Deterministic Step-Function Integrals

From a Step Integrand to Brownian Increments

A deterministic step function is one of the simplest integrands for an Ito integral: on each interval of a partition, its value is fixed in advance. The integral is then a finite weighted sum of Brownian increments. This representation makes its expectation and variance accessible directly from the mean, variance, and independence of those increments.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$, with increments independent of the past filtration. Fix a finite horizon $T>0$. An integrand for an Ito integral on $[0,T]$ is taken from $\mathcal{L}^2$: it is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Every stochastic integral below is an Ito integral, written $\int H_s\,dB_s$, not a Stratonovich integral, written $\int H_s\circ dB_s$.

Definition: Choose a partition $0=t_0<t_1<\cdots<t_n=T$ and deterministic real numbers $c_1,\ldots,c_n$. The associated deterministic step function $H$ is $H_s=c_i$ for $s\in(t_{i-1},t_i]$. Its value at $s=0$ may be set arbitrarily, since a single time has zero length in the integral. The function is progressively measurable and belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^T H_s^2\,ds=\sum_{i=1}^n c_i^2(t_i-t_{i-1})<\infty$.

For a step integrand, the Ito integral is defined by the corresponding finite sum. Write $\Delta t_i=t_i-t_{i-1}$ and $\Delta B_i=B_{t_i}-B_{t_{i-1}}$. Then

\int_0^T H_s\,dB_s = \sum_{i=1}^n c_i\Delta B_i = \sum_{i=1}^n c_i(B_{t_i}-B_{t_{i-1}}).

Each increment has mean zero and variance $\Delta t_i$, and increments over disjoint partition intervals are independent. The coefficients $c_i$ are deterministic, so they do not introduce any additional randomness into the weights.

The Finite-Sum Formula

Proposition: For the deterministic step function above, the Ito integral is the weighted-increment sum $\sum_{i=1}^n c_i(B_{t_i}-B_{t_{i-1}})$. If an interval is subdivided into smaller intervals while its coefficient remains unchanged, the resulting sum is unchanged.

Proof. The formula for the integral is the defining construction of the Ito integral for a simple predictable process: on $(t_{i-1},t_i]$, the value of the integrand is $c_i$, so that interval contributes $c_i(B_{t_i}-B_{t_{i-1}})$. Summing these contributions gives the stated expression. For the refinement claim, subdivide one interval $(a,b]$ at points $a=u_0<u_1<\cdots<u_m=b$ and keep the coefficient $c$ on every subinterval. Its refined contribution is $\sum_{j=1}^m c(B_{u_j}-B_{u_{j-1}})=c(B_b-B_a)$, because all intermediate Brownian values cancel. Thus refinement does not change the integral. Any two representations of the same step function agree after passing to their common refinement, apart from values at finitely many endpoints, which do not affect the integral. $\square$

The same representation gives the integral up to any time $t\in[0,T]$: include the full contributions from intervals already completed and, if $t$ lies inside an interval, include the increment only up to $t$. For example, when $t\in[t_{k-1},t_k]$,

\int_0^t H_s\,dB_s = \sum_{i=1}^{k-1}c_i(B_{t_i}-B_{t_{i-1}}) +c_k(B_t-B_{t_{k-1}}).

This process is square-integrable. In fact, the finite sum has a finite second moment because each increment has a finite second moment and there are only finitely many terms.

Worked Example: A Three-Interval Integral

Let $T=2$, with partition $0<\tfrac12<\tfrac32<2$, and coefficients $c_1=2$, $c_2=-1$, and $c_3=3$. Write the integral as a Brownian-increment sum, then compute its expectation and variance.

The interval lengths are $\Delta t_1=\tfrac12$, $\Delta t_2=1$, and $\Delta t_3=\tfrac12$. Therefore

\int_0^2 H_s\,dB_s = 2(B_{1/2}-B_0) -(B_{3/2}-B_{1/2}) +3(B_2-B_{3/2}).

Each increment has mean zero, so linearity of expectation gives $\mathbb{E}\int_0^2 H_s\,dB_s=0$. The increments are independent, so the variance is the sum of the variances of the three weighted terms:

\operatorname{Var}\left(\int_0^2 H_s\,dB_s\right) = 2^2\cdot\tfrac12+(-1)^2\cdot1+3^2\cdot\tfrac12 = 2+1+\tfrac92 = \tfrac{15}{2}.

Check: the mean is zero, and the second moment is the variance, $\tfrac{15}{2}$. Directly, the three weighted increments have variances $4\cdot\tfrac12=2$, $1\cdot1=1$, and $9\cdot\tfrac12=\tfrac92$; their sum is again $\tfrac{15}{2}$.

Mean and Variance from Independence

Theorem: Let $H_s=c_i$ on $(t_{i-1},t_i]$ for a deterministic partition of $[0,T]$. Then, almost surely as a finite-sum identity, $$ \int_0^T H_s\,dB_s=\sum_{i=1}^n c_i\Delta B_i. $$ Moreover, $$ \mathbb{E}\left[\int_0^T H_s\,dB_s\right]=0, \qquad \operatorname{Var}\left(\int_0^T H_s\,dB_s\right) = \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right] = \sum_{i=1}^n c_i^2\Delta t_i. $$

Proof. The integral identity follows from the definition for step processes. Set $I=\sum_{i=1}^n c_i\Delta B_i$. Since $\mathbb{E}[\Delta B_i]=0$ for every $i$, finite linearity gives $\mathbb{E}[I]=0$. Expanding the square gives $\mathbb{E}[I^2]=\sum_i c_i^2\mathbb{E}[(\Delta B_i)^2]+2\sum_{i<j}c_ic_j\mathbb{E}[\Delta B_i\Delta B_j]$. For $i<j$, the increments are independent and have mean zero, so $\mathbb{E}[\Delta B_i\Delta B_j]=\mathbb{E}[\Delta B_i]\mathbb{E}[\Delta B_j]=0$. Also, $\mathbb{E}[(\Delta B_i)^2]=\operatorname{Var}(\Delta B_i)=\Delta t_i$. Hence $\mathbb{E}[I^2]=\sum_i c_i^2\Delta t_i$. Because $\mathbb{E}[I]=0$, this second moment equals $\operatorname{Var}(I)$. If all coefficients are zero, or if $T=0$ and the integral is interpreted over the empty interval, both sides are zero as well. $\square$

This is the finite-step version of the Ito isometry: the second moment equals the time integral of the squared integrand. Here that integral is simply $\sum_i c_i^2\Delta t_i$, and the proof needs only Brownian increment independence and the increment variances. It does not require identifying the full probability distribution of $I$.

Worked Example: Refining a Step Interval

Let $T=3$, with $H_s=-2$ on $(0,1]$ and $H_s=\tfrac12$ on $(1,3]$. First use the original partition; then refine the second interval at time $2$.

On the original partition, the integral is $-2(B_1-B_0)+\tfrac12(B_3-B_1)$. Its mean is zero. The interval lengths are $1$ and $2$, so the variance is $(-2)^2\cdot1+(\tfrac12)^2\cdot2=4+\tfrac12=\tfrac92$.

On the refined partition $0<1<2<3$, the same step function has coefficients $-2,\tfrac12,\tfrac12$. The integral becomes $-2(B_1-B_0)+\tfrac12(B_2-B_1)+\tfrac12(B_3-B_2)$. The last two terms add to $\tfrac12(B_3-B_1)$, so this is exactly the original expression. The refined variance calculation is $(-2)^2\cdot1+(\tfrac12)^2\cdot1+(\tfrac12)^2\cdot1=4+\tfrac14+\tfrac14=\tfrac92$.

Check: both representations give the same increment sum, and both variance calculations give $\tfrac92$. The split intervals contribute $\tfrac14+\tfrac14=\tfrac12$, equal to the original second-interval contribution $(\tfrac12)^2\cdot2=\tfrac12$.

The Integral Process Is a Martingale

For deterministic step integrands, the same independence argument also verifies a conditional-mean property. Define $M_t=\int_0^t H_s\,dB_s$ for $0\le t\le T$, using the partial-interval formula above.

Proposition: The process $M_t$ is a square-integrable martingale with respect to $(\mathcal{F}_t)$: it is adapted, $\mathbb{E}[|M_t|]<\infty$, and $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $0\le s\le t\le T$.

Proof. The partial-sum representation shows that $M_t$ depends only on Brownian values up to time $t$, so it is $\mathcal{F}_t$-measurable. The finite-sum second-moment calculation, applied to the portions of the partition up to $t$, gives $\mathbb{E}[M_t^2]=\int_0^t H_u^2\,du<\infty$. In particular, $M_t$ is integrable. For $s\le t$, split $(s,t]$ at the deterministic step boundaries. On each resulting interval $(u,v]$, the coefficient is deterministic and the contribution to $M_t-M_s$ is $c(B_v-B_u)$. Brownian increments after $u$ have conditional mean zero given $\mathcal{F}_u$; since $\mathcal{F}_s\subseteq\mathcal{F}_u$, the tower property gives $\mathbb{E}[B_v-B_u\mid\mathcal{F}_s]=0$. Adding the finitely many contributions yields $\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0$, or $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$. Thus the martingale property is verified. $\square$

Worked Example: A Step Function with a Zero Coefficient

On $[0,2]$, take coefficients $0$, $-3$, and $2$ on the intervals $(0,\tfrac12]$, $(\tfrac12,1]$, and $(1,2]$, respectively. Compute the integral and its mean and variance.

The first interval contributes zero. The integral is $-3(B_1-B_{1/2})+2(B_2-B_1)$. Its mean is zero. The interval lengths for the nonzero coefficients are $\tfrac12$ and $1$, so

\operatorname{Var}\left(\int_0^2 H_s\,dB_s\right) = (-3)^2\cdot\tfrac12+2^2\cdot1 = \tfrac92+4 = \tfrac{17}{2}.

Check: expanding the square of the two nonzero terms gives two squared terms and a cross term. The expected squared terms are $\tfrac92$ and $4$; the cross term has expectation $-12\,\mathbb{E}[(B_1-B_{1/2})(B_2-B_1)]=0$ by independence and zero means. The second moment is therefore $\tfrac{17}{2}$, as required.

Common Mistake: Forgetting the Interval Length

A Brownian increment over an interval of length $\Delta t_i$ has variance $\Delta t_i$, not $1$. The coefficient is squared in the variance formula, but the interval length must also be included.

For the first worked example, an incorrect calculation might assign variance $2^2+(-1)^2+3^2=14$, as though each increment had variance $1$. That is wrong because the intervals have lengths $\tfrac12$, $1$, and $\tfrac12$. The correct calculation is $2^2\cdot\tfrac12+(-1)^2\cdot1+3^2\cdot\tfrac12=\tfrac{15}{2}$. A useful check is that the variance changes if an interval length changes, even when its coefficient stays fixed.

These finite-sum calculations matter because they show concretely how the Ito integral is built from Brownian increments and why its second moment is controlled by the squared integrand. They also clarify that a deterministic step integral is an Ito construction, rather than an ordinary Riemann integral or an application of the ordinary chain rule. In stochastic differential notation, the Brownian increment variance is reflected by $dB_t\,dB_t=dt$; no Stratonovich conversion is being made here.

Check Your Understanding

Use the finite-sum representation and the increment mean, variance, and independence properties in each question.

For $H_s=1$ on $(0,1]$ and $H_s=-2$ on $(1,3]$, write $\int_0^3 H_s\,dB_s$ as a weighted sum and find its expectation and variance.
Explain why splitting an interval of length $2$ into two intervals of length $1$, without changing its coefficient, leaves both the integral and its variance unchanged.
For a step function with coefficients $c_i$ and interval lengths $\Delta t_i$, show where independence is used when computing the second moment.
Suppose every coefficient is multiplied by a constant $a$. What happens to the mean and variance of the integral?
For an interval on which the coefficient is zero, what contribution does it make to the integral and to the variance?