Integrating Deterministic Step Functions

From a Step Integrand to a Gaussian Sum

For a deterministic step function, the Ito integral is an explicit finite sum. Each term is a fixed coefficient multiplied by a Brownian increment, and increments over disjoint time intervals are independent Gaussian random variables. This makes the distribution and moments of the integral directly computable.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $T>0$ and a partition $0=t_0<t_1<\cdots<t_n=T$. Write $\Delta t_j=t_j-t_{j-1}>0$, and let $h_1,\ldots,h_n$ be deterministic real numbers. Set $H_s=h_j$ on $(t_{j-1},t_j]$; the value at the single time $0$ can be chosen arbitrarily.

Definition: The deterministic step integrand $H$ is integrated here in the Ito sense. It is progressively measurable and belongs to $\mathcal{L}^2$, the class of progressively measurable processes satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Its Ito integral is $$ \int_0^T H_s\,dB_s=\sum_{j=1}^n h_j(B_{t_j}-B_{t_{j-1}}). $$ The integral is not a Stratonovich integral: no symmetric-sum convention or Ito-to-Stratonovich correction is being applied.

In this setting the integrability condition is finite because $\mathbb{E}\int_0^T H_s^2\,ds=\sum_{j=1}^n h_j^2\Delta t_j$, a finite sum. Since the partition intervals have positive lengths, $H$ is square-integrable precisely when its step heights are finite. Changing the values of $H$ at finitely many endpoints does not change this time integral or the displayed sum.

Distribution, Mean, and Variance

Theorem: Let $I_T=\int_0^T H_s\,dB_s$ for the deterministic step integrand above, and put $V=\sum_{j=1}^n h_j^2\Delta t_j$. If $V>0$, then $I_T$ has distribution $N(0,V)$, so $\mathbb{E}[I_T]=0$ and $\operatorname{Var}(I_T)=V$. If $V=0$, then $I_T=0$ almost surely, with mean and variance both zero.

Proof. Let $\Delta B_j=B_{t_j}-B_{t_{j-1}}$. Brownian independent increments give independent random variables $\Delta B_j$, each with distribution $N(0,\Delta t_j)$. For any real $u$, independence and the characteristic function of a centered normal variable give

\mathbb{E}\!\left[e^{iuI_T}\right] =\prod_{j=1}^n\mathbb{E}\!\left[e^{iu h_j\Delta B_j}\right] =\prod_{j=1}^n\exp\!\left(-\frac{u^2h_j^2\Delta t_j}{2}\right) =\exp\!\left(-\frac{u^2V}{2}\right).

When $V>0$, this is the characteristic function of $N(0,V)$, proving the claimed distribution. Its mean is zero and its variance is $V$. When $V=0$, every nonnegative summand $h_j^2\Delta t_j$ must be zero. Since $\Delta t_j>0$, every $h_j=0$, and the finite-sum definition gives $I_T=0$ almost surely. This also verifies the stated moments in the degenerate case. $\square$

The mean and variance can also be seen directly from the sum: each increment has mean zero and variance $\Delta t_j$, and distinct increments are independent. Thus cross-covariances vanish and the variance is the sum of the individual variances $h_j^2\Delta t_j$. The characteristic-function argument supplies the stronger conclusion that the entire law is Gaussian.

Worked Example: Two Different Step Heights

Take $T=2$, with $H_s=2$ on $(0,\tfrac12]$ and $H_s=-1$ on $(\tfrac12,2]$. Compute the integral’s distribution, mean, and variance.

The two time lengths are $\tfrac12$ and $2-\tfrac12=\tfrac32$. The integral is

I_2=2(B_{1/2}-B_0)-(B_2-B_{1/2}) =3B_{1/2}-B_2,

where $B_0=0$. The variance parameter from the theorem is

V=2^2\left(\frac12\right)+(-1)^2\left(\frac32\right) =2+\frac32=\frac72.

Therefore $I_2\sim N(0,\tfrac72)$, $\mathbb{E}[I_2]=0$, and $\operatorname{Var}(I_2)=\tfrac72$.

Check: In the endpoint expression $3B_{1/2}-B_2$, the covariance is $\operatorname{Cov}(B_{1/2},B_2)=\tfrac12$. Hence its variance is $9(\tfrac12)+2-6(\tfrac12)=\tfrac92+2-3=\tfrac72$, agreeing with the increment calculation.

Worked Example: A Step Function That Vanishes on One Interval

On $[0,1]$, set $H_s=1$ on $(0,\tfrac14]$, $H_s=-2$ on $(\tfrac14,\tfrac34]$, and $H_s=0$ on $(\tfrac34,1]$. Find the integral’s law and moments.

The integral is the sum of the three weighted increments:

I_1=(B_{1/4}-B_0)-2(B_{3/4}-B_{1/4})+0(B_1-B_{3/4}) =3B_{1/4}-2B_{3/4}.

The interval lengths are $\tfrac14$, $\tfrac34-\tfrac14=\tfrac12$, and $1-\tfrac34=\tfrac14$. Thus

V=1^2\left(\frac14\right)+(-2)^2\left(\frac12\right)+0^2\left(\frac14\right) =\frac14+2+0=\frac94.

It follows that $I_1\sim N(0,\tfrac94)$, with mean $0$ and variance $\tfrac94$. The zero-height interval contributes neither an increment to the sum nor any variance.

Check: Since $\operatorname{Cov}(B_{1/4},B_{3/4})=\tfrac14$, the endpoint expression has variance $9(\tfrac14)+4(\tfrac34)-12(\tfrac14)=\tfrac94+3-3=\tfrac94$. This matches the variance computed from the independent increments.

The Integral at an Intermediate Time

Proposition: For $t\in[t_{k-1},t_k]$, the integral process $I_t=\int_0^t H_s\,dB_s$ satisfies $$ I_t=\sum_{j=1}^{k-1}h_j(B_{t_j}-B_{t_{j-1}})+h_k(B_t-B_{t_{k-1}}). $$ It is Gaussian with mean zero and variance $$ V(t)=\sum_{j=1}^{k-1}h_j^2\Delta t_j+h_k^2(t-t_{k-1}). $$ If $V(t)>0$, its law is $N(0,V(t))$; if $V(t)=0$, it is zero almost surely.

Proof. Up to time $t$, the completed intervals are $(t_{j-1},t_j]$ for $j<k$, and the remaining part of the current interval is $(t_{k-1},t]$. Applying the finite-sum definition to these pieces gives the displayed expression. The Brownian increments in these disjoint time intervals are independent; the last one is centered normal with variance $t-t_{k-1}$, including the degenerate zero increment when $t=t_{k-1}$. Multiplication by each deterministic height gives independent centered normal terms. Their variances add to the stated $V(t)$, so their sum is centered normal when that variance is positive. If $V(t)=0$, all contributing terms have zero height or zero duration, and the displayed sum is zero almost surely. $\square$

Worked Example: Distribution Halfway Through an Interval

Let $H_s=1$ on $(0,1]$ and $H_s=3$ on $(1,2]$. Find the distribution of $I_{3/2}=\int_0^{3/2}H_s\,dB_s$.

The first interval is complete and the second contributes only from $1$ to $\tfrac32$. Therefore

I_{3/2}=B_1+3(B_{3/2}-B_1)=3B_{3/2}-2B_1.

The accumulated variance is $1^2(1-0)+3^2(\tfrac32-1)=1+9(\tfrac12)=\tfrac{11}{2}$. Thus $I_{3/2}\sim N(0,\tfrac{11}{2})$, its mean is $0$, and its variance is $\tfrac{11}{2}$.

Check: Using the endpoint expression, $\operatorname{Cov}(B_{3/2},B_1)=1$. Its variance is $9(\tfrac32)+4(1)-12(1)=\tfrac{27}{2}+4-12=\tfrac{11}{2}$, as required.

Why the Independent-Increment Form Matters

The step-integrand formula gives more than a way to evaluate a particular integral. It identifies exactly which Brownian increments contribute and how their coefficients affect uncertainty. A step height $h_j$ multiplies the standard deviation of its interval’s increment by $|h_j|$, so that interval contributes $h_j^2\Delta t_j$ to the total variance. The sign of the height changes the sum but not this variance contribution.

These computations apply to deterministic coefficients. A sum with random, adapted coefficients is a different situation: one cannot generally treat the coefficient-weighted increments as independent Gaussian variables merely by using the same variance formula. The deterministic Gaussian-law conclusion proved here relies on the coefficients being fixed numbers. The calculations also concern the Ito integral; they do not replace the conversion rule between Ito and Stratonovich integrals, which, for a continuous semimartingale integrand $X$, includes the correction $\tfrac12[X,B]$.

Common Pitfall: Treating Brownian Values as Independent

Brownian increments on disjoint intervals are independent, but Brownian values at different times generally are not. In the first example, $B_{1/2}$ and $B_2$ are not independent: their covariance is $\tfrac12$. A wrong calculation would omit the covariance term and write $\operatorname{Var}(3B_{1/2}-B_2)=9(\tfrac12)+2=\tfrac{13}{2}$. The correction includes the cross term:

\operatorname{Var}(3B_{1/2}-B_2) =9\operatorname{Var}(B_{1/2})+\operatorname{Var}(B_2) -6\operatorname{Cov}(B_{1/2},B_2) =\frac92+2-6\left(\frac12\right) =\frac72.

The independent objects in the original sum are $B_{1/2}-B_0$ and $B_2-B_{1/2}$, not the endpoint values $B_{1/2}$ and $B_2$. Whenever the sum is rewritten using Brownian values, covariance must be retained.

Check Your Understanding

Use the finite-sum representation and the variance formula to organize each calculation.

For $H_s=2$ on $(0,1]$ and $H_s=-1$ on $(1,3]$, find the integral’s distribution, mean, and variance.
State the distribution of the integral when every step height is zero, and explain why the positive-variance Gaussian case does not apply.
For $H_s=1$ on $(0,1]$ and $H_s=2$ on $(1,2]$, find the variance of $I_{5/4}$.
Explain why deterministic step heights allow the Gaussian characteristic-function calculation to factor into a product.
In the expression $B_1-2(B_2-B_1)$, identify the independent Brownian increments and explain why $B_1$ and $B_2$ themselves are not independent.