Computing Initial Picard Iterates for Constant Coefficients

Constant Coefficients Make the Iteration Explicit

Picard iteration replaces the unknown process inside an SDE’s drift and diffusion coefficients by the preceding iterate. For state-dependent coefficients, this creates successively more involved stochastic integrals. With constant coefficients, however, the coefficients do not change when the process is replaced: after the first update, every subsequent iterate is the same process. Computing this simple case makes the role of the initial guess and the iteration index especially clear.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ with respect to the filtration. Fix a finite horizon $T>0$, an initial value $x_0\in\mathbb{R}$, and constants $\mu,\sigma\in\mathbb{R}$. Consider the scalar SDE $dX_t=\mu\,dt+\sigma\,dB_t$, with $X_0=x_0$. Here $\mu$ is the drift coefficient and $\sigma$ is the diffusion coefficient.

Definition: For this equation, take the zeroth Picard iterate to be $X^{(0)}_t=x_0$. Recursively, for $n\ge0$, define $$ X^{(n+1)}_t=x_0+\int_0^t\mu\,ds+\int_0^t\sigma\,dB_s,\qquad 0\le t\le T. $$ The stochastic integral is an Ito integral: its integrand must be progressively measurable and belong to $\mathcal{L}^2$, the class satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The constant integrand $H_s=\sigma$ satisfies this condition because $\mathbb{E}\int_0^T\sigma^2\,ds=\sigma^2T<\infty$.

In the usual Picard map, the coefficients are evaluated at the previous iterate. In this constant-coefficient equation that dependence disappears: $\mu(s,X^{(n)}_s)=\mu$ and $\sigma(s,X^{(n)}_s)=\sigma$, for every $n$. Thus the displayed recursion is exactly the Picard update for this SDE.

Evaluating the First Iterates

The deterministic integral is $\int_0^t\mu\,ds=\mu t$. The Ito integral of the constant integrand is $\int_0^t\sigma\,dB_s=\sigma B_t$. This follows directly from the definition of the Ito integral: for a constant integrand, every step-process approximation gives $\sigma$ times the sum of the Brownian increments, which telescopes to $\sigma(B_t-B_0)=\sigma B_t$. Therefore the first update is $x_0+\mu t+\sigma B_t$. Applying the update again gives the same expression, not a new term involving the first iterate.

Worked Example: First and Second Iterates

Take $x_0=-1$, $\mu=2$, and $\sigma=3$. Compute the first two iterates at time $t=2$, and find each iterate’s mean and variance.

The initial iterate is constant: $X^{(0)}_2=-1$. Its mean is $\mathbb{E}[X^{(0)}_2]=-1$, and its variance is $0$.

For the first update, evaluate both integrals: $\int_0^2 2\,ds=2\cdot2=4$, and $\int_0^2 3\,dB_s=3B_2$. Hence $X^{(1)}_2=-1+4+3B_2=3+3B_2$. Since $B_2$ has mean $0$ and variance $2$, $\mathbb{E}[X^{(1)}_2]=3+3\cdot0=3$, and $\operatorname{Var}(X^{(1)}_2)=3^2\cdot2=18$.

For the second update, the coefficients are still the constants $2$ and $3$: $X^{(2)}_2=-1+\int_0^2 2\,ds+\int_0^2 3\,dB_s=-1+4+3B_2=3+3B_2$. Thus its mean is $3$ and its variance is $18$, just as for the first update.

Check: Since $B_2$ has distribution $N(0,2)$, $3+3B_2$ has distribution $N(3,18)$: the mean is $3$, and the variance is $3^2\cdot2=18$. This agrees with the direct moment calculation for both positive-index iterates.

Theorem: For every $n\ge1$ and $0\le t\le T$, $$ X^{(n)}_t=x_0+\mu t+\sigma B_t. $$ Consequently, $X^{(n)}_t$ has distribution $N(x_0+\mu t,\sigma^2t)$ when $\sigma\ne0$ and $t>0$, with mean $x_0+\mu t$ and variance $\sigma^2t$. If $\sigma=0$ or $t=0$, the distribution is degenerate at $x_0+\mu t$, and the same mean and variance formulas apply.

Proof. For $n=1$, substitute the two evaluated integrals into the recursion to obtain $X^{(1)}_t=x_0+\mu t+\sigma B_t$. Suppose $X^{(n)}_t=x_0+\mu t+\sigma B_t$ for some $n\ge1$. The recursion defining the next iterate uses the constant coefficients $\mu$ and $\sigma$, so $X^{(n+1)}_t=x_0+\int_0^t\mu\,ds+\int_0^t\sigma\,dB_s=x_0+\mu t+\sigma B_t$. Induction proves the formula for every $n\ge1$. Brownian motion has $B_t\sim N(0,t)$; adding $x_0+\mu t$ and multiplying $B_t$ by $\sigma$ gives the stated normal distribution when $\sigma\ne0$ and $t>0$. If $\sigma=0$ or $t=0$, the random Brownian term is zero almost surely, so the iterate is the stated constant. These cases also give variance $\sigma^2t=0$. $\square$

The mean and variance can also be checked directly from the integral. For each $t\le T$, the constant integrand is in $\mathcal{L}^2$ on $[0,t]$, since $\mathbb{E}\int_0^t\sigma^2\,ds=\sigma^2t<\infty$. The Ito isometry gives $\mathbb{E}[(\int_0^t\sigma\,dB_s)^2]=\sigma^2t$. The integral has mean zero, so adding the deterministic quantity $x_0+\mu t$ gives mean $x_0+\mu t$ and variance $\sigma^2t$, including the zero-variance cases.

Checking the Moments at Specific Times

Worked Example: Negative Drift and a Nonzero Initial Value

Let $x_0=1$, $\mu=-\tfrac12$, and $\sigma=2$. Find the first two iterates at $t=4$, and compare their means and variances with the initial iterate.

The zeroth iterate is $X^{(0)}_4=1$, so its mean is $1$ and its variance is $0$. For the first update, the drift integral is $\int_0^4-\tfrac12\,ds=-\tfrac12\cdot4=-2$, while the stochastic integral is $\int_0^4 2\,dB_s=2B_4$. Therefore $X^{(1)}_4=1-2+2B_4=-1+2B_4$. Its mean is $-1+2\cdot0=-1$, and its variance is $2^2\cdot4=16$.

The next update has the same constant integrands, so $X^{(2)}_4=1+\int_0^4-\tfrac12\,ds+\int_0^4 2\,dB_s=-1+2B_4$. It has mean $-1$ and variance $16$, while the zeroth iterate still has mean $1$ and variance $0$.

Check: $B_4\sim N(0,4)$, so $-1+2B_4\sim N(-1,2^2\cdot4)=N(-1,16)$. The second moment about zero is $16+(-1)^2=17$, and subtracting the squared mean gives variance $17-1=16$, in agreement with the direct calculation.

Worked Example: Negative Diffusion Coefficient

Take $x_0=2$, $\mu=1$, and $\sigma=-1$. Compute the first and second iterates at $t=1$, including their means and variances.

The zeroth iterate is $X^{(0)}_1=2$, with mean $2$ and variance $0$. The drift integral in the first update is $\int_0^1 1\,ds=1$, and the stochastic integral is $\int_0^1(-1)\,dB_s=-B_1$. Thus $X^{(1)}_1=2+1-B_1=3-B_1$. Since $\mathbb{E}[B_1]=0$ and $\operatorname{Var}(B_1)=1$, its mean is $3$ and its variance is $(-1)^2\cdot1=1$.

The second update is $X^{(2)}_1=2+\int_0^1 1\,ds+\int_0^1(-1)\,dB_s=3-B_1$, so it also has mean $3$ and variance $1$.

Check: $B_1\sim N(0,1)$, and the negative sign changes the direction of the Brownian fluctuation but not its variance. In particular, $3-B_1$ has mean $3$ and variance $1$, matching both iterates’ calculated moments.

Why the Stochastic Term Has These Moments

The stochastic term is an Ito integral, not an ordinary integral along a differentiable path. In this constant-integrand case it equals $\sigma B_t$, whose mean is zero and whose variance is $\sigma^2t$. This also illustrates the Ito isometry: for a progressively measurable $H\in\mathcal{L}^2$, it states $\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds$. Here $H_s=\sigma$, so the right side is $\int_0^t\sigma^2\,ds=\sigma^2t$. The isometry concerns the second moment; the zero mean follows from the Ito integral’s martingale property.

For clarity, that martingale property can be verified directly here. The process $M_t=\int_0^t\sigma\,dB_s=\sigma B_t$ is square-integrable because $\mathbb{E}[M_t^2]=\sigma^2t<\infty$. For $0\le s\le t$, Brownian increments are independent of $\mathcal{F}_s$ and have mean zero, so $\mathbb{E}[M_t\mid\mathcal{F}_s]=\sigma\mathbb{E}[B_s+(B_t-B_s)\mid\mathcal{F}_s]=\sigma B_s=M_s$. Thus $M_t$ is a true martingale, not merely a local martingale. Its conditional-expectation property justifies its zero mean.

Proposition: For the constant-coefficient Picard recursion, $X^{(n+1)}_t=X^{(n)}_t$ for every $n\ge1$ and $t\in[0,T]$. This is equality of the processes defined by the recursion, and hence also holds almost surely at each fixed time.

Proof. By the theorem, both $X^{(n+1)}_t$ and $X^{(n)}_t$ equal $x_0+\mu t+\sigma B_t$ for each $n\ge1$. Therefore they are equal for every $t$ on every outcome where the Brownian path is defined. In particular, they are equal almost surely at each fixed time. The initial iterate $X^{(0)}_t=x_0$ need not equal this common process; for example, if $\mu\ne0$ or $\sigma\ne0$, the first update can differ from the initial constant. $\square$

What Immediate Stabilization Does—and Does Not—Show

For this equation, the common positive-index iterate is itself a solution of the SDE in integral form: $X_t=x_0+\int_0^t\mu\,ds+\int_0^t\sigma\,dB_s=x_0+\mu t+\sigma B_t$. It is adapted and continuous because $B_t$ is adapted and continuous. This direct verification is specific to the constant coefficients. It does not establish a general convergence theorem for Picard iteration, nor does it show that iterates for state-dependent coefficients stabilize after one step.

Common mistake: inserting the previous iterate where there is no dependence. A mistaken calculation might write $X^{(2)}_t=x_0+\int_0^t\mu X^{(1)}_s\,ds+\int_0^t\sigma X^{(1)}_s\,dB_s$. That expression corresponds to coefficients depending on the state, not to the constant-coefficient equation $dX_t=\mu\,dt+\sigma\,dB_t$. The correct update uses $\mu$ and $\sigma$ themselves: $X^{(2)}_t=x_0+\int_0^t\mu\,ds+\int_0^t\sigma\,dB_s=x_0+\mu t+\sigma B_t$. The distinction is whether the coefficients depend on $X^{(n)}_s$, not whether the Picard method is being used.

A second easy error is to report $\sigma t$ as the variance of the stochastic term. The term is $\sigma B_t$, so its variance is $\sigma^2\operatorname{Var}(B_t)=\sigma^2t$; the coefficient is squared. A negative $\sigma$ reverses the sign of the Brownian fluctuation but cannot make a variance negative.

Check Your Understanding

Use the stated recursion and the Brownian mean and variance to check each answer.

For $x_0=0$, $\mu=3$, and $\sigma=0$, find $X^{(0)}_t$ and $X^{(1)}_t$, and give their means and variances at a fixed $t\in[0,T]$.
For $x_0=-2$, $\mu=0$, and $\sigma=4$, compute the mean and variance of $X^{(1)}_3$ and $X^{(2)}_3$.
Explain why the Ito integrand $\sigma$ satisfies the required square-integrability condition on every finite time interval.
Verify the conditional-expectation identity for $M_t=\sigma B_t$ when $0\le s\le t$, and state why square-integrability matters here.
In the constant-coefficient recursion, identify which iterate can differ from its successor and explain why the later updates agree.