Gaussianity of Deterministic Wiener Integrals

Why Deterministic Integrands Produce Gaussian Variables

Brownian increments are Gaussian, and the Ito integral of a deterministic step function is a finite weighted sum of those increments. Such a sum is Gaussian. The less immediate question is whether this remains true when the deterministic integrand is merely square-integrable, rather than a step function. The answer follows by approximating in \(L^2\) and controlling characteristic functions under the resulting mean-square convergence.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) with respect to the filtration. Fix \(T>0\). All integrals in this tutorial are Ito integrals on \([0,T]\). The integrand \(H_s=f(s)\) is deterministic and progressively measurable, with \(\mathbb{E}\int_0^T H_s^2\,ds=\int_0^T f(s)^2\,ds<\infty\); thus \(H\) belongs to \(\mathcal{L}^2\). We identify functions that agree almost everywhere with respect to Lebesgue measure.

Definition: For a deterministic step function \(f=\sum_{j=1}^m c_j\mathbf{1}_{(t_{j-1},t_j]}\), where \(0=t_0<t_1<\cdots<t_m=T\), its Wiener integral is \(\int_0^T f(s)\,dB_s=\sum_{j=1}^m c_j(B_{t_j}-B_{t_{j-1}})\). For a general deterministic \(f\in L^2([0,T])\), the integral is the \(L^2(\mathbb{P})\) limit of the integrals of deterministic step functions converging to \(f\) in \(L^2([0,T])\). The Ito isometry ensures this limit exists and is independent of the approximating sequence.

The density of interval step functions in \(L^2([0,T])\) is the standard \(L^2\) density theorem. The Ito isometry, established for Ito integrals earlier in the course, says that for these deterministic integrands \(\mathbb{E}\left[\left(\int_0^T f(s)\,dB_s-\int_0^T g(s)\,dB_s\right)^2\right]=\int_0^T(f(s)-g(s))^2\,ds\). In particular, approximation of the functions in \(L^2([0,T])\) gives mean-square approximation of their integrals.

The Step-Function Case

Take a deterministic step function on the partition \(0=t_0<\cdots<t_m=T\). Its integral is a sum of independent Brownian increments multiplied by constants. Brownian increments over disjoint time intervals are independent, and \(B_{t_j}-B_{t_{j-1}}\) has law \(N(0,t_j-t_{j-1})\). For a real number \(u\), the characteristic function of the integral is therefore

\mathbb{E}\left[\exp\left(iu\int_0^T f(s)\,dB_s\right)\right] =\prod_{j=1}^m\exp\left(-\frac{u^2}{2}c_j^2(t_j-t_{j-1})\right) =\exp\left(-\frac{u^2}{2}\int_0^T f(s)^2\,ds\right).

This is the characteristic function of a centered normal random variable with variance \(\int_0^T f(s)^2\,ds\). In particular, the mean is zero. If that variance is zero, the step function is zero almost everywhere and its integral is zero almost surely; this is the degenerate centered Gaussian law.

Worked Example: A Constant Integrand

Let \(f(s)=2\) on \([0,3]\). This is a deterministic \(\mathcal{L}^2\) integrand: its squared norm is \(\int_0^3 2^2\,ds=4\cdot3=12\). From the step-function definition, \(\int_0^3 2\,dB_s=2(B_3-B_0)=2B_3\). Since \(B_0=0\) and \(B_3\sim N(0,3)\), the integral has law \(N(0,12)\).

Check: its characteristic function is \(\mathbb{E}[e^{iu2B_3}]=\exp(-\tfrac12(2u)^2\cdot3)=\exp(-6u^2)\). The centered Gaussian formula with variance \(12\) gives \(\exp(-\tfrac12u^2\cdot12)=\exp(-6u^2)\), as required. Its mean is \(0\), and its variance, also given by the Ito isometry, is \(12\).

Passing Characteristic Functions to an \(L^2\) Limit

Lemma: If \(X_n\to X\) in \(L^2(\mathbb{P})\), then for every real \(u\), \(\mathbb{E}[e^{iuX_n}]\to\mathbb{E}[e^{iuX}]\).

Proof. For real \(x,y\), the fundamental theorem of calculus applied to \(r\mapsto e^{ir}\) gives \(|e^{ix}-e^{iy}|\le|x-y|\). Consequently,

\left|\mathbb{E}[e^{iuX_n}]-\mathbb{E}[e^{iuX}]\right| \le \mathbb{E}\left[|e^{iuX_n}-e^{iuX}|\right] \le |u|\mathbb{E}[|X_n-X|] \le |u|\big(\mathbb{E}[(X_n-X)^2]\big)^{1/2}\longrightarrow0.

The last inequality is Cauchy–Schwarz, and the limit follows from \(L^2\) convergence. If \(u=0\), both characteristic functions are identically \(1\), so the conclusion holds in that case as well. \(\square\)

Gaussianity for Every Deterministic \(L^2\) Integrand

Theorem: Let \(f\in L^2([0,T])\) be deterministic and let \(I=\int_0^T f(s)\,dB_s\), with the integral defined as above. Then \(I\) has the centered Gaussian law with variance \(v=\int_0^T f(s)^2\,ds\). When \(v=0\), this means \(I=0\) almost surely.

Proof. Choose deterministic step functions \(f_n\) such that \(\int_0^T(f_n(s)-f(s))^2\,ds\to0\), using the density of interval step functions in \(L^2([0,T])\). Write \(I_n=\int_0^T f_n(s)\,dB_s\). By the definition of the general Ito integral and the Ito isometry, \(\mathbb{E}[(I_n-I)^2]=\int_0^T(f_n(s)-f(s))^2\,ds\to0\).

For each \(n\), the step-function calculation gives \(\mathbb{E}[e^{iuI_n}]=\exp(-\tfrac12u^2v_n)\), where \(v_n=\int_0^T f_n(s)^2\,ds\). The norms converge: the triangle inequality yields \(\big|\|f_n\|_2-\|f\|_2\big|\le\|f_n-f\|_2\to0\), and therefore \(v_n=\|f_n\|_2^2\to\|f\|_2^2=v\). By the lemma, for every real \(u\),

\mathbb{E}[e^{iuI}] =\lim_{n\to\infty}\mathbb{E}[e^{iuI_n}] =\lim_{n\to\infty}\exp\left(-\frac{u^2}{2}v_n\right) =\exp\left(-\frac{u^2}{2}v\right).

The uniqueness theorem for characteristic functions, a standard result of probability theory, identifies this as the law \(N(0,v)\) when \(v>0\), and as the point mass at zero when \(v=0\). The Ito isometry also gives \(\mathbb{E}[I^2]=v\); the identified Gaussian law has mean zero and variance \(v\), including the degenerate case. \(\square\)

The approximation argument is useful because it transfers a distributional property without requiring a pointwise limit of the random variables. The step-function integrals are all Gaussian, but that observation alone is not enough: the proof also shows that their variances converge and that their characteristic functions converge to the characteristic function of the actual Ito integral.

Worked Example: A Piecewise Constant Integrand

On \([0,2]\), let \(f(s)=1\) for \(0<s\le1\) and \(f(s)=-1\) for \(1<s\le2\). The integrand is deterministic and \(\int_0^2 f(s)^2\,ds=1+1=2\). Its Wiener integral is

\int_0^2 f(s)\,dB_s=(B_1-B_0)-(B_2-B_1)=2B_1-B_2.

The increments \(B_1-B_0\) and \(B_2-B_1\) are independent, each with law \(N(0,1)\). Changing the sign of the second increment preserves its centered normal law, so their sum has law \(N(0,2)\).

Check: the characteristic function calculated from the increments is \(\exp(-u^2/2)\exp(-u^2/2)=\exp(-u^2)\). The theorem gives \(\exp(-\tfrac12u^2\cdot2)=\exp(-u^2)\). The mean is \(0\), and the variance is \(1+1=2\), in agreement with the squared \(L^2\) norm.

Worked Example: The Integrand \(f(s)=s\)

Take \(f(s)=s\) on \([0,1]\). Its squared norm is \(\int_0^1 s^2\,ds=\tfrac13\), so the theorem says that \(\int_0^1 s\,dB_s\) has law \(N(0,\tfrac13)\). To see the approximation explicitly, set \(f_n(s)=k/n\) on \(((k-1)/n,k/n]\) for \(k=1,\dots,n\). On each interval, \(0\le f_n(s)-s\le1/n\), so

\int_0^1(f_n(s)-s)^2\,ds =\sum_{k=1}^n\int_{(k-1)/n}^{k/n}(k/n-s)^2\,ds =n\cdot\frac{1}{3n^3} =\frac{1}{3n^2}\longrightarrow0.

The variance of the step-function integral is

\int_0^1 f_n(s)^2\,ds =\sum_{k=1}^n\frac{k^2}{n^2}\cdot\frac1n =\frac{n(n+1)(2n+1)}{6n^3} \longrightarrow\frac13.

Thus its characteristic function \(\exp(-\tfrac12u^2\int_0^1 f_n(s)^2\,ds)\) converges to \(\exp(-u^2/6)\), the characteristic function of \(N(0,\tfrac13)\). Check: the Ito isometry gives \(\mathbb{E}[(\int_0^1 s\,dB_s)^2]=\int_0^1s^2\,ds=\tfrac13\), matching the limiting variance.

A Common Mistake and the Scope of the Result

A deterministic Wiener integral is not the ordinary integral of its integrand. For the piecewise constant function in the earlier example, \(\int_0^2 f(s)\,ds=1-1=0\). It would be wrong to infer that \(\int_0^2 f(s)\,dB_s=0\): that integral is \(2B_1-B_2\), with variance \(2\), and is not almost surely zero. The ordinary integral adds signed areas in time; the Wiener integral weights random Brownian increments. Cancellation in one does not imply cancellation in the other.

The theorem applies to deterministic square-integrable integrands. Its proof uses that the step-function coefficients are constants, so the integral is a weighted sum of independent Gaussian increments with a directly computed characteristic function. It does not assert that every Ito integral with a random integrand is Gaussian. For deterministic \(f\), the variance is exactly its squared \(L^2\) norm; the integral is degenerate precisely when \(f=0\) almost everywhere.

Check Your Understanding

Use the step-function calculation and the approximation argument to explain each answer.

If \(f(s)=3\) on \([0,2]\), what are the mean, variance and characteristic function of its Wiener integral?
For a deterministic step function with values \(c_j\) on intervals of lengths \(t_j-t_{j-1}\), express the variance of its integral as a sum.
Why does \(L^2\) convergence of integrands imply \(L^2(\mathbb{P})\) convergence of their Wiener integrals?
What law does the theorem assign to the integral if \(\int_0^T f(s)^2\,ds=0\)?
Why does \(\int_0^T f(s)\,ds=0\) not imply \(\int_0^T f(s)\,dB_s=0\)?