Covariance of Deterministic Wiener Integrals

Covariance as an Inner Product

For deterministic Wiener integrals, covariance is determined by how the integrands overlap in $L^2$. Portions of time where the integrands have the same sign contribute positively; portions where their signs differ contribute negatively. This gives a direct way to compute second moments of several Wiener integrals without constructing their full joint distribution.

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 progressively measurable process $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, its Ito integral $\int_0^T H_s\,dB_s$ is square-integrable. In this tutorial the integrands are deterministic functions. We take them to be Borel-measurable and square-integrable on $[0,T]$; then $H_s=f(s)$ is progressively measurable and satisfies the required integrability condition.

Lebesgue-measurable $L^2$ functions are understood up to equality almost everywhere. Each such equivalence class has a Borel-measurable representative, which can be used to define the progressively measurable process. The Ito isometry shows that changing the representative on a Lebesgue-null set does not change the integral in $L^2(\mathbb{P})$.

Definition: For $f\in L^2([0,T])$, write $I_T(f)=\int_0^T f(s)\,dB_s$. For square-integrable random variables $X,Y$, their covariance is $\operatorname{Cov}(X,Y)=\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)]$. The deterministic $L^2$ inner product is $\langle f,g\rangle=\int_0^T f(s)g(s)\,ds$.

The integral $I_T(f)$ is obtained as an $L^2(\mathbb{P})$ limit of integrals of deterministic step functions approximating $f$. We use the Ito isometry, established in the earlier material on the Ito integral: for square-integrable integrands $H$, $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. In particular, deterministic step-function approximations converge to $I_T(f)$ in $L^2(\mathbb{P})$. Step functions are dense in $L^2([0,T])$.

The Covariance Formula

Theorem: If $f,g\in L^2([0,T])$ are deterministic, then $$ \mathbb{E}[I_T(f)]=\mathbb{E}[I_T(g)]=0,\qquad \operatorname{Cov}(I_T(f),I_T(g))=\int_0^T f(s)g(s)\,ds. $$ Consequently, $\operatorname{Var}(I_T(f))=\int_0^T f(s)^2\,ds$.

Proof. First let $f$ and $g$ be step functions on a common partition $0=t_0<t_1<\cdots<t_n=T$. Write $f=c_i$ and $g=d_i$ on $(t_{i-1},t_i]$, where $c_i,d_i$ are real constants. Their Ito integrals are $$ I_T(f)=\sum_{i=1}^n c_i(B_{t_i}-B_{t_{i-1}}),\qquad I_T(g)=\sum_{i=1}^n d_i(B_{t_i}-B_{t_{i-1}}). $$ Each increment has mean zero, so both integrals have mean zero. Brownian increments over disjoint intervals are independent and centered, so the expectation of the product of two distinct increments is zero. The variance of the $i$th increment is $t_i-t_{i-1}$. Expanding the product therefore gives $$ \mathbb{E}[I_T(f)I_T(g)] =\sum_{i=1}^n c_i d_i(t_i-t_{i-1}) =\int_0^T f(s)g(s)\,ds. $$ This proves the assertion for step functions.

For general $f,g\in L^2([0,T])$, choose deterministic step functions $f_n,g_n$ with $\|f_n-f\|_{L^2}\to0$ and $\|g_n-g\|_{L^2}\to0$. By the Ito isometry, $I_T(f_n)\to I_T(f)$ and $I_T(g_n)\to I_T(g)$ in $L^2(\mathbb{P})$. Since each approximating integral has mean zero, $L^2$ convergence implies $L^1$ convergence and hence $\mathbb{E}[I_T(f)]=\mathbb{E}[I_T(g)]=0$. Moreover, the $L^2$ norms of $I_T(g_n)$ are bounded, because they converge in $L^2$. Cauchy–Schwarz gives $$ \left|\mathbb{E}[I_T(f_n)I_T(g_n)]-\mathbb{E}[I_T(f)I_T(g)]\right| \le \|I_T(f_n)-I_T(f)\|_{L^2(\mathbb{P})}\|I_T(g_n)\|_{L^2(\mathbb{P})} +\|I_T(f)\|_{L^2(\mathbb{P})}\|I_T(g_n)-I_T(g)\|_{L^2(\mathbb{P})}, $$ which tends to zero. On the deterministic side, Cauchy–Schwarz in $L^2([0,T])$ gives $$ |\langle f_n,g_n\rangle-\langle f,g\rangle| \le \|f_n-f\|_{L^2}\|g_n\|_{L^2}+\|f\|_{L^2}\|g_n-g\|_{L^2}\longrightarrow0. $$ The step-function identity thus passes to the limit. Taking $g=f$ gives the variance formula. $\square$

The formula identifies covariance with an inner product, not with a product of ordinary integrals. Its sign depends on the signed overlap of $f$ and $g$, and its value is unchanged by modifying either integrand on a set of Lebesgue measure zero.

Computing with Piecewise-Constant Integrands

Worked Example: Overlapping Intervals with Different Signs

On $[0,3]$, let $f=2$ on $[0,1]$ and $f=-1$ on $(1,3]$. Let $g=1$ on $[0,2]$ and $g=3$ on $(2,3]$. Compute the two variances and the covariance of $I_3(f)$ and $I_3(g)$.

Split the inner products at the points where either function changes. For $f$, its squared values are $4$ for one unit of time and $1$ for two units. Thus $$ \operatorname{Var}(I_3(f))=\int_0^3 f(s)^2\,ds=4\cdot1+1\cdot2=6. $$ For $g$, its squared values are $1$ for two units of time and $9$ for one unit. Thus $$ \operatorname{Var}(I_3(g))=\int_0^3 g(s)^2\,ds=1\cdot2+9\cdot1=11. $$ For the covariance, the product $fg$ equals $2$ on $[0,1]$, $-1$ on $(1,2]$, and $-3$ on $(2,3]$. Therefore $$ \operatorname{Cov}(I_3(f),I_3(g))=2\cdot1+(-1)\cdot1+(-3)\cdot1=-2. $$

Check: Set $A=B_1-B_0$, $C=B_2-B_1$, and $D=B_3-B_2$. Then $I_3(f)=2A-C-D$ and $I_3(g)=A+C+3D$. The three increments are independent, centered, and each has variance $1$. Expanding the covariance gives $2\operatorname{Var}(A)-\operatorname{Var}(C)-3\operatorname{Var}(D)=2-1-3=-2$, agreeing with the integral calculation. The variances also check directly: $4+1+1=6$ and $1+1+9=11$.

Covariances for Smooth Integrands

Worked Example: A Linear Function and Its Complement

On $[0,1]$, take $f(s)=s$ and $g(s)=1-s$. Compute the covariance and both variances, then check the result using $f+g$.

Both functions are Borel-measurable and square-integrable on $[0,1]$. Direct integration gives $$ \operatorname{Cov}(I_1(f),I_1(g)) =\int_0^1 s(1-s)\,ds =\int_0^1(s-s^2)\,ds =\frac12-\frac13=\frac16. $$ Their variances are $$ \operatorname{Var}(I_1(f))=\int_0^1s^2\,ds=\frac13,\qquad \operatorname{Var}(I_1(g))=\int_0^1(1-s)^2\,ds =\left[s-s^2+\frac{s^3}{3}\right]_0^1=\frac13. $$

Check: Since $f+g=1$, linearity of the Ito integral gives $I_1(f)+I_1(g)=I_1(1)=B_1$. The variance of this sum, computed from the individual variances and covariance, is $$ \frac13+\frac13+2\cdot\frac16=1. $$ This equals $\operatorname{Var}(B_1)=1$, as required.

Worked Example: Zero Covariance from Disjoint Signed Contributions

On $[0,1]$, let $f(s)=1$ throughout, and let $g(s)=1$ on $[0,\tfrac12]$ and $g(s)=-1$ on $(\tfrac12,1]$. Compute the covariance and variances.

The product $fg=g$, so $$ \operatorname{Cov}(I_1(f),I_1(g)) =\int_0^{1/2}1\,ds+\int_{1/2}^1(-1)\,ds =\frac12-\frac12=0. $$ For the variances, $$ \operatorname{Var}(I_1(f))=\int_0^1 1\,ds=1,\qquad \operatorname{Var}(I_1(g))=\int_0^{1/2}1\,ds+\int_{1/2}^1 1\,ds =\frac12+\frac12=1. $$

Check: With $A=B_{1/2}-B_0$ and $C=B_1-B_{1/2}$, the integrals are $I_1(f)=A+C$ and $I_1(g)=A-C$. The covariance is $\operatorname{Var}(A)-\operatorname{Var}(C)=\tfrac12-\tfrac12=0$, and each variance is $\tfrac12+\tfrac12=1$. Zero covariance is the conclusion here; covariance alone does not establish independence.

A Bound and Its Equality Cases

Proposition: For deterministic $f,g\in L^2([0,T])$, $$ |\operatorname{Cov}(I_T(f),I_T(g))| \le \sqrt{\operatorname{Var}(I_T(f))\operatorname{Var}(I_T(g))}. $$ If both variances are positive, equality holds exactly when $f=cg$ almost everywhere for some nonzero constant $c$. If either variance is zero, both sides are zero.

Proof. The covariance theorem identifies the covariance with $\langle f,g\rangle$ and the variances with $\|f\|_{L^2}^2$ and $\|g\|_{L^2}^2$. The Cauchy–Schwarz inequality in $L^2([0,T])$ therefore gives the stated bound. If $\|g\|_{L^2}>0$, write $$ f=\frac{\langle f,g\rangle}{\|g\|_{L^2}^2}g+h,\qquad \langle h,g\rangle=0. $$ Taking squared norms yields $$ \|f\|_{L^2}^2=\frac{\langle f,g\rangle^2}{\|g\|_{L^2}^2}+\|h\|_{L^2}^2. $$ Equality in the bound holds exactly when $\|h\|_{L^2}^2=0$, that is, when $f$ is a constant multiple of $g$ almost everywhere. If both variances are positive, that constant is nonzero. If either variance is zero, the corresponding integrand is zero almost everywhere, so the covariance and both sides of the inequality are zero. $\square$

Common Mistake: Multiplying the Integrals of the Functions

A tempting but incorrect calculation is to replace $\operatorname{Cov}(I_T(f),I_T(g))$ by $(\int_0^T f(s)\,ds)(\int_0^T g(s)\,ds)$. For the first worked example, $\int_0^3 f(s)\,ds=2\cdot1-1\cdot2=0$, while $\int_0^3 g(s)\,ds=1\cdot2+3\cdot1=5$. This incorrect rule would give $0\cdot5=0$, whereas the correct covariance is $\int_0^3 fg\,ds=-2$.

The error confuses ordinary integrals of deterministic functions with the second moment of stochastic integrals. The correct calculation pairs the two integrands at the same time $s$, multiplies them, and then integrates that product. This also explains why the covariance can be negative, zero, or positive. In particular, zero covariance means the integrands are orthogonal in $L^2([0,T])$; it does not, by itself, state that the random variables are independent.

Check Your Understanding

Use the covariance identity and show the intermediate inner-product calculations.

For $f=2$ on $[0,1]$, $f=-1$ on $(1,2]$, and $g=1$ on $[0,2]$, compute both variances and the covariance.
On $[0,1]$, compute $\operatorname{Cov}(I_1(s),I_1(2s))$ and check it using the variance of $I_1(3s)$.
Give an example of two nonzero deterministic $L^2$ functions with zero covariance of their Wiener integrals.
State the equality condition in the covariance bound when both variances are positive.
Explain why changing a deterministic integrand on a Lebesgue-null set does not change its Wiener integral in $L^2(\mathbb{P})$.