Wiener Integrals of Smooth Deterministic Functions

From a Smooth Function to a Wiener Integral

A deterministic function can be integrated against Brownian motion even when its paths are too irregular for an ordinary pathwise Riemann–Stieltjes integral. For a smooth deterministic integrand, the Ito isometry turns the variance of the resulting random variable into an ordinary integral of the squared function. Smoothness also gives quantitative control of the error when the integrand is replaced by a time-step function.

Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Every stochastic integral below is an Ito integral. Its integrand is deterministic and measurable, hence progressively measurable, and belongs to \(\mathcal{L}^2\) on \([0,T]\): that is, it satisfies \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The integral is defined as the \(L^2(\mathbb{P})\) limit of integrals of suitable time-step approximations. In particular, for deterministic \(f\in L^2([0,T])\), \(\int_0^T f(t)\,dB_t\) is well-defined.

Definition: For a deterministic \(f\in L^2([0,T])\), the Wiener integral \(I(f)\) is the Ito integral \(I(f)=\int_0^T f(t)\,dB_t\). Its variance is \(\operatorname{Var}(I(f))=\mathbb{E}[(I(f)-\mathbb{E}[I(f)])^2]\). The integral is an Ito integral, not a Stratonovich integral or an ordinary integral with respect to time.

The earlier material on moments and covariances of Ito integrals gives \(\mathbb{E}[I(f)]=0\) and the Ito isometry \(\mathbb{E}[I(f)^2]=\int_0^T f(t)^2\,dt\). The earlier result on Gaussianity of deterministic Wiener integrals gives that \(I(f)\) is Gaussian. We use these established results below; no new Gaussianity theorem is needed here.

Variance and Distribution

Proposition: If \(f\in L^2([0,T])\) is deterministic, then \(\mathbb{E}[I(f)]=0\), \(\operatorname{Var}(I(f))=\int_0^T f(t)^2\,dt\), and \(I(f)\) has distribution \(N(0,\int_0^T f(t)^2\,dt)\). If the integral of \(f^2\) is zero, this denotes the degenerate distribution concentrated at zero.

Proof. Since \(I(f)\) is a deterministic-integrand Ito integral, the earlier mean and isometry results apply. Thus \(\mathbb{E}[I(f)]=0\), and \(\operatorname{Var}(I(f))=\mathbb{E}[I(f)^2]-(\mathbb{E}[I(f)])^2=\int_0^T f(t)^2\,dt\). The earlier Gaussianity result says that \(I(f)\) is Gaussian, so its mean and variance determine its distribution. If \(\int_0^T f(t)^2\,dt=0\), the isometry gives \(\mathbb{E}[I(f)^2]=0\), hence \(I(f)=0\) almost surely; the distribution is then concentrated at zero. \(\square\)

Worked Example: Integrating a Power of Time

Let \(f(t)=t^p\) on \([0,T]\), where \(p\) is a nonnegative integer and \(T>0\). Compute the mean, variance, and distribution of \(I(f)\).

The integrand is continuous and square-integrable on this finite interval. Its mean is zero. Using the proposition,

\operatorname{Var}\left(\int_0^T t^p\,dB_t\right) =\int_0^T t^{2p}\,dt =\frac{T^{2p+1}}{2p+1}.

Therefore \(\int_0^T t^p\,dB_t\) has distribution \(N(0,T^{2p+1}/(2p+1))\). For example, when \(p=2\) and \(T=2\), the variance is \(2^5/5=32/5\), so the distribution is \(N(0,32/5)\).

Check: at \(p=0\), the integral is \(\int_0^T 1\,dB_t=B_T\). The formula gives variance \(T^1/1=T\), agreeing with \(B_T\sim N(0,T)\). For \(p=2,T=2\), direct integration gives \(\int_0^2 t^4\,dt=[t^5/5]_0^2=32/5\), as above.

Worked Example: An Affine Integrand

Let \(f(t)=1+2t\) on \([0,1]\). Find the distribution of its Wiener integral and check the variance by expanding the square.

The mean is zero. The variance is

\int_0^1(1+2t)^2\,dt =\int_0^1(1+4t+4t^2)\,dt =1+2+\frac43 =\frac{13}{3}.

Thus \(\int_0^1(1+2t)\,dB_t\sim N(0,13/3)\). The three contributions \(1\), \(2\), and \(4/3\) come respectively from integrating \(1\), \(4t\), and \(4t^2\) over \([0,1]\).

Check: integrating the expanded expression directly gives \([t+2t^2+\tfrac43t^3]_0^1=1+2+\tfrac43=13/3\). In particular, the variance is positive, as required for this nonzero integrand.

Approximating by Time-Step Functions

To approximate a smooth integrand on a partition \(0=t_0<t_1<\cdots<t_n=T\), take its left-endpoint time-step approximation \(g(t)=f(t_{i-1})\) for \(t\in[t_{i-1},t_i)\). The value at the single endpoint \(T\) does not affect the integral. Both \(f\) and \(g\) are deterministic square-integrable integrands, so their Ito integrals and the integral of \(f-g\) are defined in the same \(L^2\) class.

Theorem: Suppose \(f\in C^1([0,T])\), and let \(M=\sup_{0\le t\le T}|f'(t)|\). For a partition \(\Pi\) with mesh \(|\Pi|=\max_i(t_i-t_{i-1})\), define \(g(t)=f(t_{i-1})\) on \([t_{i-1},t_i)\). Then \(\mathbb{E}[(I(f)-I(g))^2]\le \frac{M^2}{3}\sum_{i=1}^n(t_i-t_{i-1})^3\le\frac{M^2T}{3}|\Pi|^2\). Thus the root-mean-square error is at most \(M\sqrt{T/3}\,|\Pi|\).

Proof. Linearity of the Ito integral gives \(I(f)-I(g)=I(f-g)\). Apply the Ito isometry to the deterministic square-integrable function \(f-g\): the mean-square error is \(\int_0^T(f(t)-g(t))^2\,dt\). Write \(h_i=t_i-t_{i-1}\). By the fundamental theorem of calculus, for \(t\in[t_{i-1},t_i]\), \(|f(t)-f(t_{i-1})|=|\int_{t_{i-1}}^t f'(s)\,ds|\le M(t-t_{i-1})\). Squaring and integrating on each interval gives \(\int_{t_{i-1}}^{t_i}(f(t)-g(t))^2\,dt\le M^2\int_{t_{i-1}}^{t_i}(t-t_{i-1})^2\,dt=M^2h_i^3/3\). Summing proves the first inequality. Since \(h_i\le|\Pi|\), \(h_i^3\le|\Pi|^2h_i\); summing and using \(\sum_i h_i=T\) proves the second. Taking square roots gives the stated root-mean-square bound. \(\square\)

The estimate is an \(L^2(\mathbb{P})\) error bound for the random integrals. It does not assert that the error is bounded by the same quantity on every Brownian path. For a uniform partition with \(n\) intervals, \(|\Pi|=T/n\), so the theorem gives an error of order \(1/n\) in root mean square.

Worked Example: Exact Error on a Nonuniform Partition

Approximate \(I(f)=\int_0^2 t\,dB_t\) using the left-endpoint step function on the partition \(0<\tfrac12<1<\tfrac32<2\). Compute the exact mean-square error and verify it using variances and covariance.

On the four intervals, the step values are respectively \(0,\tfrac12,1,\tfrac32\). The integrand error \(t-g(t)\) rises from \(0\) to \(\tfrac12\) on each interval. Since each interval has length \(h=\tfrac12\), its squared error integral is \(\int_0^{1/2}u^2\,du=(1/2)^3/3=1/24\). Summing four equal contributions gives \(\mathbb{E}[(I(f)-I(g))^2]=4/24=1/6\).

For a second calculation, the variance of \(I(f)\) is \(\int_0^2t^2\,dt=8/3\), and the variance of \(I(g)\) is \(\tfrac12(0^2+(\tfrac12)^2+1^2+(\tfrac32)^2)=\tfrac12(0+\tfrac14+1+\tfrac94)=7/4\). The covariance formula for deterministic Wiener integrals, established in the earlier material on their covariances, gives \(\operatorname{Cov}(I(f),I(g))=\int_0^2t g(t)\,dt\). The four interval contributions are \(0\), \(\tfrac12\int_{1/2}^1t\,dt=\tfrac12\cdot\tfrac38=\tfrac3{16}\), \(\int_1^{3/2}t\,dt=\tfrac58\), and \(\tfrac32\int_{3/2}^2t\,dt=\tfrac32\cdot\tfrac78=\tfrac{21}{16}\). Their sum is \(\tfrac3{16}+\tfrac{10}{16}+\tfrac{21}{16}=\tfrac{17}{8}\).

Consequently the variance of the difference is \(\tfrac83+\tfrac74-2\cdot\tfrac{17}{8}=\tfrac83+\tfrac74-\tfrac{17}{4}=\tfrac83-\tfrac52=\tfrac16\), agreeing with the direct calculation. The theorem also applies: \(M=1\), \(T=2\), and \(|\Pi|=\tfrac12\), giving the upper bound \((1)(2)(1/4)/3=1/6\). Equality holds here because \(f(t)=t\) has constant derivative.

Worked Example: Error Rate for Uniform Steps

For \(f(t)=t\) on \([0,1]\), take \(n\) equal intervals and compute the exact error variance. Then evaluate it for \(n=4\).

Each interval has length \(1/n\), and on each interval the difference between \(f(t)\) and its left-endpoint value is the distance from the left endpoint. The contribution to the error variance from one interval is \(\int_0^{1/n}u^2\,du=1/(3n^3)\). There are \(n\) intervals, so the total is \(1/(3n^2)\). For \(n=4\), the exact variance is \(1/(3\cdot16)=1/48\), and the root-mean-square error is \(1/\sqrt{48}=1/(4\sqrt3)\).

Check: here \(M=1\), \(T=1\), and \(|\Pi|=1/4\). The theorem's mean-square bound is \(1\cdot1\cdot(1/4)^2/3=1/48\), matching the exact value. As \(n\) increases, the variance decreases as \(n^{-2}\), while the root-mean-square error decreases as \(n^{-1}\).

Common Pitfalls and Practical Meaning

Do not confuse the variance with its square root. The Ito isometry yields \(\mathbb{E}[(I(f)-I(g))^2]\), which is the mean-square error. The root-mean-square error is its square root. In the uniform example, \(1/(3n^2)\) is the variance of the error, whereas \(1/(\sqrt3\,n)\) is its root-mean-square size.

Use the squared integrand. The variance is \(\int_0^T f(t)^2\,dt\), not \((\int_0^T f(t)\,dt)^2\). For \(f(t)=t\) on \([0,1]\), these expressions are \(\int_0^1t^2\,dt=1/3\) and \((\int_0^1t\,dt)^2=1/4\), respectively, and are not equal.

The derivative bound explains why smooth deterministic integrands are computationally convenient: refining the time grid controls the stochastic-integral approximation in mean square. The Ito isometry reduces this stochastic error analysis to estimating an ordinary integral. The distribution result then also identifies the approximation error as Gaussian when considered as the integral of the deterministic function \(f-g\), with variance given by its squared \(L^2\) norm.

Check Your Understanding

Use the isometry and the approximation estimate to check your reasoning.

What are the mean and variance of \(\int_0^3 t\,dB_t\), and what is its distribution?
For \(f(t)=\cos(t)\) on \([0,T]\), what value of \(M\) can be used in the theorem?
On \([0,1]\), what mean-square error bound does the theorem give for left-endpoint steps of width \(1/5\) when \(f(t)=t^2\)?
Why is the distribution of \(I(f)\) degenerate when \(\int_0^T f(t)^2\,dt=0\)?
For \(f(t)=t\) on \([0,1]\) with \(n\) equal steps, how do the variance and root-mean-square approximation error depend on \(n\)?