Passing the Picard Limit Through the SDE Integrals

From Convergent Iterates to an SDE Solution

The Picard construction produces a sequence of adapted continuous processes by repeatedly substituting the previous process into the drift and diffusion coefficients. Convergence of those processes alone does not yet establish that the limit solves the SDE: both integrals in the defining equation must also converge. The drift term is controlled by an ordinary integral inequality, while the stochastic term requires the Itô isometry. These estimates are the key step connecting the Picard approximation to an actual solution.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix a finite horizon \(T>0\). Let \(\xi\) be \(\mathcal{F}_0\)-measurable with \(\mathbb{E}|\xi|^2<\infty\), and consider the SDE with drift \(\mu(t,x)\) and diffusion coefficient \(\sigma(t,x)\). Assume both coefficients are measurable in \(t\), globally Lipschitz in \(x\) uniformly in \(t\), and of linear growth. More precisely, for constants \(L,C<\infty\), assume \(|\mu(t,x)-\mu(t,y)|\le L|x-y|\), \(|\sigma(t,x)-\sigma(t,y)|\le L|x-y|\), and \(|\mu(t,x)|+|\sigma(t,x)|\le C(1+|x|)\).

Define the Picard iterates by \(X^0_t=\xi\) and \(X^{n+1}_t=\xi+\int_0^t\mu(s,X^n_s)\,ds+\int_0^t\sigma(s,X^n_s)\,dB_s\). The second integral is an Itô integral. The earlier Picard convergence estimates and completeness result provide convergence \(X^n\to X\) in \(S^2\); this tutorial establishes that the limit passes through the two integrals. Linear growth and the finite second moments ensure that each coefficient process used as an Itô integrand belongs to \(\mathcal{L}^2\). In particular, \(\mathbb{E}\int_0^T|\sigma(s,X^n_s)|^2ds\le C_1\int_0^T(1+\mathbb{E}|X^n_s|^2)ds<\infty\) for a finite constant \(C_1\), and the same bound applies to \(X\). The processes are progressively measurable because they are adapted and continuous and the coefficients are measurable.

Continuity of the Drift and Stochastic Integrals

Proof. Put \(\Delta^n=Y^n-Y\). For every \(t\le T\), the Lipschitz condition and Cauchy–Schwarz give \[ \left|\int_0^t\big(\mu(s,Y^n_s)-\mu(s,Y_s)\big)\,ds\right|^2 \le t\int_0^t L^2|\Delta^n_s|^2\,ds \le L^2T^2\sup_{0\le s\le T}|\Delta^n_s|^2. \] Taking the supremum over \(t\), then expectation, shows that the squared \(S^2\) error for the drift is at most \(L^2T^2\|Y^n-Y\|_{S^2}^2\), which tends to zero.

For the stochastic terms, \(H^n_s=\sigma(s,Y^n_s)-\sigma(s,Y_s)\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T|H^n_s|^2ds\le L^2T\|Y^n-Y\|_{S^2}^2<\infty\). Thus it belongs to \(\mathcal{L}^2\). The Itô integral \(M^n_t=\int_0^tH^n_s\,dB_s\) is a square-integrable martingale: for \(u\le t\), its defining simple-integrand approximations have conditional mean-zero increments after time \(u\), and the Itô-isometry limit preserves the conditional expectation. The Itô isometry gives \(\mathbb{E}|M^n_T|^2=\mathbb{E}\int_0^T|H^n_s|^2ds\). Apply the standard Doob \(L^2\) maximal inequality, which states that \(\mathbb{E}\sup_{t\le T}|M_t|^2\le4\mathbb{E}|M_T|^2\) for a square-integrable martingale starting at zero. This named inequality is used here in its standard form; its proof is beyond this tutorial. We obtain \[ \mathbb{E}\sup_{t\le T}|M^n_t|^2 \le4L^2T\|Y^n-Y\|_{S^2}^2\longrightarrow0. \] Both integral terms therefore converge in \(S^2\), as claimed. \(\square\)

The maximal inequality is useful because it controls the entire time-indexed stochastic integral with one estimate. The Itô isometry alone controls its value at a fixed time; it does not, by itself, bound the supremum over all times. The integrability hypothesis matters as well: it ensures the integrands are in \(\mathcal{L}^2\), so the Itô isometry and square-integrable martingale property apply.

The Picard Limit Satisfies the Integral Equation

Proof. For each \(n\), the Picard definition gives \[ X^{n+1}_t=\xi+\int_0^t\mu(s,X^n_s)\,ds+\int_0^t\sigma(s,X^n_s)\,dB_s. \] Since \(X^n\to X\) in \(S^2\), the proposition shows that the drift and stochastic-integral terms on the right converge in \(S^2\) to the corresponding terms evaluated at \(X\). Also \(X^{n+1}\to X\) in \(S^2\), because deleting the first term of a convergent sequence does not change its limit. Consequently, subtracting the proposed limiting identity from the Picard identity and using the triangle inequality in \(S^2\) gives \[ \left\|X-\xi-\int_0^\cdot\mu(s,X_s)\,ds-\int_0^\cdot\sigma(s,X_s)\,dB_s\right\|_{S^2}=0. \] A nonnegative random variable with expectation zero is zero almost surely. Therefore the supremum over \(t\in[0,T]\) of the absolute residual is zero almost surely. The equation holds simultaneously for every \(t\in[0,T]\) outside one null set, which is precisely indistinguishability of the two continuous processes. This proves that the Picard limit satisfies the integral equation. \(\square\)

This argument proves the existence part of the Picard method once the stated convergence in \(S^2\) has been established. It does not prove uniqueness; that is a separate claim. In particular, existence here means a strong solution on the given filtered probability space: the process is adapted to the filtration carrying the given Brownian motion. The proof has not established any uniqueness property.

Worked Examples of Passing the Limit

A Common Mistake

A stochastic integral cannot be evaluated as though it were an ordinary time integral. For example, for the constant integrand \(H_s=1/2\) on \([0,2]\), an incorrect calculation would write \(\int_0^2(1/2)\,dB_s=(1/2)\cdot2=1\). The correct Itô integral is \((1/2)B_2\), which has mean zero and second moment \((1/4)\cdot2=1/2\). The deterministic integral \(\int_0^2(1/2)\,ds=1\) is a different object. In the limit argument, the stochastic-integral difference is controlled by the Itô isometry, not by ordinary integration of its integrand.

The central logic is therefore precise: convergence of the inputs in \(S^2\), together with global Lipschitz bounds, makes both integral maps continuous in \(S^2\). Applying that continuity to the Picard recursion and taking its limit yields the integral equation, with equality for all times outside a single null set.