Mean-Square Continuity in Time

What Continuity Means for an Itô Integral

For a stochastic process, continuity can refer to different kinds of convergence. Mean-square continuity concerns the distance between random variables at nearby times, measured by their $L^2(\mathbb{P})$ norm. It does not say that every sample path is continuous. The distinction matters because the Itô integral is first constructed as a family of random variables, while questions about choosing a pathwise continuous version require additional arguments.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix a finite horizon $T>0$. Throughout, an integrand $H$ is progressively measurable and satisfies $\mathbb{E}\int_0^T H_u^2\,du<\infty$; that is, $H\in\mathcal{L}^2$ on $[0,T]$. All integrals below are Itô integrals with respect to $B$. The Itô isometry, established in the earlier material on the $L^2$ Itô integral, states that for such an integrand $G$, $\mathbb{E}[(\int_a^b G_u\,dB_u)^2]=\mathbb{E}\int_a^bG_u^2\,du$ whenever $0\le a\le b\le T$.

Definition: A process $M_t$ is continuous in mean square at $s$ if $\mathbb{E}[|M_t-M_s|^2]\to0$ as $t\to s$ within its time domain. Equivalently, the $L^2(\mathbb{P})$ distance $\|M_t-M_s\|_2=(\mathbb{E}[|M_t-M_s|^2])^{1/2}$ tends to zero.

The Increment Formula and Energy Bound

Theorem: Let $H\in\mathcal{L}^2$ on $[0,T]$ and set $M_t=\int_0^tH_u\,dB_u$ for $0\le t\le T$. For $0\le s\le t\le T$, $$ \mathbb{E}[|M_t-M_s|^2]=\mathbb{E}\int_s^t H_u^2\,du. $$ In particular, $M$ is continuous in mean square at every $s\in[0,T]$, with one-sided limits at the endpoints.

Proof. Time-interval additivity of the Itô integral gives $M_t-M_s=\int_s^tH_u\,dB_u$ as an equality in $L^2(\mathbb{P})$. The restricted integrand remains progressively measurable and square-integrable, since $\mathbb{E}\int_s^tH_u^2\,du\le\mathbb{E}\int_0^TH_u^2\,du<\infty$. Apply the Itô isometry on $[s,t]$ to obtain the asserted equality. To prove continuity, define $g(u)=\mathbb{E}[H_u^2]$, which is nonnegative and integrable on $[0,T]$ by Tonelli's theorem and the assumption on $H$. If $t\ge s$, then $$ \mathbb{E}[|M_t-M_s|^2]=\int_s^t g(u)\,du. $$ The integral of an $L^1$ function over an interval tends to zero as the interval length tends to zero. If $t\le s$, interchange $s$ and $t$ in the same argument. Thus the mean-square increment tends to zero from either available side, proving the claim at interior points and the corresponding one-sided claim at $0$ and $T$. $\square$

The equality is more informative than an estimate alone: it identifies the mean-square size of an increment with the expected energy of the integrand over precisely that time interval. No bound on $H_u$ at an individual time is needed. In particular, an integrand may have large values on some times, provided its expected squared energy over the whole horizon is finite.

Corollary: Under the theorem's assumptions, $t\mapsto M_t$ is uniformly continuous in mean square on $[0,T]$: for every $\varepsilon>0$, some $\delta>0$ satisfies $\mathbb{E}[|M_t-M_s|^2]<\varepsilon$ whenever $s,t\in[0,T]$ and $|t-s|<\delta$.

Proof. Let $g(u)=\mathbb{E}[H_u^2]\in L^1([0,T])$, as in the theorem. The absolute continuity of the Lebesgue integral says that for every $\varepsilon>0$, there is $\delta>0$ such that $\int_Ag(u)\,du<\varepsilon$ for every measurable $A\subseteq[0,T]$ with Lebesgue measure less than $\delta$. For $s\le t$ with $t-s<\delta$, the increment formula gives $\mathbb{E}[|M_t-M_s|^2]=\int_{[s,t]}g(u)\,du<\varepsilon$. For $t<s$, apply the same reasoning to $[t,s]$. The choice of $\delta$ does not depend on the locations of $s$ and $t$, which proves uniform mean-square continuity. $\square$

A useful stronger estimate is available when the second moments of the integrand are bounded. If $\mathbb{E}[H_u^2]\le C$ for almost every $u\in[0,T]$, where $C<\infty$, then for $s\le t$, $$ \mathbb{E}[|M_t-M_s|^2]=\int_s^t\mathbb{E}[H_u^2]\,du\le C(t-s). $$ This estimate gives a linear bound in the interval length. The general theorem does not require such a $C$; its bound is the actual integrated energy, which need not be proportional to $t-s$.

Worked Calculations

Worked Example: A Deterministic Linear Integrand

Take $H_u=u$ on $[0,2]$. This is progressively measurable, and $\mathbb{E}\int_0^2H_u^2\,du=\int_0^2u^2\,du=8/3<\infty$, so it is admissible. Calculate the mean-square increment from $s=1/2$ to $t=3/2$.

By the increment formula, the expected squared increment is the integral of $u^2$ over this interval:

\mathbb{E}[|M_{3/2}-M_{1/2}|^2] =\int_{1/2}^{3/2}u^2\,du =\frac{(3/2)^3-(1/2)^3}{3} =\frac{27/8-1/8}{3} =\frac{13}{12}.

Thus the $L^2$ distance is $\sqrt{13/12}$. The time interval has length $1$, and its average value of $u^2$ is $13/12$, consistent with the computed energy. As a check, if the endpoints coincide, the integral is over an interval of length zero and the increment has mean-square size zero.

Worked Example: The Brownian Motion Integrand

On $[0,2]$, let $H_u=B_u$. Brownian motion is progressively measurable and $\mathbb{E}\int_0^2B_u^2\,du=\int_0^2u\,du=2<\infty$, since $\mathbb{E}[B_u^2]=u$. Hence $H$ is admissible. For $M_t=\int_0^tB_u\,dB_u$, compute the mean-square increment from $s=1$ to $t=2$.

Apply the increment formula and then use the Brownian second moment:

\mathbb{E}[|M_2-M_1|^2] =\mathbb{E}\int_1^2B_u^2\,du =\int_1^2u\,du =\frac{2^2-1^2}{2} =\frac32.

This is an equality, not merely an upper bound. It also checks against the general estimate obtained from $\mathbb{E}[B_u^2]=u\le2$ on $[1,2]$: that estimate gives $\mathbb{E}[|M_2-M_1|^2]\le2(2-1)=2$, which is consistent with $3/2$. At the single endpoint $u=0$, the second moment is zero; the integral remains finite on the entire horizon.

Worked Example: An Exponentially Decaying Integrand

Take the deterministic integrand $H_u=e^{-u}$ on $[0,2]$. It is progressively measurable and square-integrable because $\int_0^2e^{-2u}\,du=(1-e^{-4})/2<\infty$. Evaluate the mean-square increment from $s=0$ to $t=\ln 2$, which lies in $[0,2]$.

The increment formula gives $$ \mathbb{E}[|M_{\ln2}-M_0|^2] =\int_0^{\ln2}e^{-2u}\,du =\frac{1-e^{-2\ln2}}{2} =\frac{1-1/4}{2} =\frac38. $$ The identity $e^{-2\ln2}=1/4$ follows from $e^{\ln2}=2$. The value is positive, as expected for an interval of positive length with a nonzero integrand. For an interval $[a,a]$, the same integral is zero, agreeing with $M_a-M_a=0$.

What the Bound Does—and Does Not—Say

The exact formula makes continuity local in time: increments become small when the expected energy accumulated between the two times becomes small. It also separates the role of the integrand from the stochastic integration. Once the Itô isometry applies, the bound is an ordinary integral of the nonnegative function $u\mapsto\mathbb{E}[H_u^2]$.

Mean-square continuity is not the same assertion as continuity of sample paths. The result proves that the random variables $M_t$ approach one another in $L^2(\mathbb{P})$; it does not by itself prove that, for almost every $\omega$, the function $t\mapsto M_t(\omega)$ is continuous. Establishing a continuous version of an $L^2$ Itô integral is a separate topic. Nor should a mean-square statement be silently replaced by an almost-sure statement: they describe different modes of convergence.

Common mistake: using a pointwise integrand value in place of the interval energy. For $H_u=u$, an incorrect shortcut might assert $\mathbb{E}[|M_t-M_s|^2]=(t-s)\mathbb{E}[H_t^2]$. At $s=0,t=1$, this shortcut gives $1\cdot1^2=1$. The increment formula instead gives $$ \mathbb{E}[|M_1-M_0|^2]=\int_0^1u^2\,du=\frac13. $$ The error is treating the integrand as if it were constant at its endpoint value. In general, the correct quantity is $\int_s^t\mathbb{E}[H_u^2]\,du$. The shortcut becomes valid only when $H_u$ has the same second moment throughout the interval, and even then it is the common second moment—not an arbitrary endpoint value—that is multiplied by the interval length.

The estimate is useful whenever one needs to compare integral values at nearby times or establish convergence in $L^2$. For example, if $t_n\to t$, the theorem directly gives $\mathbb{E}[|M_{t_n}-M_t|^2]\to0$. The uniform version on a finite horizon is stronger: one choice of time tolerance controls every pair of times in that horizon. Neither conclusion requires a pointwise bound on $H$, and both rely on the square-integrability condition that makes the Itô isometry applicable.

Check Your Understanding

Use the increment formula and state the integrability condition needed for each Itô integral.

For $H_u=1+u$ on $[0,1]$, compute $\mathbb{E}[|M_1-M_0|^2]$.
For $H_u=B_u$ on $[0,3]$, compute $\mathbb{E}[|M_3-M_2|^2]$.
If $\mathbb{E}[H_u^2]\le 5$ for almost every $u\in[0,T]$, what bound follows for the mean-square increment over $[s,t]$?
Explain why the absolute continuity of $u\mapsto\int_0^u\mathbb{E}[H_v^2]\,dv$ gives uniform mean-square continuity on a finite horizon.
Does mean-square continuity alone assert that almost every sample path of $M$ is continuous? State the distinction precisely.

What You'll Learn