Deterministic Wiener Integrals

From a Deterministic Function to a Wiener Integral

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $B_t$ adapted to this filtration. Fix $T>0$. The Ito integral on $[0,T]$ is defined for progressively measurable integrands $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this class is denoted by $\mathcal{L}^2$. A deterministic function $f$ with $\int_0^T f(s)^2\,ds<\infty$ should therefore give an Ito integral, but a small measurability point matters: a Lebesgue-measurable representative need not be Borel measurable in time.

The space $L^2([0,T])$ consists of equivalence classes of Lebesgue-measurable functions, where two functions are identified if they agree almost everywhere with respect to Lebesgue measure. Every such class has a Borel-measurable representative agreeing with it almost everywhere. Choose one and denote it by $f$. The process $H_s(\omega)=f(s)$ is progressively measurable: on each $[0,t]\times\Omega$, it is measurable with respect to $\mathcal{B}([0,t])\otimes\mathcal{F}_t$, since it is Borel in time and constant in $\omega$. It also satisfies $\mathbb{E}\int_0^T H_s^2\,ds=\int_0^T f(s)^2\,ds<\infty$.

Definition: For a deterministic $L^2([0,T])$ equivalence class $f$, its Wiener integral over $[0,T]$ is the Ito integral $$ I_T(f)=\int_0^T f(s)\,dB_s, $$ formed using any Borel-measurable representative of the class. It is a random variable defined up to almost-sure equality.

The Ito isometry, used below, says that for $H\in\mathcal{L}^2$, $\mathbb{E}\big[(\int_0^T H_s\,dB_s)^2\big]=\mathbb{E}\int_0^T H_s^2\,ds$. The integral has mean zero as well. For simple predictable integrands this follows by conditioning each Brownian increment on its starting-time sigma-field; for general $\mathcal{L}^2$ integrands it follows by approximation in $\mathcal{L}^2$, the Ito isometry, and convergence in $L^2(\mathbb{P})$. These facts ensure both that the integral is square-integrable and that its value is determined by the $L^2$ class, not by pointwise choices.

Why the Representative Does Not Matter

Proposition: If $f$ and $g$ are Borel-measurable representatives of the same element of $L^2([0,T])$, then $\int_0^T f(s)\,dB_s=\int_0^T g(s)\,dB_s$ almost surely.

Proof. Because $f$ and $g$ represent the same equivalence class, $\int_0^T(f(s)-g(s))^2\,ds=0$. Both deterministic processes are progressively measurable and square-integrable, so their difference is an admissible Ito integrand. Linearity for Ito integrals, or the construction by $L^2$ limits, gives $\int_0^T f(s)\,dB_s-\int_0^T g(s)\,dB_s=\int_0^T(f(s)-g(s))\,dB_s$. Applying the Ito isometry to the right side yields $\mathbb{E}[(\int_0^T(f-g)\,dB_s)^2]=\int_0^T(f-g)^2\,ds=0$. A nonnegative random variable with expectation zero is zero almost surely, so the two integrals agree almost surely. $\square$

The choice of a Borel representative is a way to place the deterministic function in the admissible integrand class; it does not add extra information to the resulting integral. In particular, one cannot argue that every Lebesgue-measurable representative automatically gives a progressively measurable process. Choosing a Borel version resolves this issue while leaving the $L^2$ class unchanged.

Worked Example: Changing a Function at One Point

On $[0,1]$, let $f(s)=1$ for every $s$, and let $g(s)=1$ for $0\le s<1$ but $g(1)=7$. Compute the two Wiener integrals and verify that changing the endpoint value has no effect.

The functions differ only at the singleton $\{1\}$, which has Lebesgue measure zero. Therefore $\int_0^1(f(s)-g(s))^2\,ds=0$. The representative-independence proposition gives $I_1(f)=I_1(g)$ almost surely.

For the constant integrand, the Ito integral is $I_1(f)=\int_0^1 1\,dB_s=B_1-B_0=B_1$. Consequently $I_1(g)=B_1$ almost surely as well. Check: the Ito isometry applied to the difference gives $\mathbb{E}[(I_1(f)-I_1(g))^2]=\int_0^1(f-g)^2\,ds=0$, which independently confirms equality almost surely.

Linearity of the Wiener Integral

Theorem: For $f,g\in L^2([0,T])$ and real constants $a,b$, $$ I_T(af+bg)=aI_T(f)+bI_T(g)\quad\text{almost surely}. $$

Proof. Choose Borel representatives of $f$ and $g$. By density of simple predictable integrands in $\mathcal{L}^2$, there are simple predictable integrands $H_n$ and $K_n$ such that $\mathbb{E}\int_0^T(H_n-f)^2\,ds\to0$ and $\mathbb{E}\int_0^T(K_n-g)^2\,ds\to0$. Here $f$ and $g$ also denote their deterministic processes. The sum $aH_n+bK_n$ is simple predictable and converges in $\mathcal{L}^2$ to $af+bg$, since \[ \mathbb{E}\int_0^T\big(a(H_n-f)+b(K_n-g)\big)^2\,ds \le 2a^2\mathbb{E}\int_0^T(H_n-f)^2\,ds +2b^2\mathbb{E}\int_0^T(K_n-g)^2\,ds\longrightarrow0. \] For simple predictable integrands, the Ito integral is defined by finite sums of coefficient times Brownian increment, and this definition is linear. Thus $\int_0^T(aH_n+bK_n)\,dB_s=a\int_0^T H_n\,dB_s+b\int_0^T K_n\,dB_s$. The Ito isometry implies convergence of each integral in $L^2(\mathbb{P})$: for example, $\mathbb{E}[(\int_0^T(H_n-f)\,dB_s)^2]=\mathbb{E}\int_0^T(H_n-f)^2\,ds\to0$. The same argument applies to $K_n$ and $aH_n+bK_n$. Taking these $L^2$ limits in the finite-sum identity proves the asserted equality almost surely. This argument includes $a=0$, $b=0$, and the zero equivalence class: in each case the displayed convergence and the simple-integrand identity still hold. $\square$

Linearity is an equality of random variables up to almost-sure equality, not necessarily a pointwise equality for every outcome under independently chosen versions. The distinction is standard for Ito integrals, which are themselves defined as equivalence classes of random variables. When finitely many such equalities are combined, their exceptional null sets can be united into a single null set, so the resulting finite collection holds simultaneously almost surely.

Worked Example: A Constant Integrand and Its Scaling

On $[0,2]$, take $f(s)=3$. Compute $I_2(f)$, its expectation, and its second moment.

The constant process is an admissible deterministic integrand because $\mathbb{E}\int_0^2 3^2\,ds=9\cdot2=18<\infty$. The simple-integrand definition gives \[ I_2(f)=\int_0^2 3\,dB_s=3(B_2-B_0)=3B_2. \] Since $B_0=0$ and $B_2$ has mean zero, $\mathbb{E}[I_2(f)]=3\mathbb{E}[B_2]=0$. The Ito isometry gives $\mathbb{E}[I_2(f)^2]=\int_0^2 9\,ds=18$.

Check: directly, $I_2(f)^2=9B_2^2$, and $\mathbb{E}[B_2^2]=2$, so $\mathbb{E}[I_2(f)^2]=9\cdot2=18$, matching the isometry. If the integrand is scaled by a real number $a$, linearity gives $I_2(af)=aI_2(f)=3aB_2$; its second moment is $9a^2\cdot2=18a^2$, also equal to $\int_0^2(3a)^2\,ds=18a^2$.

Worked Example: A Singular but Square-Integrable Integrand

On $[0,1]$, define $f(0)=0$ and $f(s)=s^{-1/4}$ for $s>0$, and let $g(s)=1$. Verify that $f$ and $f+g$ define Wiener integrals, then compute the second moment of $I_1(f+g)$ and express it using linearity.

Both functions are Borel measurable. Although $f$ is unbounded near zero, its squared norm is finite: \[ \int_0^1 f(s)^2\,ds=\int_0^1s^{-1/2}\,ds =\left[2s^{1/2}\right]_0^1=2. \] Also $\int_0^1g(s)^2\,ds=1$. For the sum, expand the square and integrate each term: \[ \int_0^1(f(s)+g(s))^2\,ds =\int_0^1\big(s^{-1/2}+2s^{-1/4}+1\big)\,ds =2+2\left[\frac{4}{3}s^{3/4}\right]_0^1+1 =\frac{17}{3}. \] These finite integrals verify admissibility. The linearity theorem gives $I_1(f+g)=I_1(f)+I_1(g)$ almost surely. The Ito isometry and the mean-zero property give $\mathbb{E}[I_1(f+g)]=0$ and $\mathbb{E}[I_1(f+g)^2]=17/3$.

Check: applying the isometry directly to the sum gives the same second moment, because its integrand norm was computed as $2+8/3+1=17/3$. No pointwise value at the singular endpoint is needed: the definition uses the $L^2$ class, and changing $f(0)$ would not change that class or its Wiener integral.

What the Definition Does—and Does Not—Mean

The Wiener integral is constructed from Ito integrals, which are limits in $L^2(\mathbb{P})$ of integrals of simple predictable processes. It is therefore controlled by the squared-integral distance between integrands. For deterministic functions $f$ and $g$, the Ito isometry reads \[ \mathbb{E}\big[(I_T(f)-I_T(g))^2\big] =\int_0^T(f(s)-g(s))^2\,ds. \] This formula both explains independence from the representative and shows that approximation of deterministic integrands in $L^2([0,T])$ gives approximation of their Wiener integrals in $L^2(\mathbb{P})$.

This construction should not be confused with an ordinary pathwise Riemann–Stieltjes integral. An $L^2$ equivalence class does not specify pointwise values, and an arbitrary representative need not have the regularity needed for Riemann–Stieltjes integration. Some particular integrands, such as constants, do yield simple pathwise expressions. That special case does not turn the general definition into a pathwise Riemann–Stieltjes construction.

Common mistake: using pointwise sampling instead of the $L^2$ class. On $[0,1]$, let $f(s)=1$ when $s$ is rational and $f(s)=0$ when $s$ is irrational. This is a Borel-measurable function that is zero almost everywhere, so its Wiener integral is zero almost surely: the Ito isometry gives $\mathbb{E}[I_1(f)^2]=\int_0^1 f(s)^2\,ds=0$. A wrong calculation takes left endpoints $i/n$, for $i=0,\ldots,n-1$, in the sums $\sum_{i=0}^{n-1}f(i/n)(B_{(i+1)/n}-B_{i/n})$. Every endpoint $i/n$ is rational, so this sum equals $\sum_{i=0}^{n-1}(B_{(i+1)/n}-B_{i/n})=B_1$. Since $\mathbb{P}(B_1=0)=0$, these sums do not give the Wiener integral of $f$. The correction is to use the Ito construction in the $\mathcal{L}^2$ norm: $f$ and the zero function represent the same $L^2$ class, so their integrals agree almost surely. These point-sampling sums are not a valid substitute for that construction.

A related measurability error is to claim that every deterministic process is progressively measurable regardless of its time measurability. A deterministic process generated by a Borel function is progressively measurable, as shown above; mere Lebesgue measurability is not enough for that argument. Choosing a Borel representative is the precise correction. These points are particularly important when defining an integral from an equivalence class rather than from a preferred pointwise formula.

Check Your Understanding

Use the definition, the Ito isometry, and the linearity result to answer the following.

Why must a deterministic representative be chosen Borel measurable when establishing progressive measurability?
If two Borel functions on $[0,T]$ differ only on a set of Lebesgue measure zero, what does the Ito isometry say about the difference of their Wiener integrals?
For $f(s)=c$ on $[0,T]$, compute $I_T(f)$ and verify its second moment using the Ito isometry.
Explain why a finite collection of linearity identities can be taken to hold simultaneously almost surely, even though each identity is defined only up to almost-sure equality.
For the rational-indicator example, why do the uniform left-endpoint sums fail to identify the Wiener integral?

What You'll Learn