horizon is consistent with a global RMS rate of order \(h^{1/2}\).
The Mean-Square Error Bound
Proof. Set \(e_t=X_t-Y_t\). Subtract the integral equations for \(X\) and \(Y\). By Cauchy–Schwarz for the drift integral and the Ito isometry for the stochastic integral,
Use the Lipschitz conditions and \(|X_s-Y_{\eta(s)}|^2\le2|e_s|^2+2|Y_s-Y_{\eta(s)}|^2\). Since \(t\le T\), the right-hand side is at most \(4L^2(T+1)\int_0^t\big(\mathbb{E}|e_s|^2+\mathbb{E}|Y_s-Y_{\eta(s)}|^2\big)\,ds\). Put \(A=4L^2(T+1)\). The lemma bounds the local term by \(Rh\), so \[ u(t):=\mathbb{E}|e_t|^2\le A\int_0^t u(s)\,ds+ARhT. \]
For completeness, the integral form of Gronwall’s inequality used here says that a nonnegative continuous function satisfying \(u(t)\le a+b\int_0^t u(s)\,ds\), for \(a,b\ge0\), obeys \(u(t)\le ae^{bt}\). If \(a>0\), set \(v(t)=a+b\int_0^t u(s)\,ds\); then \(u\le v\), \(v(0)=a\), and \(v'\le bv\), so differentiating \(e^{-bt}v(t)\) shows it is nonincreasing. If \(a=0\), apply the result with any positive \(\varepsilon\) in place of \(a\), then let \(\varepsilon\) decrease to zero. Applying this inequality with \(a=ARhT\) and \(b=A\) gives \(u(t)\le ARhT e^{AT}\). Substituting \(A=4L^2(T+1)\), \(R=4KM_0\), and \(t\le T\) gives the stated constant. If \(L=0\), both coefficients are constant in space, and the Euler values at grid points equal the exact solution there; the bound with \(C=0\) agrees. Finally, at each grid point \(Y_{t_k}=Y_k\), so \(\max_k(\mathbb{E}|X_{t_k}-Y_k|^2)^{1/2}\le\sqrt{C}\,h^{1/2}\). \(\square\)
Worked Example: Checking the Theorem for an Ornstein–Uhlenbeck SDE
Take \(dX_t=-X_t\,dt+dB_t\), \(x_0=0\), and \(T=1\). The drift and diffusion are globally Lipschitz with \(L=1\). Also \(|\mu(x)|^2+|\sigma(x)|^2=x^2+1\), so \(K=1\) works. The theorem’s constants are \(M_0=e^{(1+3)\cdot1}(1+0^2)=e^4\) and \(C=16\cdot1^2\cdot1\cdot e^4\cdot1\cdot(1+1)\cdot e^{4\cdot1^2\cdot(1+1)\cdot1}=32e^{12}\). Thus, for \(h\le1\), the theorem gives an RMS error at each grid point of at most \(\sqrt{32e^{12}h}\). With \(h=\tfrac1{16}\), the squared-error bound is \(32e^{12}/16=2e^{12}\), and the RMS bound is \(\sqrt{2}\,e^6\).
Check: substituting \(h=\tfrac1{16}\) into the squared bound \(Ch\) gives \(2e^{12}\), as above. This is a deliberately conservative uniform bound, not a prediction of the actual error’s size.
Worked Example: Constant Coefficients Give Zero Grid Error
Let \(dX_t=2\,dt+3\,dB_t\), \(x_0=1\), on any fixed horizon. Here \(\mu(x)=2\) and \(\sigma(x)=3\), so \(L=0\). The exact solution is \(X_t=1+2t+3B_t\). Euler’s recursion gives \(Y_k=Y_{k-1}+2h+3\Delta B_{k-1}\). Summing from \(0\) to \(k-1\) yields \(Y_k=1+2kh+3\sum_{j=0}^{k-1}\Delta B_j=1+2t_k+3B_{t_k}=X_{t_k}\).
Check: at \(k=0\), both values are \(1\); at every \(k\), the Brownian increments telescope to \(B_{t_k}-B_0=B_{t_k}\). Thus the grid error is exactly zero. Strong order one-half is an upper bound valid under the hypotheses; it need not be the best rate for every SDE.
Why the Rate Matters—and a Common Mistake
The estimate separates two ideas: the within-step interpolation error is controlled by Brownian increments, and the global error is controlled by stability of the dynamics. The Lipschitz conditions let coefficient differences be bounded by state differences; Gronwall’s inequality prevents those differences from growing faster than the derived bound over a finite horizon. In practice, halving the step size therefore reduces the guaranteed RMS scale by a factor of \(1/\sqrt{2}\), though the actual reduction can be better or worse at a particular resolution.
Common mistake: treating Brownian increments as order \(h\). For a step of length \(h\), \(\mathbb{E}[(\Delta B_k)^2]=h\), so the RMS increment is \(\sqrt{h}\), not \(h\). A wrong argument might write \(|\sigma(Y_k)\Delta B_k|=O(h)\) and conclude that the global error is order \(h\) without further analysis. The correction is to use the second moment: the stochastic increment contributes \(\mathbb{E}|\sigma(Y_k)\Delta B_k|^2=h\,\mathbb{E}|\sigma(Y_k)|^2\). The theorem controls its accumulated effect in mean square, producing an RMS order of one-half. This is a statement under \(\mathbb{P}\) for the same Brownian motion, not a claim about an error in distribution under another measure.
Check Your Understanding
Use the assumptions and the proof structure above to answer the following.
- Which part of the local increment estimate uses independence of the Brownian increment from the past?
- Why does the linear-growth condition help establish that the Euler stochastic integrand is square-integrable?
- In the theorem, what quantity is bounded by \(Ch\): the RMS error or its square?
- What changes in the grid error when both drift and diffusion are constant in space?
- Does the proved bound control \(\mathbb{E}[\max_k|X_{t_k}-Y_k|^2]\)? Explain the distinction.