Exponential and Trigonometric Wiener Integrals

From a Kernel to a Probability Law

A deterministic Wiener integral turns a function of time into a random variable. For a deterministic kernel $h$, the Ito isometry says that its variance is the ordinary integral of $h^2$, provided that $h$ is square-integrable. When the kernel is deterministic, the integral is also Gaussian. Thus the main work in many examples is to evaluate an ordinary integral accurately and then use the resulting variance to specify the probability law.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a finite horizon $T>0$. Throughout, $\int_0^T h(t)\,dB_t$ denotes the Ito integral of the deterministic, hence progressively measurable, integrand $H_t=h(t)$. We require $h\in L^2([0,T])$, meaning $\int_0^T h(t)^2\,dt<\infty$. Equivalently, $H$ belongs to the class $\mathcal{L}^2$ on this horizon because $\mathbb{E}\int_0^T H_t^2\,dt=\int_0^T h(t)^2\,dt<\infty$.

Definition: For a deterministic $h\in L^2([0,T])$, write $I_T(h)=\int_0^T h(t)\,dB_t$. The Ito isometry gives $\mathbb{E}[I_T(h)]=0$ and $\mathbb{E}[I_T(h)^2]=\int_0^T h(t)^2\,dt$, so $\operatorname{Var}(I_T(h))=\int_0^T h(t)^2\,dt$. These are standard properties of the Ito integral established in the earlier material on deterministic Wiener integrals.

The zero mean follows because the Ito integral of a square-integrable adapted process is a square-integrable martingale starting at zero. In particular, the variance calculation does not require finding the distribution first. The Gaussian law then completes the description: if the squared-kernel integral is $v$, the result has law $N(0,v)$. When $v=0$, this notation means the degenerate law concentrated at zero.

Exponential Kernels

For $a>0$, the kernel $h(t)=e^{-at}$ is square-integrable on every finite interval. Squaring it changes the decay rate from $a$ to $2a$, so the relevant ordinary integral is not $\int_0^T e^{-at}\,dt$, but $\int_0^T e^{-2at}\,dt$.

Proposition: Let $a>0$ and $T>0$. The deterministic Wiener integral $I_T(e^{-a\cdot})=\int_0^T e^{-at}\,dB_t$ is Gaussian with mean zero and variance $\dfrac{1-e^{-2aT}}{2a}$.

Proof. The kernel is continuous on the compact interval $[0,T]$, so its squared integral is finite and the Ito integral is defined. By the Ito isometry and the Gaussian-law property for deterministic Wiener integrals, it is centered Gaussian with variance

\int_0^T (e^{-at})^2\,dt =\int_0^T e^{-2at}\,dt =\left[-\frac{e^{-2at}}{2a}\right]_{0}^{T} =\frac{1-e^{-2aT}}{2a}.

Since $a>0$ and $T>0$, $e^{-2aT}<1$, so the variance is positive. Therefore the law is $N(0,(1-e^{-2aT})/(2a))$. $\square$

Worked Example: An Exponential Kernel on a Logarithmic Interval

Find the variance and law of $X=\int_0^{\ln 2} e^{-t}\,dB_t$. The integrand is deterministic and continuous, so it is progressively measurable, and $\int_0^{\ln 2}e^{-2t}\,dt$ is finite; thus it is an admissible $\mathcal{L}^2$ integrand.

Here $a=1$ and $T=\ln 2$. The squared kernel is $e^{-2t}$, and direct integration gives

\operatorname{Var}(X) =\int_0^{\ln 2}e^{-2t}\,dt =\left[-\frac12e^{-2t}\right]_0^{\ln 2} =\frac12\left(1-e^{-2\ln 2}\right) =\frac12\left(1-\frac14\right) =\frac38.

The deterministic-integrand Gaussian law therefore gives $X\sim N(0,\tfrac38)$. Check: the mean is zero, and the variance is positive because the integrand is not zero on the interval. As $T$ increases, the formula $(1-e^{-2T})/2$ increases toward $1/2$, consistent with accumulating squared kernel over a longer interval.

The same computation also covers the constant kernel by taking $a=0$ directly rather than substituting into a formula with denominator $2a$: $\int_0^T 1^2\,dt=T$, so $\int_0^T 1\,dB_t=B_T$ has law $N(0,T)$. This limiting case is a useful check on the scale of the formula.

Sine and Cosine Kernels

For frequency $b>0$, the sine and cosine kernels are bounded and therefore square-integrable on any finite interval. Their variances and covariance follow from the elementary identities $\sin^2 x=(1-\cos(2x))/2$, $\cos^2 x=(1+\cos(2x))/2$, and $\sin x\cos x=\sin(2x)/2$.

Proposition: For $b>0$ and $T>0$, let $S=\int_0^T\sin(bt)\,dB_t$ and $C=\int_0^T\cos(bt)\,dB_t$. These are jointly Gaussian, centered random variables, with $$ \operatorname{Var}(S)=\frac{T}{2}-\frac{\sin(2bT)}{4b},\qquad \operatorname{Var}(C)=\frac{T}{2}+\frac{\sin(2bT)}{4b},\qquad \operatorname{Cov}(S,C)=\frac{1-\cos(2bT)}{4b}. $$

Proof. Both kernels are deterministic and bounded on $[0,T]$, so each is progressively measurable and satisfies the $\mathcal{L}^2$ condition. The Ito isometry and its covariance form give the variances as the integrals of the squared kernels and the covariance as the integral of their product. Evaluating each one yields

\begin{aligned} \operatorname{Var}(S) &=\int_0^T\sin^2(bt)\,dt =\int_0^T\frac{1-\cos(2bt)}{2}\,dt =\frac{T}{2}-\frac{\sin(2bT)}{4b},\\ \operatorname{Var}(C) &=\int_0^T\cos^2(bt)\,dt =\int_0^T\frac{1+\cos(2bt)}{2}\,dt =\frac{T}{2}+\frac{\sin(2bT)}{4b},\\ \operatorname{Cov}(S,C) &=\int_0^T\sin(bt)\cos(bt)\,dt =\frac12\int_0^T\sin(2bt)\,dt =\frac{1-\cos(2bT)}{4b}. \end{aligned}

Each linear combination $uS+vC$, for real $u,v$, is the deterministic Wiener integral with kernel $u\sin(bt)+v\cos(bt)$. The deterministic-integrand Gaussian-law property makes every such combination Gaussian; hence $(S,C)$ is jointly Gaussian. This proves the asserted joint law parameters as well as the individual variances. $\square$

Worked Example: A Sine-Cosine Pair Over One Quarter-Period

Take $b=2$ and $T=\pi/2$. Compute the two variances, their covariance, and whether the two integrals are independent. Both kernels are bounded and deterministic, so they meet the required progressive measurability and finite-energy conditions.

For $S=\int_0^{\pi/2}\sin(2t)\,dB_t$, the variance is

\operatorname{Var}(S) =\int_0^{\pi/2}\sin^2(2t)\,dt =\frac{\pi/2}{2}-\frac{\sin(2\cdot2\cdot\pi/2)}{4\cdot2} =\frac{\pi}{4}-\frac{\sin(2\pi)}{8} =\frac{\pi}{4}.

For $C=\int_0^{\pi/2}\cos(2t)\,dB_t$,

\operatorname{Var}(C) =\int_0^{\pi/2}\cos^2(2t)\,dt =\frac{\pi/2}{2}+\frac{\sin(2\cdot2\cdot\pi/2)}{4\cdot2} =\frac{\pi}{4}+\frac{\sin(2\pi)}{8} =\frac{\pi}{4}.

The covariance is

\operatorname{Cov}(S,C) =\int_0^{\pi/2}\sin(2t)\cos(2t)\,dt =\frac{1-\cos(2\cdot2\cdot\pi/2)}{4\cdot2} =\frac{1-\cos(2\pi)}{8} =0.

The pair is jointly Gaussian by the proposition. For a jointly Gaussian pair, zero covariance implies independence: its joint characteristic function factors into the two marginal characteristic functions when the covariance is zero. Thus $S$ and $C$ are independent, each with law $N(0,\pi/4)$. Check: the two variances sum to $\pi/2=T$, matching $\int_0^T(\sin^2(2t)+\cos^2(2t))\,dt=\int_0^T1\,dt=T$.

Zero covariance does not in general imply independence for arbitrary random variables. The joint Gaussian property is essential to the conclusion in this example. Also, changing the endpoint changes the covariance: the formula contains $1-\cos(2bT)$, which need not vanish at an arbitrary $T$.

An Exponentially Damped Sine

Products of exponential and trigonometric kernels occur naturally when a signal oscillates while decaying. The integral is still computed from the squared kernel, but now the trigonometric identity must be combined with an exponential integral. The next calculation gives one reusable formula and then applies it at parameters that simplify exactly.

Proposition: For $a>0$, $b>0$, and $T>0$, the integral $Y=\int_0^T e^{-at}\sin(bt)\,dB_t$ is centered Gaussian with variance $$ \operatorname{Var}(Y) =\frac{1-e^{-2aT}}{4a} -\frac{a+e^{-2aT}\big(-a\cos(2bT)+b\sin(2bT)\big)}{4(a^2+b^2)}. $$

Proof. The deterministic kernel $e^{-at}\sin(bt)$ is continuous on $[0,T]$, and its squared integral is finite, so it defines an Ito integral with an $\mathcal{L}^2$ integrand. By the isometry, the variance is $\int_0^T e^{-2at}\sin^2(bt)\,dt$. Use $\sin^2(bt)=(1-\cos(2bt))/2$. The first resulting integral is $(1-e^{-2aT})/(4a)$. For the second, direct differentiation verifies the antiderivative $$ \int e^{-2at}\cos(2bt)\,dt =\frac{e^{-2at}(-a\cos(2bt)+b\sin(2bt))}{2(a^2+b^2)}. $$

Evaluating this antiderivative at $T$ and $0$ gives $$ \int_0^T e^{-2at}\cos(2bt)\,dt =\frac{a+e^{-2aT}(-a\cos(2bT)+b\sin(2bT))}{2(a^2+b^2)}. $$ The cosine integral is multiplied by $-1/2$ in the squared-sine identity, producing the displayed variance. The deterministic-integrand Gaussian law then gives the centered Gaussian distribution with that variance. $\square$

Worked Example: A Damped Sine Over a Full Oscillation

Let $Y=\int_0^\pi e^{-t}\sin(t)\,dB_t$. Its kernel is deterministic and continuous, hence progressively measurable, and its squared integral is finite on $[0,\pi]$. Use the proposition with $a=1$, $b=1$, and $T=\pi$. Then $e^{-2aT}=e^{-2\pi}$, $\cos(2bT)=\cos(2\pi)=1$, and $\sin(2bT)=\sin(2\pi)=0$. Substitution gives

\begin{aligned} \operatorname{Var}(Y) &=\frac{1-e^{-2\pi}}{4} -\frac{1+e^{-2\pi}(-1\cdot1+1\cdot0)}{4(1^2+1^2)}\\ &=\frac{1-e^{-2\pi}}{4}-\frac{1-e^{-2\pi}}{8} =\frac{1-e^{-2\pi}}{8}. \end{aligned}

Consequently, $Y\sim N(0,(1-e^{-2\pi})/8)$. Check: the variance is positive because $e^{-2\pi}<1$. If the damping parameter $a$ is increased while $b,T$ are fixed, the integrand decays faster; the original variance integral $\int_0^T e^{-2at}\sin^2(bt)\,dt$ accordingly decreases.

Why the Squared Kernel Matters

A common mistake is to integrate the kernel itself and treat the result as a variance. For example, the incorrect calculation $$ \operatorname{Var}\left(\int_0^T e^{-t}\,dB_t\right) \mathrel{\color{red}{=}}\int_0^T e^{-t}\,dt=1-e^{-T} $$ uses the wrong integrand on the right-hand side. The Ito isometry squares the integrand before integrating. The correction is $$ \operatorname{Var}\left(\int_0^T e^{-t}\,dB_t\right) =\int_0^T(e^{-t})^2\,dt =\frac{1-e^{-2T}}{2}. $$ For $T=\ln 2$, the wrong value would be $1/2$, while the correct value is $3/8$, as computed above. The difference is consequential: it changes the probability law, not merely its presentation.

For a collection of deterministic kernels $h_1,\ldots,h_n\in L^2([0,T])$, the same principle gives $$ \operatorname{Cov}\left(\int_0^T h_i(t)\,dB_t,\int_0^T h_j(t)\,dB_t\right) =\int_0^T h_i(t)h_j(t)\,dt. $$ Each integral is centered, and the vector is jointly Gaussian because every linear combination is a Wiener integral with a deterministic square-integrable kernel. Thus explicit ordinary integrals of products determine both the marginal variances and the dependence structure. If the covariance matrix is diagonal, the coordinates of this jointly Gaussian vector are independent.

These calculations are useful whenever deterministic time profiles weight Brownian noise. Exponential decay controls the energy accumulated over time; sine and cosine kernels encode oscillations; and products of the two combine decay with periodic behavior. In each case, the reliable order is to check square-integrability, square the kernel, evaluate the resulting ordinary integral, and then state the Gaussian law. Keeping that order prevents confusing the deterministic kernel's integral with the variance of the random integral.

Check Your Understanding

For each question, treat the integral as an Ito integral on the stated finite interval with a deterministic kernel.

Find the variance and law of $\int_0^T e^{-3t}\,dB_t$ for $T>0$.
Compute $\operatorname{Cov}(\int_0^{\pi/4}\sin(2t)\,dB_t,\int_0^{\pi/4}\cos(2t)\,dB_t)$.
For what positive endpoints $T$ are the sine and cosine integrals with the same frequency $b>0$ uncorrelated?
Explain why zero covariance in the sine-cosine example implies independence, and name the condition on their joint law that is needed.
Write the squared-kernel integral that gives the variance of $\int_0^T e^{-at}\cos(bt)\,dB_t$, for $a>0$ and $b>0$.