Deterministic Singularities and Local Square Integrability

Singular Integrands: Local Versus Endpoint Integrability

A deterministic function can become unbounded near an endpoint and still define an Ito integral up to that endpoint. The deciding quantity is not whether the function itself is bounded, but whether its square has finite integral. For a power singularity, this produces a sharp threshold: an exponent below one half gives finite energy, while the borderline exponent one half gives a logarithmic divergence.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. All stochastic integrals below are Ito integrals. On a fixed time interval $[0,T]$, the Ito integral $\int_0^t H_s\,dB_s$ is defined for progressively measurable integrands $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. A deterministic Borel function is progressively measurable; changing its value at one time does not change its integral. The discussion concerns Ito integrals, not Stratonovich integrals.

Definition: A progressively measurable integrand is locally square integrable on $[0,T)$ if $\mathbb{E}\int_0^u H_s^2\,ds<\infty$ for every $u<T$. Its Ito integral is then defined on every such shorter interval. It is square integrable up to $T$ if $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; in that case the Ito integral is defined on the closed interval, including its terminal value.

For a deterministic integrand $h$, the expectation in these conditions reduces to the ordinary energy $\int h(s)^2\,ds$. Thus a singularity may be harmless on every interval bounded away from its endpoint but still prevent the integral from being defined at the endpoint. The distinction is important: defining $\int_0^t h(s)\,dB_s$ for every $t<T$ does not, by itself, define $\int_0^T h(s)\,dB_s$.

The Power-Law Threshold

Proposition: Let $b>a$, and let $h(t)=(b-t)^{-\alpha}$ for $a\le t<b$, where $\alpha$ is real. The function is locally square integrable on every $[a,u]$ with $u<b$. It is square integrable up to $b$ if and only if $\alpha<\tfrac12$. For $t<b$, its energy is $$ \int_a^t h(s)^2\,ds= \begin{cases} \dfrac{(b-a)^{1-2\alpha}-(b-t)^{1-2\alpha}}{1-2\alpha},&\alpha\ne\tfrac12,\\[6pt] \log\!\left(\dfrac{b-a}{b-t}\right),&\alpha=\tfrac12. \end{cases} $$

Proof. On any $[a,u]$ with $u<b$, the quantity $b-s$ is at least $b-u>0$, so $h(s)^2=(b-s)^{-2\alpha}$ is bounded there and has finite integral. To test the endpoint, substitute $x=b-s$; the energy near $b$ is an integral of $x^{-2\alpha}$ from $0$ to a positive constant. If $2\alpha\ne1$, an antiderivative is $x^{1-2\alpha}/(1-2\alpha)$. Its limit at zero is finite exactly when $1-2\alpha>0$, or $\alpha<\tfrac12$. If $2\alpha=1$, the integral is $\int dx/x$, which diverges logarithmically. This proves both directions of the criterion. Evaluating the antiderivative between $b-t$ and $b-a$ gives the displayed formula, including the logarithmic case. $\square$

If the integrand is $c(b-t)^{-\alpha}$ for a constant $c$, the energy is multiplied by $c^2$, so the threshold is unchanged. The same test applies at a left endpoint: $h(t)=t^{-\alpha}$ on $(0,T]$ has finite energy from zero precisely when $\alpha<\tfrac12$. A value can be assigned arbitrarily at $t=0$; it does not affect the integral.

Why Finite Energy Gives a Gaussian Terminal Value

Theorem: Let $h$ be a deterministic Borel function on $[0,T]$ with $\int_0^T h(s)^2\,ds<\infty$. Then $M_t=\int_0^t h(s)\,dB_s$ is a square-integrable martingale on $[0,T]$, and $M_T$ has the normal distribution with mean $0$ and variance $\int_0^T h(s)^2\,ds$. In particular, $\mathbb{E}[M_T^2]=\int_0^T h(s)^2\,ds$.

Proof. Since $h$ is deterministic and square integrable, it belongs to $\mathcal{L}^2$, and the Ito isometry gives $\mathbb{E}[M_t^2]=\int_0^t h(s)^2\,ds<\infty$. For $s\le t$, the increment $M_t-M_s=\int_s^t h(r)\,dB_r$ is independent of $\mathcal{F}_s$ and has mean zero. This follows first for deterministic step functions from the independent, centered Brownian increments, and then for $h$ by square-integrable approximation. Therefore $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$, verifying the martingale property and its required integrability.

For the distribution, approximate $h$ in $L^2([0,T])$ by deterministic step functions $h_n$. The corresponding integrals $M_T^{(n)}$ are finite sums of independent centered normal Brownian increments, so they are centered normal with variance $\int_0^T h_n(s)^2\,ds$. By the Ito isometry, $M_T^{(n)}\to M_T$ in $L^2(\mathbb{P})$, hence in probability, while $\int_0^T h_n(s)^2\,ds\to\int_0^T h(s)^2\,ds$. The characteristic functions of $M_T^{(n)}$ are $\exp(-u^2\int h_n^2/2)$; convergence in probability implies convergence of these characteristic functions to that of $M_T$. The limit is $\exp(-u^2\int h^2/2)$, the characteristic function of the stated centered normal distribution. The variance also follows directly from the Ito isometry. $\square$

When the energy is finite on $[0,T)$ but diverges at $T$, the same construction gives an Ito integral on every $[0,u]$, $u<T$. Each such restricted integrand belongs to $\mathcal{L}^2$, and the integral is a square-integrable martingale on that shorter interval. The theorem does not supply a terminal value at $T$: its finite-energy hypothesis fails there.

Worked Classifications

Worked Example: A Singularity at the Initial Endpoint

Let $h(t)=t^{-1/3}$ for $0<t\le1$. Determine whether $\int_0^1 h(s)\,dB_s$ exists and compute its variance.

The squared integrand is $h(s)^2=s^{-2/3}$. For $0<t\le1$, its accumulated energy is $$ \int_0^t s^{-2/3}\,ds=\left[3s^{1/3}\right]_0^t=3t^{1/3}. $$ At the endpoint $t=1$, the energy is $3$, which is finite. Thus $h$ belongs to $\mathcal{L}^2$ on $[0,1]$, and the endpoint Ito integral exists. By the theorem, it is normal with mean $0$ and variance $3$.

Check: the Ito isometry gives $\mathbb{E}\big[(\int_0^1 s^{-1/3}\,dB_s)^2\big]=\int_0^1 s^{-2/3}\,ds=3$, agreeing with the stated variance. The integrand is unbounded near zero, but its square has finite integral.

Worked Example: The Borderline Endpoint

Let $h(t)=(2-t)^{-1/2}$ for $0\le t<2$. Determine where the Ito integral exists and compute its energy at $t=1$ and $t=\tfrac32$.

Here $\alpha=\tfrac12$, so the endpoint criterion is at equality and the integral up to $2$ diverges. For $t<2$, direct integration gives $$ \int_0^t h(s)^2\,ds=\int_0^t\frac{1}{2-s}\,ds =\left[-\log(2-s)\right]_0^t =\log\!\left(\frac{2}{2-t}\right). $$ At $t=1$, this is $\log(2/(2-1))=\log 2$. At $t=\tfrac32$, it is $\log(2/(2-\tfrac32))=\log 4$. Both are finite, so the Ito integral is defined up to each of these times. As $t$ increases to $2$, the denominator $2-t$ decreases to zero and the energy diverges.

Check: differentiating $\log(2/(2-t))$ gives $1/(2-t)=h(t)^2$, and its value at zero is $\log 1=0$, as required. This verifies both the antiderivative and the initial value.

Worked Example: An Unbounded Integrand with a Terminal Value

Let $h(t)=(1-t)^{-1/4}$ for $0\le t<1$. Determine whether the Ito integral exists up to $1$, and find its terminal variance.

Squaring gives $h(t)^2=(1-t)^{-1/2}$. For $t<1$, $$ \int_0^t(1-s)^{-1/2}\,ds =\left[-2\sqrt{1-s}\right]_0^t =2\bigl(1-\sqrt{1-t}\bigr). $$ Taking $t$ up to $1$, the energy tends to $2(1-0)=2$, so it is finite. Define $h(1)$ arbitrarily, for example as $0$; this does not change its energy or Ito integral. The theorem now gives a terminal Ito integral with distribution $N(0,2)$.

Check: the variance computed from the energy is $\int_0^1(1-s)^{-1/2}\,ds=2$, and the mean is zero. The integrand itself grows without bound as $t$ approaches $1$, but the exponent $1/4$ is below the threshold $1/2$.

A Common Error and Its Correction

A common mistake is to infer endpoint integrability from integrability on every shorter interval. For the borderline example, one might calculate that $\int_0^t(2-s)^{-1}\,ds=\log(2/(2-t))$ is finite for every $t<2$, then incorrectly conclude that the integral up to $2$ exists. The correction is to test the limiting energy: $\lim_{t\uparrow2}\log(2/(2-t))=+\infty$. The integrand is locally square integrable before the endpoint, but it is not square integrable up to it. In contrast, finiteness of the limiting energy is sufficient: the restricted Ito integrals converge in $L^2(\mathbb{P})$ to the terminal integral by the Ito isometry.

The practical test is therefore to square the integrand first, integrate its energy up to a variable time, and examine the endpoint limit. For $h(t)=(b-t)^{-\alpha}$, the critical exponent is $\alpha=\tfrac12$; for $h(t)=t^{-\alpha}$ at the initial endpoint, the same threshold applies. Once the energy is finite, it is also the variance of the deterministic Ito integral.

Check Your Understanding

For each power singularity, test the energy at the endpoint rather than only at times strictly before it.

For $h(t)=(3-t)^{-1/3}$ on $0\le t<3$, does the Ito integral exist up to $3$? Give its variance.
For $h(t)=t^{-2/5}$ on $0<t\le1$, compute the energy up to $1$ and state whether the endpoint integral exists.
For $h(t)=(4-t)^{-3/4}$ on $0\le t<4$, is the integral locally defined before $4$, and is it defined up to $4$?
Explain why finite energy on every interval $[0,u]$, $u<T$, does not by itself imply finite energy on $[0,T]$.
If a deterministic integrand has finite energy $A$ up to its endpoint, what are the mean and variance of its terminal Ito integral?