Random-Amplitude Integrands

When the Amplitude Is Random but Fixed in Time

A deterministic Wiener integral $\int_0^T f(t)\,dB_t$ is built from Brownian increments with deterministic weights. If the weights are all multiplied by a random variable known at time zero, the resulting integral has a useful factorization. The random variable does not change over time, but it can change the integral’s distribution and its second moment.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to this filtration. In particular, Brownian increments after time $s$ are independent of $\mathcal{F}_s$. Fix $T>0$, let $Z$ be a real-valued, square-integrable, $\mathcal{F}_0$-measurable random variable, and let $f:[0,T]\to\mathbb{R}$ be deterministic and measurable with $\int_0^T f(t)^2\,dt<\infty$. All stochastic integrals below are Itô integrals. Their integrands belong to $\mathcal{L}^2$: they are progressively measurable and have finite expected integrated square.

Definition: The random-amplitude integrand is $H_t=Zf(t)$. It is progressively measurable because $Z$ is $\mathcal{F}_0$-measurable and $f$ is deterministic and measurable. It belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^T H_t^2\,dt=\mathbb{E}[Z^2]\int_0^T f(t)^2\,dt<\infty$.

The integral’s factorization is not an appeal to an ordinary chain rule or to a Stratonovich convention. It follows from the construction of the Itô integral and the independence between $\mathcal{F}_0$ and Brownian increments. The same independence is what allows the second moment to separate into a factor involving $Z$ and a factor involving $f$.

Factorization, Moments, and Approximation

Theorem: Under the assumptions above, for the Itô integral with integrand $H_t=Zf(t)$, $$ \int_0^T Zf(t)\,dB_t=Z\int_0^T f(t)\,dB_t $$ in $L^2(\mathbb{P})$. Writing $I=\int_0^T Zf(t)\,dB_t$, we have $$ \mathbb{E}[I]=0,\qquad \mathbb{E}[I^2]=\mathbb{E}[Z^2]\int_0^T f(t)^2\,dt,\qquad \mathbb{E}[I^2\mid\mathcal{F}_0]=Z^2\int_0^T f(t)^2\,dt. $$ If $g$ is any deterministic square-integrable step function on $[0,T]$ and $I_g=\int_0^T Zg(t)\,dB_t$, then $$ \mathbb{E}\big[(I-I_g)^2\big]=\mathbb{E}[Z^2]\int_0^T(f(t)-g(t))^2\,dt. $$

Proof. Choose deterministic step functions $f_n$ converging to $f$ in $L^2([0,T])$, and put $J_n=\int_0^T f_n(t)\,dB_t$ and $J=\int_0^T f(t)\,dB_t$. Each $J_n$ is a finite sum of deterministic multiples of Brownian increments. The Brownian increments are independent of $\mathcal{F}_0$, so $J_n$ is independent of $\mathcal{F}_0$. The Itô isometry gives $\mathbb{E}[(J_n-J)^2]=\int_0^T(f_n-f)^2\,dt\to0$. Thus $J$ is measurable with respect to the completed sigma-field generated by Brownian motion on $[0,T]$, which is independent of $\mathcal{F}_0$; in particular, $J_n-J$ is independent of $Z$. By the definition of the integral for step integrands, $\int_0^T Zf_n(t)\,dB_t=ZJ_n$. Also, $$ \mathbb{E}\big[(ZJ_n-ZJ)^2\big] =\mathbb{E}[Z^2]\mathbb{E}[(J_n-J)^2] =\mathbb{E}[Z^2]\int_0^T(f_n-f)^2\,dt\longrightarrow0. $$ The Itô isometry applied to $Z(f_n-f)$ shows that the left-hand sequence of integrals converges in $L^2(\mathbb{P})$ to $\int_0^T Zf(t)\,dB_t$. Uniqueness of an $L^2$ limit proves the factorization. For the moment identities, $J$ is centered and independent of $Z$, and the Itô isometry gives $\mathbb{E}[J^2]=\int_0^T f^2\,dt$. Therefore $\mathbb{E}[ZJ]=\mathbb{E}[Z]\mathbb{E}[J]=0$ and $\mathbb{E}[(ZJ)^2]=\mathbb{E}[Z^2]\mathbb{E}[J^2]$. Independence also gives $\mathbb{E}[J^2\mid\mathcal{F}_0]=\mathbb{E}[J^2]$, so $\mathbb{E}[Z^2J^2\mid\mathcal{F}_0]=Z^2\mathbb{E}[J^2]$. Finally, apply the same factorization to $f-g$: the difference of the two integrals is $Z\int_0^T(f-g)\,dB_t$. Its second moment, by independence and the Itô isometry, is $\mathbb{E}[Z^2]\int_0^T(f-g)^2\,dt$. All expressions remain valid in degenerate cases: if $f=0$ almost everywhere or $\mathbb{E}[Z^2]=0$, the integral and the stated second moments are zero. $\square$

The approximation formula gives more than convergence: it is the exact squared $L^2$ error, not just an upper bound. If $g_n$ approaches $f$ in deterministic $L^2([0,T])$, then the random-amplitude integral of $g_n$ approaches $I$ in $L^2(\mathbb{P})$. The error is scaled by $\mathbb{E}[Z^2]$, not by $(\mathbb{E}[Z])^2$.

The Integral Process Is a Martingale

For $0\le t\le T$, define $M_t=\int_0^t Zf(u)\,dB_u$, using the same $\mathcal{L}^2$ assumptions on each interval $[0,t]$. This is a true martingale, not merely a local martingale. Indeed, the Itô isometry gives $\mathbb{E}[M_t^2]=\mathbb{E}[Z^2]\int_0^t f(u)^2\,du<\infty$, so $M_t$ is integrable. For $0\le s\le t$, the increment has the factorization $M_t-M_s=Z\int_s^t f(u)\,dB_u$. The latter deterministic Wiener integral is independent of $\mathcal{F}_s$, is centered, and is square-integrable. Since $Z$ is $\mathcal{F}_s$-measurable, it follows that $\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0$, hence $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$.

Worked Example: A Two-Level Amplitude

Let $T=1$, let $\mathbb{P}(Z=1)=\mathbb{P}(Z=3)=\tfrac12$, and set $f(t)=1$ for $0\le t<\tfrac12$ and $f(t)=2$ for $\tfrac12\le t\le1$. Compute the integral and its second moment.

First compute the amplitude’s second moment: $\mathbb{E}[Z^2]=\tfrac12(1^2)+\tfrac12(3^2)=\tfrac12+\tfrac92=5$. The deterministic Wiener integral is a weighted sum of two Brownian increments: $$ J=\int_0^1 f(t)\,dB_t =(B_{1/2}-B_0)+2(B_1-B_{1/2}). $$ The factorization gives $I=ZJ$. Since the increments are independent and centered, and their variances are both $\tfrac12$, $$ \mathbb{E}[J^2]=\tfrac12+2^2\cdot\tfrac12=\tfrac12+2=\tfrac52. $$ Therefore $\mathbb{E}[I]=0$, and $\mathbb{E}[I^2]=5\cdot\tfrac52=\tfrac{25}{2}$.

Check: Directly expanding $J^2$, the cross term has expectation zero because the increments are independent and centered. Thus $\mathbb{E}[J^2]=\tfrac12+4\cdot\tfrac12=\tfrac52$, and multiplying by the independently distributed $Z^2$ again gives $\tfrac{25}{2}$.

Worked Example: Error from a Left-Endpoint Approximation

Let $T=1$, let $Z$ have values $1$ and $3$, each with probability $\tfrac12$, and let $f(t)=t$. Approximate $I=\int_0^1 Zt\,dB_t$ using four equal intervals and left endpoints, and compute the exact mean-square error.

Here $g(t)=0,\tfrac14,\tfrac12,\tfrac34$ on the successive intervals $[0,\tfrac14),[\tfrac14,\tfrac12),[\tfrac12,\tfrac34),[\tfrac34,1]$. Thus $$ I_g=Z\left(0\cdot(B_{1/4}-B_0)+\tfrac14(B_{1/2}-B_{1/4}) +\tfrac12(B_{3/4}-B_{1/2})+\tfrac34(B_1-B_{3/4})\right). $$ On each interval, write $t=a+u$, where $a$ is its left endpoint and $0\le u\le\tfrac14$. Then $f(t)-g(t)=u$, and the error integral on that interval is $\int_0^{1/4}u^2\,du=(\tfrac14)^3/3=\tfrac1{192}$. There are four intervals, so $\int_0^1(f-g)^2\,dt=4\cdot\tfrac1{192}=\tfrac1{48}$. As $\mathbb{E}[Z^2]=\tfrac12(1)+\tfrac12(9)=5$, the theorem gives $$ \mathbb{E}[(I-I_g)^2]=5\cdot\tfrac1{48}=\tfrac5{48}. $$ For $n$ equal intervals, the same calculation gives $\mathbb{E}[(I-I_g)^2]=5/(3n^2)$, which tends to zero.

Check: For one interval, the left-endpoint approximation is zero. Its error is $I$, whose second moment is $\mathbb{E}[Z^2]\int_0^1t^2\,dt=5\cdot\tfrac13=\tfrac53$. The formula for $n=1$ gives $5/(3\cdot1^2)=\tfrac53$, as required.

Random Amplitudes Need Not Preserve Gaussianity

A deterministic Wiener integral with a square-integrable deterministic integrand is Gaussian. Multiplication by an independent random amplitude generally produces a mixture of Gaussian distributions rather than a Gaussian distribution. The factorization theorem still gives the second moment exactly, but that moment alone does not determine the distribution.

Worked Example: A Non-Gaussian Random-Amplitude Integral

Let $T=1$, $f(t)=1$, and let $Z=1$ or $Z=2$, each with probability $\tfrac12$. Then $I=ZB_1$, with $Z$ independent of $B_1$. Compute the second and fourth moments and test whether $I$ is Gaussian.

Since $\mathbb{E}[Z^2]=\tfrac12(1)+\tfrac12(4)=\tfrac52$ and $\mathbb{E}[B_1^2]=1$, we have $\mathbb{E}[I^2]=\tfrac52$. Also, $\mathbb{E}[Z^4]=\tfrac12(1)+\tfrac12(16)=\tfrac{17}{2}$, while a standard normal variable has fourth moment $3$. Independence gives $$ \mathbb{E}[I^4]=\mathbb{E}[Z^4]\mathbb{E}[B_1^4] =\tfrac{17}{2}\cdot3=\tfrac{51}{2}. $$ A centered Gaussian variable with variance $\tfrac52$ would have fourth moment $3(\tfrac52)^2=\tfrac{75}{4}$, which differs from $\tfrac{51}{2}=\tfrac{102}{4}$. Therefore $I$ is not Gaussian.

Check: Conditional on $Z=1$, $I$ has variance $1$; conditional on $Z=2$, it has variance $4$. The unconditional variance is their probability-weighted average, $\tfrac12(1)+\tfrac12(4)=\tfrac52$, matching the second-moment calculation.

Common Pitfall: Using the Mean of the Amplitude

A frequent error is to replace $\mathbb{E}[Z^2]$ in the second-moment formula by $(\mathbb{E}[Z])^2$. Those quantities need not agree. For the amplitude taking values $1$ and $3$ with equal probabilities, $\mathbb{E}[Z]=2$, so the incorrect multiplier would be $4$; the correct multiplier is $\mathbb{E}[Z^2]=5$. In the first worked example, this mistake would give the incorrect second moment $4\cdot\tfrac52=10$, rather than $\tfrac{25}{2}$. The correct computation squares the random amplitude first and then takes expectation.

The initial measurability assumption matters as well. The factorization argument uses that $Z$ is known at time zero and independent of the Brownian path’s increments. It does not license pulling an arbitrary time-dependent random process outside an Itô integral. Here the integrand is specifically $Zf(t)$, with $Z\in\mathcal{F}_0$ and deterministic $f$.

Check Your Understanding

Use the factorization theorem and the Itô isometry to answer the following.

If $\mathbb{E}[Z^2]=4$ and $\int_0^T f(t)^2\,dt=\tfrac32$, what are the mean and second moment of $\int_0^T Zf(t)\,dB_t$?
For $f(t)=t$ on $[0,1]$ and left-endpoint approximations on $n$ equal intervals, derive the deterministic squared approximation error.
Why is $Z$ independent of $\int_0^T f(t)\,dB_t$ under the stated assumptions?
Give an example of a nonzero $Z$ with $\mathbb{E}[Z^2]=0$, or explain why none exists. What does the integral become in that degenerate case?
For $Z=1$ or $2$, each with probability $\tfrac12$, and $f(t)=1$ on $[0,1]$, compare the fourth moment of the integral with the fourth moment of a Gaussian having the same variance.

What You'll Learn