Two Different Questions About an SDE Approximation
When an SDE is simulated on a time grid, there are at least two natural ways to ask whether the approximation is good. A pathwise question compares the approximate value with the exact value using the same Brownian noise. An expectation-based question compares averages of functions of the two values. These questions measure different errors: a small average bias need not mean that individual simulated paths are close.
Fix a finite time horizon \(T>0\). Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider the scalar SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), with deterministic initial value \(X_0=x_0\). The first integral is an ordinary time integral. The second is an Ito integral: on each finite interval, its integrand is adapted and progressively measurable, and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For the SDE, this condition applies to \(H_s=\sigma(s,X_s)\).
Assume, for the comparisons below, that the coefficients are globally Lipschitz in the state variable and satisfy linear growth, uniformly over \(0\le t\le T\). In particular, suppose there is a constant \(C\) such that \(|\mu(t,x)-\mu(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le C|x-y|\) and \(|\mu(t,x)|+|\sigma(t,x)|\le C(1+|x|)\). The standard existence-and-uniqueness theorem for Lipschitz SDEs then gives a unique strong solution with finite second moments on finite time intervals; its proof is not needed for defining the error measures here.
Take a uniform grid \(t_k=kh\), \(k=0,\ldots,N\), where \(h=T/N\). The Euler–Maruyama approximation, using the same Brownian motion as the exact solution, is defined recursively by \(X^h_0=x_0\) and \(X^h_{k+1}=X^h_k+\mu(t_k,X^h_k)h+\sigma(t_k,X^h_k)(B_{t_{k+1}}-B_{t_k})\). Thus the exact and approximate values are coupled: they are constructed on the same probability space and use the same Brownian increments. Other schemes can be compared in the same way, provided their coupling is specified.
What the Rates Say
The order is the exponent of the grid width in the relevant error measure. For example, strong order \(1/2\) means an \(L^q\) path error of size at most a constant times \(h^{1/2}\); halving \(h\) then reduces that bound by a factor \(2^{-1/2}\). Weak order \(1\) means a test-function expectation error of size at most a constant times \(h\); halving \(h\) reduces that bound by a factor \(1/2\).
Under suitable regularity and moment assumptions, Euler–Maruyama commonly has strong order \(1/2\) and weak order \(1\) for sufficiently smooth test functions. These are not unconditional guarantees: the coefficient regularity, time dependence, moment bounds, and test-function class matter. The statements also refer to different quantities, so the exponents should not be compared as if they were rates for one error. The definitions alone do not prove either convergence theorem.
| Feature | Strong convergence | Weak convergence |
|---|---|---|
| What is compared? | Exact and approximate random variables on a specified coupling | Expectations of functions of the exact and approximate variables |
| Typical error | \((\mathbb{E}|X_T-X^h_N|^q)^{1/q}\) | \(|\mathbb{E}\varphi(X_T)-\mathbb{E}\varphi(X^h_N)|\) |
| Depends on coupling? | Yes | No, if the marginal laws are fixed |
| Common Euler–Maruyama rate | \(1/2\), under suitable assumptions | \(1\), for suitable smooth tests and assumptions |
Worked Comparisons
Worked Example: Euler Error for a Deterministic Equation
Consider \(dX_t=X_t\,dt\), \(X_0=1\), on \([0,1]\). This is an SDE with diffusion coefficient zero. Compute the exact and Euler–Maruyama values on a grid with \(N=2\), and compare the errors for \(\varphi(x)=x\).
The exact solution is \(X_t=e^t\), so \(X_1=e\). Here \(h=1/2\) and each update is \(X^h_{k+1}=X^h_k(1+h)\). Starting from \(X^h_0=1\), the first value is \(X^h_1=1(1+1/2)=3/2\); the second is \(X^h_2=(3/2)(3/2)=9/4\). Because this equation has no randomness, the strong terminal error in every \(L^q\) is \(|e-9/4|=e-9/4\). The sign is positive since \(e>1+1+1/2=5/2>9/4\).
For \(\varphi(x)=x\), the weak error is also \(|e-9/4|=e-9/4\), because both expectations are just the corresponding deterministic values. Check: for this deterministic example, the terminal random variables are constants, so taking an expectation does not change either value. The equality of these two errors here is special; it is not a general identity between strong and weak errors.
Worked Example: A Strong Error and a Weak Bias
Let \(X_t=B_t\) on \([0,1]\), and approximate the terminal value by stopping one grid interval early: \(Y_h=B_{1-h}\), with \(h=1/4\). Compute the strong terminal error in \(L^2\) and the weak error for \(\varphi(x)=x^2\).
The coupled error is \(X_1-Y_h=B_1-B_{3/4}\). This increment has mean zero and variance \(1-3/4=1/4\), so \((\mathbb{E}|X_1-Y_h|^2)^{1/2}=(1/4)^{1/2}=1/2\). For the weak comparison, \(\mathbb{E}[X_1^2]=1\), while \(\mathbb{E}[Y_h^2]=3/4\), since Brownian motion at time \(3/4\) has variance \(3/4\). Thus the absolute weak bias for this test is \(|1-3/4|=1/4\).
Check: the increment calculation gives \(\mathbb{E}[(B_1-B_{3/4})^2]=1/4\), matching the squared \(L^2\) strong error. Also \(B_1=B_{3/4}+(B_1-B_{3/4})\), where the two terms are independent and the increment has mean zero. Expanding the square therefore gives \(\mathbb{E}[B_1^2]=3/4+1/4=1\), as required. This example compares two particular error values; the weak bias is not itself the strong error.
Worked Example: Identical Laws, Different Coupled Paths
On the same Brownian probability space, take \(X=B_1\) and \(Y=-B_1\). Show that the weak errors vanish for every integrable test function, even though the coupled \(L^2\) error is nonzero.
Both \(B_1\) and \(-B_1\) have the \(N(0,1)\) distribution. Consequently, for every test function \(\varphi\) for which the expectations exist, \(\mathbb{E}[\varphi(X)]=\mathbb{E}[\varphi(Y)]\), so the weak error is zero. But the coupled difference is \(X-Y=2B_1\), and hence \((\mathbb{E}|X-Y|^2)^{1/2}=(4\mathbb{E}[B_1^2])^{1/2}=2\). The paths differ except on the event \(B_1=0\), which has probability zero because \(B_1\) has a continuous normal distribution.
Check: for \(\varphi(x)=x^2\), both expectations equal \(1\), while the mean squared coupled difference is \(4\). This is an illustration of the distinction between the measures, not a proposed SDE time-stepping scheme: changing a coupling can change a strong error without changing either marginal law.
How Strong Error Controls Some Weak Errors
Proof. The Lipschitz condition gives \(|\varphi(X)-\varphi(Y)|\le L|X-Y|\) almost surely. The difference of expectations is bounded by the expectation of the absolute difference, so \(|\mathbb{E}\varphi(X)-\mathbb{E}\varphi(Y)|\le\mathbb{E}|\varphi(X)-\varphi(Y)|\le L\mathbb{E}|X-Y|\). Since \(q\ge1\), Jensen's inequality applied to the convex function \(u\mapsto u^q\) on nonnegative values gives \(\mathbb{E}|X-Y|\le(\mathbb{E}|X-Y|^q)^{1/q}\). Combining the inequalities proves the claim. If \(L=0\), then \(\varphi\) is constant and both sides of the desired weak-error inequality are zero. \(\square\)
This result shows that a strong rate \(p\) implies at least the same weak rate for Lipschitz tests under the same coupling and moment assumptions. It does not establish a higher weak rate for smoother tests; that requires additional analysis. Nor does a small weak error generally imply a small strong error, as the equal-law example demonstrates.
Common Mistake: Treating Bias as Path Error
A frequent error is to identify weak error with the expectation of the absolute path difference. In general, \(|\mathbb{E}[\varphi(X)]-\mathbb{E}[\varphi(Y)]|\) is not \(\mathbb{E}|X-Y|\), nor is it the \(L^2\) strong error. For the preceding example with \(X=B_1\), \(Y=-B_1\), and \(\varphi(x)=x\), the weak error is \(|\mathbb{E}B_1-\mathbb{E}(-B_1)|=|0-0|=0\). The \(L^2\) strong error is \(2\). Replacing the weak error by the strong error would therefore give \(2\) instead of \(0\) for this test. The correction is to state the measure being reported: a path-coupled norm for strong error, or a test-function expectation difference for weak error.
Both notions matter in simulation. Strong accuracy is important when the simulated path itself is the object of interest, or when path-dependent quantities must be coupled accurately. Weak accuracy concerns expected payoffs and other distributional averages. A reliable convergence claim specifies the time grid, the coupling if a strong error is used, the norm or test-function class, and the assumptions under which the rate holds.
Check Your Understanding
For each question, identify whether the quantity is a strong error, a weak error, or a statement about the assumptions needed to define a rate.
- For a coupled pair \(X,X^h\), write the terminal-time strong error in \(L^2\).
- For a specified test function \(\varphi\), write the weak error at time \(T\).
- Why can two variables have zero weak error for every integrable test function but a positive strong error under a particular coupling?
- State the Lipschitz-test inequality relating a weak error to an \(L^q\) strong error.
- What additional information must accompany a claim of weak order \(r\) if the bound is asserted uniformly over a class of test functions?