The Square-Integrable Norm for Simple Integrands

Measuring the Size of a Simple Integrand

In constructing the Ito integral, one first works with simple predictable processes: processes that remain constant on each interval of a finite time partition, with each interval's coefficient determined by information available at its left endpoint. The natural measure of their size is not the sum of the coefficient values. It is the expected time integral of the coefficient squared. This quantity determines which simple processes belong to the square-integrable class and will later provide the norm used to extend the integral.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to that filtration. Fix a finite horizon $T>0$. A process is progressively measurable if, for every $u\ge0$, its restriction to $[0,u]\times\Omega$ is measurable with respect to $\mathcal{B}([0,u])\otimes\mathcal{F}_u$. The class $\mathcal{L}^2$ on $[0,T]$ consists of progressively measurable processes $H$ such that $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The time integral here is an ordinary Lebesgue integral, not a stochastic integral.

Definition: Let $0=t_0<t_1<\cdots<t_n=T$ be a finite partition, and let each real-valued random variable $\xi_i$ be $\mathcal{F}_{t_i}$-measurable. The associated simple predictable process is $H_s=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(s)$. Its square-integrable size is $\|H\|_{2,T}=\left(\mathbb{E}\int_0^T H_s^2\,ds\right)^{1/2}$, allowed initially to take the value $+\infty$.

The half-open interval convention assigns the value $\xi_i$ immediately after $t_i$ and through $t_{i+1}$. Changing a process at the finitely many partition endpoints does not change its time integral, because a finite set of times has Lebesgue measure zero. The filtration condition $\xi_i\in\mathcal{F}_{t_i}$ expresses that the coefficient on the next interval does not use future information. It does not require $\xi_i$ to be independent of Brownian increments or of other coefficients.

The Coefficient Formula

Theorem: For the simple process above, $$ \mathbb{E}\int_0^T H_s^2\,ds =\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[\xi_i^2], $$ where either side may be infinite. Since the partition is strictly increasing, the norm is finite if and only if $\mathbb{E}[\xi_i^2]<\infty$ for every $i$.

Proof. The intervals $(t_i,t_{i+1}]$ are disjoint, so at each time $s$ other than the immaterial endpoint $0$, at most one term in the definition of $H_s$ is nonzero. Thus $H_s^2=\sum_{i=0}^{n-1}\xi_i^2\mathbf{1}_{(t_i,t_{i+1}]}(s)$. Each summand is nonnegative, so Tonelli's theorem permits interchanging expectation, the time integral, and the finite sum even when an expectation is infinite. Since $\xi_i$ is constant in $s$, integration over its interval gives $\int_0^T \xi_i^2\mathbf{1}_{(t_i,t_{i+1}]}(s)\,ds=(t_{i+1}-t_i)\xi_i^2$. Taking expectations and summing proves the formula. Every interval length is strictly positive, and every term is nonnegative. A finite sum of these terms is finite exactly when each $\mathbb{E}[\xi_i^2]$ is finite. Taking the square root gives the stated criterion for the norm. $\square$

The formula uses the second moment, not the variance. If $\xi_i$ has a nonzero mean, that mean contributes to $\mathbb{E}[\xi_i^2]$ as well. In particular, when $\mathbb{E}[\xi_i^2]<\infty$, the identity $\mathbb{E}[\xi_i^2]=\operatorname{Var}(\xi_i)+(\mathbb{E}[\xi_i])^2$ shows both sources of its size.

If one allows a representation with repeated partition points, a zero-length interval contributes zero regardless of the second moment of its coefficient. Such an interval carries no time and can be deleted without changing the process except at endpoints. The strict partition convention avoids this distraction: every coefficient is tested for square integrability.

Worked Example: A Coefficient Chosen by the Brownian Past

On $[0,1]$, take the partition $0<\tfrac12<1$, and set $\xi_0=2$ and $\xi_1=3\mathbf{1}_A-\mathbf{1}_{A^c}$, where $A=\{B_{1/2}\ge0\}$. Compute the norm.

The first coefficient is deterministic and $\mathbb{E}[\xi_0^2]=4$. Since $B_{1/2}$ has a centered normal distribution with no atom at zero, $\mathbb{P}(A)=\mathbb{P}(A^c)=\tfrac12$. The second coefficient is $\mathcal{F}_{1/2}$-measurable, as required, and $\mathbb{E}[\xi_1^2]=9\mathbb{P}(A)+1\mathbb{P}(A^c)=9\cdot\tfrac12+1\cdot\tfrac12=5$.

\|H\|_{2,1}^2 =\tfrac12\cdot4+\tfrac12\cdot5 =2+\tfrac52 =\tfrac92, \qquad \|H\|_{2,1}=\frac{3}{\sqrt{2}}.

Check: Directly, $H_s^2=4$ for $0<s\le\tfrac12$, while on the second half its value is $9$ on $A$ and $1$ on $A^c$. The expected time integral is $\tfrac12\cdot4+\tfrac12\big(\tfrac12\cdot9+\tfrac12\cdot1\big)=2+\tfrac52=\tfrac92$, agreeing with the coefficient formula.

Finiteness Is a Second-Moment Condition

A simple process can be finite at every time and still fail to belong to $\mathcal{L}^2$. Pointwise finiteness says that its coefficient values are finite almost surely; square integrability additionally controls the tails of their distributions. The next example uses a finite-valued coefficient whose square has infinite expectation.

Worked Example: A Finite-Valued Process with Infinite Norm

On $[0,2]$, let $H_s=e^{B_1^2}\mathbf{1}_{(1,2]}(s)$. This is a simple predictable process because $e^{B_1^2}$ is $\mathcal{F}_1$-measurable. Determine whether its norm is finite.

The only nonzero interval has length $2-1=1$, so $\|H\|_{2,2}^2=\mathbb{E}[e^{2B_1^2}]$. The random variable $B_1$ has density $(2\pi)^{-1/2}e^{-x^2/2}$, and therefore \[ \mathbb{E}[e^{2B_1^2}] =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{2x^2}e^{-x^2/2}\,dx =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{3x^2/2}\,dx =\infty. \] The last integral diverges because its integrand is at least $1$ for all real $x$, so its integral over the whole real line is infinite. Thus $H_s$ is finite almost surely at each time, but $\|H\|_{2,2}=\infty$ and $H\notin\mathcal{L}^2$.

Check: The coefficient is nonzero only on an interval of length $1$. The divergence therefore comes from its second moment, not from a long time horizon or an interval-length factor.

The same reasoning handles several intervals: no cancellation between coefficients can make the norm finite when one positive-length interval has an infinite second moment. The expression contains only nonnegative terms. Nor does one need independence among coefficients: each contribution is computed from its own second moment, and no cross-products arise in the time integral because the intervals do not overlap.

Worked Example: Unequal Intervals and a Brownian Coefficient

On $[0,1]$, let $H_s=1$ for $0<s\le\tfrac14$, and $H_s=B_{1/4}$ for $\tfrac14<s\le1$. Compute its norm.

The first interval has length $\tfrac14$ and coefficient second moment $1$. The second has length $1-\tfrac14=\tfrac34$, and its coefficient is measurable at the left endpoint. Since $B_{1/4}\sim N(0,\tfrac14)$, $\mathbb{E}[B_{1/4}^2]=\tfrac14$. Hence \[ \|H\|_{2,1}^2 =\tfrac14\cdot1+\tfrac34\cdot\tfrac14 =\tfrac14+\tfrac{3}{16} =\tfrac{7}{16}, \qquad \|H\|_{2,1}=\frac{\sqrt{7}}{4}. \]

Check: On the first interval, the expected integral of $H_s^2$ is $\tfrac14$. On the second, $H_s^2=B_{1/4}^2$ throughout, so that expected contribution is $\tfrac34\mathbb{E}[B_{1/4}^2]=\tfrac34\cdot\tfrac14=\tfrac{3}{16}$. The total is $\tfrac{7}{16}$, as above.

Why This Quantity Is a Norm

Strictly speaking, $\|H\|_{2,T}$ is a norm on processes identified when they agree almost everywhere with respect to the product of time measure and probability. Without that identification, two processes can differ at a time endpoint and still have distance zero. For simple processes on a strict partition, the coefficient formula makes this distinction explicit.

Proposition: On square-integrable simple processes identified when they agree almost everywhere in time and probability, $\|H\|_{2,T}$ is a norm. In particular, $\|H\|_{2,T}=0$ exactly when $H=0$ almost everywhere in that sense.

Proof. Nonnegativity follows because $H_s^2\ge0$, and homogeneity follows from $\|cH\|_{2,T}^2=c^2\|H\|_{2,T}^2$ for every real constant $c$. If the norm is zero, the coefficient formula gives $\sum_i(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]=0$. Each summand is nonnegative and each interval length is positive, so $\mathbb{E}[\xi_i^2]=0$ for every $i$. A nonnegative random variable with expectation zero is zero almost surely; hence each $\xi_i=0$ almost surely and $H=0$ almost everywhere. Conversely, if $H=0$ almost everywhere, its squared time integral is zero almost surely, so its norm is zero. For the triangle inequality, regard $H$ and $G$ as square-integrable functions on the product of time measure and probability. Cauchy–Schwarz gives $\mathbb{E}\int_0^T HG\,ds\le\|H\|_{2,T}\|G\|_{2,T}$. Expanding the square then yields $\|H+G\|_{2,T}^2\le\|H\|_{2,T}^2+2\|H\|_{2,T}\|G\|_{2,T}+\|G\|_{2,T}^2=(\|H\|_{2,T}+\|G\|_{2,T})^2$. Taking nonnegative square roots proves the triangle inequality. $\square$

A Common Mistake

Do not square the mean instead of taking the mean of the square. In the first worked example, the second coefficient satisfies $\mathbb{E}[\xi_1]=3\cdot\tfrac12-1\cdot\tfrac12=1$. Replacing $\mathbb{E}[\xi_1^2]$ by $(\mathbb{E}[\xi_1])^2=1$ would give the incorrect value $\tfrac12\cdot4+\tfrac12\cdot1=\tfrac52$ for the squared norm. The correct second contribution is $\tfrac12\mathbb{E}[\xi_1^2]=\tfrac12\cdot5=\tfrac52$, so the correct squared norm is $2+\tfrac52=\tfrac92$. The distinction matters whenever a coefficient is random; the norm measures its second moment, not just its average.

This norm is the precise way to screen simple predictable processes for square integrability. Once a process has finite norm, it belongs to the class used in constructing the Ito integral. The coefficient formula gives that conclusion directly, without yet defining or evaluating a stochastic integral.

Check Your Understanding

For each question, use a strict finite partition and state whether the coefficients satisfy the second-moment condition.

For a partition $0<\tfrac13<1$ with deterministic coefficients $2$ and $-1$, compute the squared norm and norm.
If a coefficient on an interval of length $h>0$ has mean zero and variance $v<\infty$, express that interval's contribution to the squared norm.
Explain why a finite-valued coefficient can nevertheless give an infinite norm.
What changes in the coefficient formula if a zero-length interval is included in a non-strict partition?
For $H_s=B_{1/2}$ on $(\tfrac12,1]$, compute its squared norm using the variance of $B_{1/2}$.