Mean and Variance of a Deterministic Wiener Integral
A deterministic Wiener integral is an Ito integral whose integrand is a fixed function of time rather than a random process. Its mean and variance can be found directly from the squared size of that function. The mean is zero, while the variance records the integrand’s total squared energy over the time interval.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix \(T>0\). For a deterministic Borel-measurable function \(f:[0,T]\to\mathbb{R}\) with \(\int_0^T f(t)^2\,dt<\infty\), the process \(H_t=f(t)\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). Thus \(H\) belongs to the Ito-integrand class \(\mathcal{L}^2\), and \(\int_0^T f(t)\,dB_t\) is a well-defined Ito integral. Functions equal almost everywhere give the same integral.
The integral is defined by extending integrals of step functions in the \(L^2(\mathbb{P})\) norm. For a deterministic step function, each integral is a finite sum of Brownian increments multiplied by deterministic constants. Brownian increments have mean zero, are independent on disjoint time intervals, and have variance equal to the length of their interval. Those facts give the formulas first for step functions. The Ito isometry then carries the variance formula to all deterministic square-integrable functions.
The Mean-and-Variance Formula
Proof. For a deterministic step function \(g=\sum_{j=1}^m c_j\mathbf{1}_{(t_{j-1},t_j]}\), where \(0=t_0<t_1<\cdots<t_m=T\) and each \(c_j\) is a real number, its Ito integral is \(I(g)=\sum_{j=1}^m c_j(B_{t_j}-B_{t_{j-1}})\). Each increment has mean zero, so \(\mathbb{E}[I(g)]=0\). Distinct increments are independent and centered; hence their cross-products have expectation zero. Since \(\mathbb{E}[(B_{t_j}-B_{t_{j-1}})^2]=t_j-t_{j-1}\), it follows that $$ \mathbb{E}[I(g)^2]=\sum_{j=1}^m c_j^2(t_j-t_{j-1})=\int_0^T g(t)^2\,dt. $$ Now choose deterministic step functions \(g_n\) converging to \(f\) in \(L^2([0,T])\), using the standard density of step functions in this space. The Ito isometry gives \(\mathbb{E}[(I(g_n)-I(f))^2]=\int_0^T(g_n(t)-f(t))^2\,dt\to0\). Thus \(I(g_n)\to I(f)\) in \(L^2(\mathbb{P})\), and also in \(L^1(\mathbb{P})\) because \(\mathbb{E}|I(g_n)-I(f)|\le\big(\mathbb{E}[(I(g_n)-I(f))^2]\big)^{1/2}\). Taking expectations therefore gives \(\mathbb{E}[I(f)]=\lim_n\mathbb{E}[I(g_n)]=0\). Convergence in \(L^2\) also implies convergence of the \(L^2\) norms: \(\big|\|I(g_n)\|_2-\|I(f)\|_2\big|\le\|I(g_n)-I(f)\|_2\to0\). Consequently \(\mathbb{E}[I(f)^2]=\lim_n\mathbb{E}[I(g_n)^2]=\lim_n\int_0^Tg_n(t)^2\,dt=\int_0^T f(t)^2\,dt\). The mean is zero, so this second moment is the variance. \(\square\)
This proof uses the Ito isometry in the form \(\mathbb{E}[(\int_0^T H_t\,dB_t)^2]=\mathbb{E}\int_0^T H_t^2\,dt\) for progressively measurable \(H\) with \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). Here the integrands are deterministic, so the right-hand side is simply \(\int_0^T f(t)^2\,dt\). No ordinary chain rule is involved, and no Stratonovich integral is being computed.
Worked Example: A Deterministic Step Integrand
On \([0,2]\), let \(f(t)=2\) for \(0\le t\le\tfrac12\), and \(f(t)=-1\) for \(\tfrac12<t\le2\). This deterministic step function is progressively measurable and \(\mathbb{E}\int_0^2 f(t)^2\,dt<\infty\), so its Ito integral is in the class covered by the theorem. Compute its integral, mean, and variance.
Using the step-function definition, the integral is $$ I(f)=2(B_{1/2}-B_0)-(B_2-B_{1/2})=3B_{1/2}-B_2, $$ where \(B_0=0\). Its mean is \(2(0)-1(0)=0\), since both Brownian increments are centered. The theorem gives the variance directly: $$ \operatorname{Var}(I(f))=\int_0^{1/2}2^2\,dt+\int_{1/2}^2(-1)^2\,dt =4\cdot\tfrac12+1\cdot\tfrac32=\tfrac72. $$
Check: the increment form is \(2\Delta B_1-\Delta B_2\), where the increments have variances \(\tfrac12\) and \(\tfrac32\), respectively, and are independent. Therefore its variance is \(2^2\cdot\tfrac12+(-1)^2\cdot\tfrac32=2+\tfrac32=\tfrac72\), agreeing with the Ito-isometry calculation.
Worked Example: A Continuous Integrand
Let \(f(t)=t\) on \([0,2]\). It is deterministic and progressively measurable, and \(\mathbb{E}\int_0^2 f(t)^2\,dt=\int_0^2t^2\,dt<\infty\), so \(I(f)=\int_0^2t\,dB_t\) is an Ito integral in \(\mathcal{L}^2\). Find its mean and variance.
The mean is zero. The variance is $$ \operatorname{Var}(I(f))=\int_0^2t^2\,dt=\left[\frac{t^3}{3}\right]_0^2=\frac83. $$
To check the energy calculation by step approximations, divide \([0,2]\) into \(n\) equal intervals of length \(2/n\) and take the left endpoint value \(2(i-1)/n\) on interval \(i\). The squared \(L^2\) norm of this step function is $$ \sum_{i=1}^n\left(\frac{2(i-1)}{n}\right)^2\frac{2}{n} =\frac{8}{n^3}\sum_{j=0}^{n-1}j^2 =\frac{8}{n^3}\frac{(n-1)n(2n-1)}{6} =\frac{4(n-1)(2n-1)}{3n^2}\longrightarrow\frac83. $$
Check: the approximating step functions converge to \(t\) in \(L^2([0,2])\), so the Ito isometry says their integrals converge in \(L^2(\mathbb{P})\) and their variances converge to the same limit, \(\tfrac83\). The mean remains zero by the theorem. This confirms the variance using the construction of the integral rather than treating \(t\,dB_t\) as an ordinary integral.
Worked Example: A Square-Integrable Singular Integrand
On \((0,1]\), define \(f(t)=t^{-1/4}\); its value at \(t=0\) can be set arbitrarily because a single point does not change its \(L^2\) class. The function is deterministic and progressively measurable as a time process. Its squared energy is finite: $$ \int_0^1 f(t)^2\,dt=\int_0^1t^{-1/2}\,dt=2. $$ Thus \(f\) is an admissible Ito integrand, even though it is unbounded near zero. The theorem gives \(\mathbb{E}[I(f)]=0\) and \(\operatorname{Var}(I(f))=2\).
For a check, truncate the function by \(f_n(t)=t^{-1/4}\mathbf{1}_{[1/n,1]}(t)\), for integers \(n\ge1\). Each \(f_n\) is deterministic, progressively measurable, and square-integrable. Its energy is $$ \int_0^1 f_n(t)^2\,dt=\int_{1/n}^1t^{-1/2}\,dt=2\left(1-\frac{1}{\sqrt n}\right). $$ The omitted squared energy is \(\int_0^{1/n}t^{-1/2}\,dt=2/\sqrt n\), which tends to zero. The Ito isometry therefore gives \(I(f_n)\to I(f)\) in \(L^2(\mathbb{P})\), and the variances converge from \(2(1-1/\sqrt n)\) to \(2\). The means of the truncated integrals are all zero, consistent with the limiting mean.
Zero Variance and a Common Mistake
Proof. By the theorem, \(\operatorname{Var}(I(f))=\int_0^T f(t)^2\,dt\). If \(f=0\) almost everywhere, this integral is zero, so \(I(f)\) has mean zero and second moment zero. A nonnegative random variable with expectation zero is zero almost surely; applying this to \(I(f)^2\) gives \(I(f)=0\) almost surely. Conversely, if the variance is zero, then \(\int_0^T f(t)^2\,dt=0\). A nonnegative measurable function has zero integral only if it is zero almost everywhere, so \(f(t)^2=0\), and hence \(f(t)=0\), almost everywhere. \(\square\)
A common error is to integrate \(f\) rather than \(f^2\) when computing variance. For example, with \(f(t)=t\) on \([0,2]\), the incorrect calculation \(\int_0^2t\,dt=2\) would give variance \(2\). The correct calculation squares the integrand: \(\int_0^2t^2\,dt=\tfrac83\). The mean is still zero; integrating \(f\) does not give the mean of the Wiener integral.
The mean-and-variance formulas do not, by themselves, specify the full probability distribution of \(I(f)\). They establish its first two moments from the Ito integral and its isometry. The essential condition is square integrability: an integrand may be unbounded, as in the singular example, as long as its squared integral is finite. If \(\int_0^T f(t)^2\,dt\) is infinite, this \(\mathcal{L}^2\) definition and the variance formula do not apply.
Check Your Understanding
Use the mean-and-variance theorem and the Ito isometry to answer the following.
- For \(f(t)=3\) on \([0,1]\), compute \(\mathbb{E}[I(f)]\) and \(\operatorname{Var}(I(f))\).
- For \(f(t)=1-t\) on \([0,1]\), calculate \(\int_0^1f(t)^2\,dt\) and state the variance.
- Explain why \(f(t)=t^{-1/3}\) on \((0,1]\) is an admissible deterministic Ito integrand, and find the variance of its integral.
- If \(f\) is zero almost everywhere but nonzero at finitely many times, what is the variance of \(I(f)\), and why?
- For a deterministic step function on disjoint intervals, explain why the cross terms in the second moment of its integral have expectation zero.