Proving Weak Order One for Euler-Maruyama

From a One-Step Error to a Global Weak Error

Euler–Maruyama approximates an SDE by freezing its drift and diffusion coefficients at the start of each time step. Its strong error measures the difference between simulated and exact paths; its weak error instead measures the difference between expectations of test functions of their endpoints. For sufficiently smooth coefficients and test functions, the weak error at a fixed terminal time is of order $h$, where $h$ is the time-step size.

The proof has two parts. First, compare one Euler step with the exact evolution for that step, but apply both to a suitable smooth function. The resulting conditional expectation error is of order $h^2$. Second, use the Markov property and conditional expectations to add these local errors over the grid. There are $T/h$ steps, so the total bound is of order $(T/h)h^2=Th$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider the scalar SDE, on a fixed finite interval $[0,T]$,

dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t,\qquad X_0=x_0.

The drift and diffusion are the functions $\mu$ and $\sigma$. Assume for now that both are bounded and globally Lipschitz. The exact solution is the strong solution of the corresponding integral equation; under these conditions it exists and is pathwise unique. The stochastic integrand $s\mapsto \sigma(X_s)$ is adapted and progressively measurable, and boundedness gives $\mathbb{E}\int_0^T \sigma(X_s)^2\,ds<\infty$, so it belongs to $\mathcal{L}^2$ and its Ito integral is defined.

For an integer $N\ge1$, set $h=T/N$ and $t_k=kh$. The Euler–Maruyama approximation is defined recursively by

Y_0=x_0,\qquad Y_{k+1}=Y_k+\mu(Y_k)h+\sigma(Y_k)(B_{t_{k+1}}-B_{t_k}),\quad 0\le k<N.

Here $Y_k$ is $\mathcal{F}_{t_k}$-measurable, and the next Brownian increment is independent of $\mathcal{F}_{t_k}$ with distribution $N(0,h)$. For a test function $f$, the terminal weak error is $\left|\mathbb{E}f(X_T)-\mathbb{E}f(Y_N)\right|$.

The Smoothness Assumptions and the Main Bound

Define the generator $\mathcal{L}$ by $\mathcal{L}g(x)=\mu(x)g'(x)+\tfrac12\sigma(x)^2g''(x)$. The key auxiliary function is $u(t,x)=\mathbb{E}[f(X_{T-t}^{x})]$, where $X_s^x$ denotes the solution started from $x$. It describes the expected terminal payoff when the current state is $x$ at time $t$.

Regularity assumption: Assume $f$ and the coefficients are sufficiently smooth that $u$ is a classical solution of $u_t+\mathcal{L}u=0$, $u(T,x)=f(x)$, and that $u_t,u_{tt},u_x,u_{tx},u_{xx},u_{txx},u_{xxx},u_{xxxx}$ are bounded on $[0,T]\times\mathbb{R}$. A standard Kolmogorov regularity theorem supplies this type of regularity under suitable smooth bounded coefficient and payoff hypotheses; that theorem is not proved here. The stated bounds on $u$, together with the backward equation, are the precise regularity conditions used below.

The backward equation is consistent with Ito's formula. Applied to $u(s,X_s)$, that formula uses the function $u(s,x)$, derivatives $u_t,u_x,u_{xx}$, and the correction $\tfrac12\sigma(X_s)^2u_{xx}\,ds$: its drift is $(u_t+\mathcal{L}u)(s,X_s)\,ds=0$. The remaining stochastic integral has integrand $\sigma(X_s)u_x(s,X_s)$, which is adapted and progressively measurable and belongs to $\mathcal{L}^2$ because both factors are bounded. Thus it is a true martingale, and its conditional expectation over a time step is zero.

Theorem: Under the assumptions above, there is a finite constant $C$, independent of $N$, such that $$ \left|\mathbb{E}f(X_T)-\mathbb{E}f(Y_N)\right|\le CTh. $$ In particular, the Euler–Maruyama weak error is of order one as $h\to0$ with $T$ fixed.

One-Step Expansion

We first establish the estimate that drives the proof. Fix $t$, $x$, and a step size $h$ with $0\le t\le t+h\le T$. Write $m=\mu(x)$, $s=\sigma(x)$, let $Z\sim N(0,1)$, and set $\delta=mh+s\sqrt{h}Z$. The Euler step from $x$ is $x+\delta$.

One-step estimate: There is a constant $C$, uniform in $t,x,h$, such that $$ \left|\mathbb{E}\!\left[u(t+h,x+\delta)\right]-u(t,x)\right|\le Ch^2. $$

Proof. Expand first in the spatial variable at time $t+h$, through the cubic term. Boundedness of $u_{xxxx}$ bounds the remainder in absolute value by $\|u_{xxxx}\|_\infty|\delta|^4/24$. Taking expectations gives

\mathbb{E}u(t+h,x+\delta) =u(t+h,x)+u_x(t+h,x)\mathbb{E}\delta +\tfrac12u_{xx}(t+h,x)\mathbb{E}\delta^2 +\tfrac16u_{xxx}(t+h,x)\mathbb{E}\delta^3+R,

where $|R|\le \|u_{xxxx}\|_\infty\mathbb{E}|\delta|^4/24$. The normal moments give

\mathbb{E}\delta=mh,\quad \mathbb{E}\delta^2=s^2h+m^2h^2,\quad \mathbb{E}\delta^3=3ms^2h^2+m^3h^3.

These identities follow by expanding powers of $mh+s\sqrt h Z$, using $\mathbb{E}Z=0$, $\mathbb{E}Z^2=1$, $\mathbb{E}Z^3=0$, and $\mathbb{E}Z^4=3$. Since $m$ and $s$ are bounded, $|\mathbb{E}\delta^3|\le C_1h^2$ for $h\le T$. Also, $|a+b|^4\le8(|a|^4+|b|^4)$, so $\mathbb{E}|\delta|^4\le8(m^4h^4+3s^4h^2)\le C_2h^2$ for $h\le T$.

Expand the time-dependent terms at $t$. The bounded derivatives assumed above imply $u(t+h,x)=u(t,x)+hu_t(t,x)+r_0$, with $|r_0|\le \tfrac12\|u_{tt}\|_\infty h^2$; $|u_x(t+h,x)-u_x(t,x)|\le h\|u_{tx}\|_\infty$; and $|u_{xx}(t+h,x)-u_{xx}(t,x)|\le h\|u_{txx}\|_\infty$. Substitute these expansions into the preceding expectation. The terms of order $h$ are exactly $$ h\left(u_t(t,x)+m u_x(t,x)+\tfrac12s^2u_{xx}(t,x)\right) =h(u_t+\mathcal{L}u)(t,x)=0. $$ All remaining terms are bounded by a constant times $h^2$: this follows from the three time-remainder bounds, the $m^2h^2$ part of $\mathbb{E}\delta^2$, the cubic-moment bound, and the fourth-moment bound for $R$. The constants are uniform in $t,x$, since the derivatives and coefficients are bounded. This proves the estimate. $\square$

Telescoping Conditional Expectations

Proof of the theorem. Apply the one-step estimate at each grid point, with the current state $Y_k$. Given $\mathcal{F}_{t_k}$, the increment $B_{t_{k+1}}-B_{t_k}$ has the same distribution as $\sqrt h Z$, independently of $Y_k$. Therefore

\left|\mathbb{E}\!\left[u(t_{k+1},Y_{k+1})-u(t_k,Y_k)\mid\mathcal{F}_{t_k}\right]\right|\le Ch^2 \quad\text{almost surely}.

Taking expectations preserves the bound. Sum the expectation differences over $k=0,\ldots,N-1$. The intermediate terms cancel, leaving $\left|\mathbb{E}u(T,Y_N)-u(0,x_0)\right|\le NCh^2$. By $u(T,x)=f(x)$, the first term is $\mathbb{E}f(Y_N)$. By its definition, $u(0,x_0)=\mathbb{E}f(X_T)$. Finally $Nh=T$, so the right side is $CTh$, as claimed. $\square$

This argument compares expectations, not paths. The conditional expectation at each step is essential: it lets the local estimate apply at the random state $Y_k$, while the backward function carries the remaining time-to-terminal payoff.

Worked Expansions

Worked Example: Brownian Motion and a Fourth-Power Payoff

Take $dX_t=dB_t$, $X_0=0$, and $f(x)=x^4$. Calculate the expected payoff after one step of length $h=\tfrac12$ and verify the leading terms in the expansion.

Here $\mu=0$, $\sigma=1$, and the Euler increment is $\delta=\sqrt h Z$. Directly, $\mathbb{E}f(\delta)=\mathbb{E}[\delta^4]=3h^2=3(\tfrac12)^2=\tfrac34$. For the generator, $\mathcal{L}g=\tfrac12g''$. Thus $\mathcal{L}f(x)=6x^2$ and $\mathcal{L}^2f(x)=6$. At $x=0$, the expansion through second order is $f(0)+h\mathcal{L}f(0)+\tfrac12h^2\mathcal{L}^2f(0)=0+0+\tfrac12(\tfrac14)6=\tfrac34$.

Check: $X_h=\sqrt h Z$, so its fourth moment is $3h^2=\tfrac34$, exactly the computed Euler expectation. In this constant-coefficient case the Euler step has the exact transition distribution.

Worked Example: A Drifted Brownian Increment and Its Moments

Take $dX_t=dt+dB_t$, start from $x=0$, use one step of length $h=\tfrac12$, and again take $f(x)=x^4$. Compute the exact expected payoff.

The Euler endpoint is $\delta=h+\sqrt h Z$. Expanding its fourth power and using the normal moments gives $\mathbb{E}\delta^4=h^4+6h^3+3h^2$. At $h=\tfrac12$, these three terms are $\tfrac1{16}$, $\tfrac34$, and $\tfrac34$, respectively. Hence $\mathbb{E}\delta^4=\tfrac1{16}+\tfrac{12}{16}+\tfrac{12}{16}=\tfrac{25}{16}$.

Check: the exact solution is $X_h=h+B_h$, which has the same distribution as $\delta$. Its fourth moment is therefore the same $\tfrac{25}{16}$. This check confirms the expansion of the increment, including the contributions involving the drift.

Worked Example: Two Euler Steps for a Quadratic Payoff

For the same SDE $dX_t=dt+dB_t$, start at $x_0=1$, take $T=1$, and use $N=2$, so $h=\tfrac12$. Find $\mathbb{E}[Y_2^2]$ for $f(x)=x^2$, and compare it with the exact expectation.

After two steps, $Y_2=1+\tfrac12+\tfrac12+(B_1-B_0)=2+B_1$. Since $\mathbb{E}B_1=0$ and $\mathbb{E}B_1^2=1$, $\mathbb{E}[Y_2^2]=\mathbb{E}[(2+B_1)^2]=4+0+1=5$. The exact solution at time $1$ is $X_1=1+1+B_1=2+B_1$, so $\mathbb{E}[X_1^2]=5$ as well.

Check: the exact and Euler endpoints agree here because both coefficients are constant. Their second moments are each $2^2+1=5$, consistent with the direct calculation.

Why the Order Is One—and a Common Pitfall

The one-step estimate is a weak, conditional expectation estimate of order $h^2$, not a claim that the pathwise one-step error is of that order. The total bound comes from summing over $N=T/h$ steps. This proof also shows why smoothness matters: the Taylor expansion requires bounded derivatives, and the cancellation requires the backward equation. For less regular payoffs or coefficients, this argument may not apply as stated, and a weak order-one bound needs separate justification.

A common mistake is to confuse the local and global orders. For instance, if $h=T/N$ and a local defect is bounded by $Ch^2$, writing the global bound as $Ch^2$ omits the sum over all $N$ steps. The correct calculation is $NCh^2=(T/h)Ch^2=CTh$. The conclusion is first-order global weak error, not second-order global weak error.

Takeaway: Smoothness allows a one-step expansion; the backward equation cancels its order-$h$ term; conditional telescoping accumulates $N$ errors of order $h^2$, yielding a global weak error of order $h$.

Check Your Understanding

Use the assumptions and proof steps above to answer the following questions.

Why does the term of order $h$ vanish in the one-step expansion?
For $N=T/h$ steps, what bound results from summing $N$ local errors, each bounded by $Ch^2$?
Which normal moments determine $\mathbb{E}(mh+s\sqrt h Z)^2$ and $\mathbb{E}(mh+s\sqrt h Z)^3$?
Why is the Brownian increment at step $k$ independent of the Euler state $Y_k$?
Does the theorem assert that the pathwise difference between $X_T$ and $Y_N$ is of order $h$? Explain the distinction.