Factorial Bounds for Successive Picard Differences

Why Successive Differences Matter

Picard iteration constructs a sequence of processes by repeatedly substituting the previous process into the drift and diffusion coefficients of an SDE. To prove that this sequence converges, one needs to control the differences between consecutive iterates. Under global Lipschitz assumptions, each new difference is bounded by a time integral of the preceding difference. Repeating that estimate integrates over successively nested time intervals, whose volume is $t^n/n!$. The factorial in this volume is the key decay mechanism.

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 an $\mathcal{F}_0$-measurable random variable with $\mathbb{E}|\xi|^2<\infty$. Assume the drift $\mu(t,x)$ and diffusion coefficient $\sigma(t,x)$ are measurable, adapted when evaluated along adapted processes, and globally Lipschitz in $x$, uniformly over $0\le t\le T$: for some $L\ge0$,

|\mu(t,x)-\mu(t,y)|\le L|x-y|, \qquad |\sigma(t,x)-\sigma(t,y)|\le L|x-y|.

Also assume linear growth: for some $A\ge0$, $|\mu(t,x)|^2+|\sigma(t,x)|^2\le A^2(1+|x|^2)$. These assumptions ensure the iterates below have finite second moments on $[0,T]$. All time integrals without $dB_s$ are ordinary Lebesgue integrals. Every stochastic integral is an Itô integral with progressively measurable integrand in $\mathcal{L}^2$, the class of processes $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: Set $X^0_t=\xi$ and define the Picard iterates by $$ X^{n+1}_t=\xi+\int_0^t\mu(s,X^n_s)\,ds+\int_0^t\sigma(s,X^n_s)\,dB_s,\qquad n\ge0. $$ The $n$-th successive difference is $D_n(t)=X^{n+1}_t-X^n_t$. Its squared $L^2$ supremum is $F_n(t)=\mathbb{E}\!\left[\sup_{0\le u\le t}|D_n(u)|^2\right]$.

In particular, $D_0$ is the difference between the first iterate and the constant initial process. The later differences measure how much each additional application of the Picard map changes the process.

The Recursive Supremum Estimate

Proposition: Under the assumptions above, for $0\le t\le T$ and every $n\ge0$, $$ F_{n+1}(t)\le C\int_0^t F_n(s)\,ds, \qquad C=L^2(2T+8). $$

Proof. Subtracting the formulas for $X^{n+2}$ and $X^{n+1}$ gives $$ D_{n+1}(u)=\int_0^u\big(\mu(s,X^{n+1}_s)-\mu(s,X^n_s)\big)\,ds +\int_0^u\big(\sigma(s,X^{n+1}_s)-\sigma(s,X^n_s)\big)\,dB_s. $$ The coefficient differences are progressively measurable. Their squared integrals have finite expectation by the Lipschitz condition and the finite second moments of the iterates, so the stochastic integral is a square-integrable martingale. The Itô isometry gives its second moment, and Doob’s $L^2$ maximal inequality gives $\mathbb{E}\sup_{u\le t}|M_u|^2\le4\mathbb{E}|M_t|^2$ for this martingale $M$. Using $|a+b|^2\le2|a|^2+2|b|^2$, Cauchy–Schwarz for the drift integral, and these stochastic-integral estimates, we obtain $$ F_{n+1}(t)\le 2t\,\mathbb{E}\int_0^t|\mu(s,X^{n+1}_s)-\mu(s,X^n_s)|^2\,ds +8\,\mathbb{E}\int_0^t|\sigma(s,X^{n+1}_s)-\sigma(s,X^n_s)|^2\,ds. $$ The Lipschitz bounds make each integrand at most $L^2|D_n(s)|^2$, and $\mathbb{E}|D_n(s)|^2\le F_n(s)$. Since $t\le T$, the right-hand side is at most $L^2(2T+8)\int_0^t F_n(s)\,ds$, as asserted. $\square$

The constant $C$ is not generally sharp. Its purpose is to give one uniform coefficient in the recursion for the entire time interval. The term $2T$ comes from estimating the drift, while $8$ includes both the factor from adding the two integrals and the factor $4$ in Doob’s inequality.

Repeated Integration Produces a Factorial

First the initial difference must be bounded. From $D_0(t)=\int_0^t\mu(s,\xi)\,ds+\int_0^t\sigma(s,\xi)\,dB_s$, the same drift and martingale estimates, now using linear growth, give the uniform bound

F_0(t)\le K_0,\qquad K_0=(2T^2+8T)A^2(1+\mathbb{E}|\xi|^2),\qquad 0\le t\le T.

Indeed, the drift contribution is bounded by $2t\int_0^t\mathbb{E}|\mu(s,\xi)|^2ds$, and the martingale contribution by $8\int_0^t\mathbb{E}|\sigma(s,\xi)|^2ds$. Each coefficient’s squared expectation is at most $A^2(1+\mathbb{E}|\xi|^2)$, yielding the displayed bound.

Theorem: For every integer $n\ge0$ and every $0\le t\le T$, $$ F_n(t)\le K_0\frac{C^n t^n}{n!}, \qquad C=L^2(2T+8). $$

Proof. For $n=0$, the bound is $F_0(t)\le K_0$, just established. Suppose it holds for some $n\ge0$. The recursive estimate then gives $$ F_{n+1}(t)\le C\int_0^t K_0\frac{C^n s^n}{n!}\,ds =K_0\frac{C^{n+1}}{n!}\frac{t^{n+1}}{n+1} =K_0\frac{C^{n+1}t^{n+1}}{(n+1)!}. $$ This proves the claim by induction. The calculation also covers $C=0$: the recursion forces $F_n(t)=0$ for every $n\ge1$. $\square$

The factorial has a direct integral interpretation: $n$ successive time integrations range over the ordered region $0\le s_n\le\cdots\le s_1\le t$, whose volume is $t^n/n!$. This estimate concerns consecutive differences, not yet the convergence of the whole sequence; controlling the sum of those differences is the next step in the Picard argument.

Worked Examples

Worked Example: Exact Differences for a Deterministic Linear Equation

Take $T=1$, $\xi=1$, $\mu(t,x)=2x$, and $\sigma(t,x)=0$. Compute the Picard differences and compare them with the factorial estimate.

Here $X^0_t=1$. The first iterate is $X^1_t=1+\int_0^t2\,ds=1+2t$, so $D_0(t)=2t$ and $F_0(t)=4t^2$. Since $X^{n+1}_t=1+2\int_0^tX^n_s\,ds$, each next iterate adds one term to the power series. In particular, $X^2_t=1+2t+2t^2$ and $D_1(t)=2t^2$; then $X^3_t=1+2t+2t^2+\frac43t^3$ and $D_2(t)=\frac43t^3$. In general, direct substitution gives $$ D_n(t)=\frac{2^{n+1}t^{n+1}}{(n+1)!}, \qquad F_n(1)=\frac{2^{2(n+1)}}{((n+1)!)^2}. $$

For example, $F_0(1)=4$, $F_1(1)=4$, and $F_2(1)=16/9$. To check the formula, the displayed $D_2$ has supremum $4/3$ on $[0,1]$, whose square is $16/9$. For the general bound choose $L=2$, $A=2$, and $\mathbb{E}|\xi|^2=1$. Then $C=2^2(2+8)=40$ and $K_0=(2+8)2^2(1+1)=80$. The theorem therefore gives $F_n(1)\le80\cdot40^n/n!$; for $n=0$, the exact value $4\le80$ confirms the initial estimate. The theorem is deliberately coarse here, but its factorial form holds without needing the exact iterates.

Worked Example: A Stochastic First Difference

Let $T=1$, $\xi=1$, $\mu(t,x)=0$, and $\sigma(t,x)=x$. Find the first two differences and bound the second one.

The first iterate is $X^1_t=1+B_t$, so $D_0(t)=B_t$. The second is $X^2_t=1+\int_0^t(1+B_s)\,dB_s$. Thus $$ D_1(t)=X^2_t-X^1_t=\int_0^t B_s\,dB_s=\frac12(B_t^2-t). $$ The last equality follows from Itô’s formula applied to $f(x)=x^2$: $f'(x)=2x$, $f''(x)=2$, so $d(B_t^2)=2B_t\,dB_t+\frac12(2)\,dt$.

To check the difference independently at a fixed time, the Itô isometry gives $\mathbb{E}|D_1(1)|^2=\int_0^1\mathbb{E}|B_s|^2ds=\int_0^1s\,ds=1/2$. Equivalently, $B_1$ is standard normal, so $\mathbb{E}[(B_1^2-1)^2]=3-2+1=2$, and the displayed formula again gives $\mathbb{E}|D_1(1)|^2=2/4=1/2$. Doob’s inequality applies because $D_1$ is a square-integrable Itô-integral martingale; it yields $F_1(1)\le4(1/2)=2$.

For comparison, $F_0(s)=\mathbb{E}\sup_{u\le s}|B_u|^2\le4s$ by Doob’s inequality. Here $L=1$, so $C=10$, and the recursive estimate gives $F_1(1)\le10\int_0^1 4s\,ds=20$. The direct martingale bound $2$ is sharper, while both calculations verify the required finiteness.

Worked Example: State-Independent Coefficients Stop the Differences

Let $T=1$, $\xi=0$, $\mu(t,x)=3$, and $\sigma(t,x)=2$. Compute the iterates and verify why all differences after the first vanish.

The first iterate is $X^1_t=3t+2B_t$, while $X^0_t=0$. Since neither coefficient depends on the state, substituting $X^1$ into the Picard map gives $X^2_t=3t+2B_t=X^1_t$. Repeating the same substitution shows $X^n=X^1$ for every $n\ge1$. Thus $D_0(t)=3t+2B_t$ and $D_n(t)=0$ for every $n\ge1$.

As a check at the terminal time, independence of the deterministic term and the centered Brownian term gives $\mathbb{E}|D_0(1)|^2=3^2+2^2\mathbb{E}|B_1|^2=9+4=13$, while $\mathbb{E}|D_1(1)|^2=0$. Both coefficients are globally Lipschitz with $L=0$, so $C=0$; the recursive bound predicts $F_1(t)=0$, exactly as computed.

A Common Pitfall

A tempting but incorrect shortcut is to apply the recursive bound $n$ times and write $F_n(t)\le K_0(Ct)^n$, forgetting the factorial. Already after two integrations the omitted factor is visible: starting with $F_0(s)\le K_0$, the first estimate gives $F_1(t)\le CK_0t$, and the second gives $$ F_2(t)\le C\int_0^t CK_0s\,ds=\frac{K_0C^2t^2}{2}, $$ not $K_0C^2t^2$ as an exact iterated estimate. At the $n$-th step, integrating $s^n$ divides by $n+1$, building $n!$ in the denominator. Dropping it may leave a valid but weaker bound in some settings; it does not correctly record the repeated-integration calculation and hides the source of factorial decay.

Another important distinction is between a terminal second moment and the supremum quantity $F_n(t)$. A bound on $\mathbb{E}|D_n(t)|^2$ at one time does not by itself bound $\mathbb{E}\sup_{u\le t}|D_n(u)|^2$. In the recursion, the supremum is controlled by applying the drift estimate and Doob’s maximal inequality to the entire interval, not by checking only the endpoint.

Check Your Understanding

Use the recursive estimate and the worked calculations to check your reasoning.

For $T=2$ and Lipschitz constant $L=1$, what value of $C$ follows from the proposition?
If $F_0(t)\le K_0$, compute the recursive bounds for $F_1(t)$ and $F_2(t)$.
For the deterministic example with drift $\mu(t,x)=2x$, verify the formula for $D_2(t)$ by subtracting $X^2_t$ from $X^3_t$.
In the stochastic example with $\sigma(x)=x$, why can Doob’s inequality be applied to $D_1$?
What does the factorial in the estimate represent in terms of the ordered time variables?