Finite-Energy Integrands and Failure Examples

Why Finite Expected Energy Matters

An expression such as $\int_0^T H_t\,dW_t$ is not automatically an Itô integral just because $H_t$ is adapted and its values look finite. The integrand must also satisfy conditions that make the stochastic integral well-defined. For the standard square-integrable Itô integral on a finite time interval, the central condition is finite expected energy: the expected value of $\int_0^T H_t^2\,dt$ must be finite.

This condition is easy to check for deterministic kernels, but random integrands require attention to both their paths and their distributions. A process can have finite energy on almost every path and still have infinite expected energy. That distinction determines whether the standard $L^2$ construction applies, even when a broader local stochastic integral can be defined.

Definition: Let $W_t$, $0\leq t\leq T$, be Brownian motion relative to a filtration $(\mathcal F_t)$, and assume the filtration satisfies the usual conditions. A process $H_t$ is square-integrable in the Itô sense on $[0,T]$ if it is progressively measurable and $$ E\!\left[\int_0^T H_t^2\,dt\right] < \infty. $$ Progressive measurability means that, for each $t$, the restriction of $(s,\omega)\mapsto H_s(\omega)$ to $[0,t]\times\Omega$ is measurable with respect to $\mathcal B([0,t])\otimes\mathcal F_t$. It ensures the integrand is measurable and uses no future information.

Adaptation alone says that $H_t$ is $\mathcal F_t$-measurable at each fixed time. It does not, by itself, guarantee the joint measurability needed for integration. In common models, an adapted process with continuous paths is progressively measurable, but that useful fact does not make its expected energy finite automatically.

The Itô Isometry and the Energy Test

The Itô isometry explains why expected energy is the relevant test. For an elementary adapted process, define $H_t=\sum_{i=0}^{m-1}\xi_i\mathbf 1_{(t_i,t_{i+1}]}(t)$, where $0=t_0<\cdots<t_m=T$ and each $\xi_i$ is square-integrable and $\mathcal F_{t_i}$-measurable. Its integral is the sum of the left-endpoint terms $\sum_i \xi_i(W_{t_{i+1}}-W_{t_i})$.

Theorem (Itô isometry): For a progressively measurable $H$ with $E\int_0^T H_t^2\,dt<\infty$, the standard Itô integral $I_T=\int_0^T H_t\,dW_t$ is a square-integrable random variable, and $$ E[I_T]=0, \qquad E[I_T^2]=E\!\left[\int_0^T H_t^2\,dt\right]. $$

For an elementary process, the isometry follows directly from Brownian increment properties. Write $\Delta W_i=W_{t_{i+1}}-W_{t_i}$. Each $\Delta W_i$ is independent of $\mathcal F_{t_i}$, has mean zero, and has variance $t_{i+1}-t_i$. Thus

E[(\xi_i\Delta W_i)^2] =E[\xi_i^2](t_{i+1}-t_i).

For $i<j$, the variables $\xi_i\Delta W_i$ and $\xi_j$ are $\mathcal F_{t_j}$-measurable, whereas $\Delta W_j$ is independent of $\mathcal F_{t_j}$ and has mean zero. Consequently,

E[\xi_i\Delta W_i\,\xi_j\Delta W_j] =E[\xi_i\Delta W_i\,\xi_j\,E[\Delta W_j\mid\mathcal F_{t_j}]] =0.

Adding the diagonal terms gives $E[(\sum_i\xi_i\Delta W_i)^2]=\sum_i E[\xi_i^2](t_{i+1}-t_i)=E\int_0^T H_t^2\,dt$. To extend the result, approximate a progressively measurable $H$ in $L^2(P\times dt)$ by elementary adapted processes. The isometry applied to differences shows that their integrals are Cauchy in $L^2(P)$. Completeness of $L^2(P)$ gives a limit, which defines $\int_0^T H_t\,dW_t$. Passing to the limit in the isometry gives the stated identity; the mean is zero because the approximating integrals have mean zero and converge in $L^2$, hence in $L^1$. $\square$

This also gives a practical convergence test. If $H$ satisfies the finite expected-energy condition, and $H^{(n)}$ is a sequence of progressively measurable approximations, then

E\!\left[\left(\int_0^T (H_t^{(n)}-H_t)\,dW_t\right)^2\right] =E\!\left[\int_0^T (H_t^{(n)}-H_t)^2\,dt\right].

So convergence of the integrands in $L^2(P\times dt)$ is exactly what ensures convergence of their integrals in $L^2(P)$. Checking energy is not a formality: it establishes that the integral is obtainable as a stable limit of the defining sums.

Worked Examples: Checking Candidate Integrands

Worked Example: A Brownian-Driven Integrand with Finite Energy

Let $H_t=W_t$ on $[0,T]$. Brownian motion is adapted and has continuous paths, so $H$ is progressively measurable. Since $W_t\sim N(0,t)$, its second moment is $E[W_t^2]=t$. Therefore,

E\!\left[\int_0^T W_t^2\,dt\right] =\int_0^T E[W_t^2]\,dt =\int_0^T t\,dt =\frac{T^2}{2}<\infty.

The interchange is valid for the nonnegative function $W_t^2$ by Tonelli's theorem. The standard Itô integral exists, has mean zero, and the isometry gives $E[(\int_0^T W_t\,dW_t)^2]=T^2/2$. In particular, the randomness of an integrand is not itself an obstacle; its expected energy is finite here.

Worked Example: A Deterministic Singularity with No Terminal L² Integral

Set $H_t=t^{-1/2}$ for $0<t\leq T$, assigning any value at $t=0$. It is deterministic and measurable, but its energy is

\int_0^T H_t^2\,dt =\int_0^T \frac{1}{t}\,dt =\infty.

Thus $H$ does not satisfy the finite expected-energy condition. More strongly, truncating away from zero does not produce a terminal integral limit in probability. For $0<\varepsilon<T$, let $I_\varepsilon=\int_\varepsilon^T t^{-1/2}\,dW_t$. For $0<\varepsilon<\delta<T$, the difference is a centered normal random variable with variance

E[(I_\varepsilon-I_\delta)^2] =\int_\varepsilon^\delta \frac{1}{t}\,dt =\log(\delta/\varepsilon).

Choose $\delta_n=e^{-n}$ and $\varepsilon_n=e^{-2n}$, for sufficiently large $n$ that $\delta_n<T$. Then the difference has distribution $N(0,n)$, so for any fixed $M>0$, $P(|I_{\varepsilon_n}-I_{\delta_n}|\leq M)\to 0$. The truncated integrals are not Cauchy in probability. There is therefore no terminal limit obtained by this truncation procedure.

Worked Example: Finite Pathwise Energy but Infinite Expected Energy

Fix $0<s<T$ and set $H_t=\mathbf 1_{(s,T]}(t)/|W_s|$, defining the value arbitrarily on the probability-zero event $W_s=0$. This process is progressively measurable: after time $s$, $W_s$ is already observed. Since $W_s\sim N(0,s)$, it is nonzero almost surely, and hence

\int_0^T H_t^2\,dt =\frac{T-s}{W_s^2}<\infty \quad\text{almost surely}.

But $E[1/W_s^2]=\infty$. Indeed, the density of $W_s$ is positive and bounded below on some interval around zero, and the integral of $x^{-2}$ over that interval diverges. Consequently,

E\!\left[\int_0^T H_t^2\,dt\right] =(T-s)E[1/W_s^2] =\infty.

The standard square-integrable Itô integral on $[0,T]$ is not available. Nevertheless, this example has a local Itô integral: after time $s$, the integrand is a finite, already observed random constant, and the integral can be localized by stopping when that constant exceeds a fixed bound. Thus infinite expected energy rules out the standard $L^2$ construction, not every broader notion of stochastic integration.

Truncation, Diagnostics, and Common Pitfalls

Result: If $H$ is progressively measurable and $E\int_0^T H_t^2\,dt<\infty$, then for $0\leq a<T$, the truncated integrals $I_a=\int_0^a H_t\,dW_t$ converge to $I_T$ in $L^2(P)$ as $a\uparrow T$. In fact, $$ E[(I_T-I_a)^2]=E\!\left[\int_a^T H_t^2\,dt\right]\longrightarrow 0. $$

The equality is the isometry applied to the integrand $H_t\mathbf 1_{(a,T]}(t)$. The right-hand side tends to zero because the nonnegative function $H_t^2$ is integrable on the product space with measure $P\times dt$; equivalently, the integral over the shrinking tail $(a,T]\times\Omega$ tends to zero by absolute continuity of the integral. This supplies a useful diagnostic: a claimed terminal $L^2$ integral must have vanishing expected energy on the omitted tail. $\square$

When checking a candidate, first confirm progressive measurability, then compute or bound $E\int H_t^2dt$. For nonnegative integrands, Tonelli's theorem permits exchanging expectation and time integration, even before finiteness is known. For signed expressions or more complicated bounds, justify any interchange separately. A divergent integral of expected squared values is a failure of the standard $L^2$ condition; a finite integral of expected squared values is a sufficient test under the stated measurability assumption.

Mistake: “The integrand is finite almost surely, so the Itô integral exists in $L^2$.” The random variable $1/|W_s|$ in the third example is finite almost surely but has an infinite second moment. Almost-sure finiteness is not finite expected energy.
Mistake: “Adapted means integrable.” Adaptation is about available information, not size or joint measurability. Check progressive measurability and energy separately.
Mistake: “Infinite expected energy means no stochastic integral of any kind exists.” The third example admits a local integral. State precisely whether the issue is the standard square-integrable construction or a broader local construction.
Mistake: “A finite answer from a truncated calculation proves convergence.” A single truncated integral can be finite while the truncations fail to converge. Compare differences, using the isometry when its hypotheses hold.

In pharmacokinetic modeling, a candidate diffusion coefficient or state-dependent integrand should not be accepted merely because it is meaningful at observed concentrations. The relevant check concerns the process over the full time interval and its distribution under the model. If expected energy diverges, the usual square-integrable construction and its moment calculations cannot be invoked without additional localization or a different integrability argument. This is the key distinction from calculations with deterministic kernels, for which the energy test reduces directly to an ordinary integral.

Check Your Understanding

For each question, decide whether the standard square-integrable Itô integral is justified, and compare your reasoning with the solutions.

For $H_t=2$ on $[0,T]$, compute the expected energy and the variance of the integral.
For $H_t=t^{-1/4}$ on $(0,T]$, is the finite expected-energy condition satisfied?
Can an adapted process have finite pathwise energy almost surely but fail the finite expected-energy condition? Give the reason.
What additional check is needed if a candidate integrand is adapted but is not known to have continuous paths?

Solutions. (1) The energy is $\int_0^T4\,dt=4T$, so the integral is square-integrable and has variance $4T$ by the isometry. (2) Its energy is $\int_0^T t^{-1/2}\,dt=2\sqrt T<\infty$, so the deterministic integrand satisfies the condition. (3) Yes. For example, $H_t=\mathbf 1_{(s,T]}(t)/|W_s|$ has finite energy almost surely but infinite expected energy because $E[1/W_s^2]=\infty$. (4) Verify progressive measurability; adaptation alone does not ensure the joint measurability required for the standard integral.

Finite-Energy Integrands and Failure Examples

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