Why Quantitative Stability Matters
The Itô integral is constructed by approximating admissible integrands in mean square. A stability estimate makes that construction quantitative: it tells us how much the integral can change when its integrand is replaced by an approximation. There are two useful questions. How large can the terminal error be in mean square? And how large can the error become at any time before the terminal time?
Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to this filtration. In particular, \(B\) is adapted, and for every \(s<t\), the increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). This independence condition is essential; adaptedness alone does not make \(B\) Brownian motion relative to the filtration.
Throughout, \(H\) and \(K\) are progressively measurable integrands on \([0,T]\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) and \(\mathbb{E}\int_0^T K_s^2\,ds<\infty\). These are Itô integrals, not Stratonovich integrals. Write \(I_t(H)=\int_0^t H_s\,dB_s\), and similarly for \(K\). We use their continuous versions. The Itô isometry and linearity of the integral, as well as Doob’s \(L^2\) maximal inequality, are standard results established in the earlier material on the Itô integral.
The Two Stability Bounds
Proof. Set \(D_s=H_s-K_s\). Linearity of the Itô integral gives \(I_t(H)-I_t(K)=\int_0^tD_s\,dB_s\). The process \(D\) is progressively measurable, and $$ \mathbb{E}\int_0^T D_s^2\,ds \le 2\mathbb{E}\int_0^T H_s^2\,ds+2\mathbb{E}\int_0^T K_s^2\,ds <\infty, $$ so \(D\) is an admissible square-integrable integrand. Applying the Itô isometry to this integral proves the first equality.
For the uniform estimate, \(M_t=\int_0^tD_s\,dB_s\) is a square-integrable martingale: it is integrable by the isometry, and for \(0\le s\le t\le T\), \(\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\), the conditional-mean property of the Itô integral. Doob’s \(L^2\) maximal inequality therefore applies to this martingale and gives \(\mathbb{E}[\sup_{0\le t\le T}|M_t|^2]\le4\mathbb{E}|M_T|^2\). Substitution of the isometry at \(T\) proves the second bound. The continuous versions ensure that the supremum is a supremum of the stated integral processes. \(\square\)
The terminal estimate is an equality. The uniform estimate is an upper bound: it controls the largest discrepancy over the whole interval, not just the discrepancy at \(T\). In particular, if a sequence of admissible integrands \(K^{(n)}\) satisfies \(\mathbb{E}\int_0^T|H_s-K_s^{(n)}|^2\,ds\to0\), then the corresponding integral processes converge to \(I(H)\) in mean square for the uniform norm. Indeed, the expected squared supremum of their difference is bounded by four times a quantity tending to zero.
Proof. Apply Markov’s inequality to the nonnegative random variable \(\sup_{0\le t\le T}|I_t(H)-I_t(K)|^2\) with threshold \(\varepsilon^2\), then use the uniform stability bound. The event is strict as written: Markov’s inequality also gives the bound for the event with \(\ge\varepsilon\). \(\square\)
Worked Applications
Worked Example: Replacing Brownian Motion by Zero
On \([0,1]\), take \(H_s=B_s\) and \(K_s=0\). Brownian motion is progressively measurable, and \(\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=\tfrac12\), so both integrands meet the required square-integrability condition. The terminal error is $$ I_1(H)-I_1(K)=\int_0^1B_s\,dB_s. $$
To compute it, apply Itô’s formula to \(f(x)=x^2\): \(f'(x)=2x\), \(f''(x)=2\), and the correction term is \(\tfrac12f''(B_s)\,ds=ds\). Thus \(d(B_s^2)=2B_s\,dB_s+ds\), and integration from \(0\) to \(1\) gives \(I_1(H)=\tfrac12(B_1^2-1)\). Since \(B_1\sim N(0,1)\), \(\mathbb{E}B_1^2=1\) and \(\mathbb{E}B_1^4=3\). Therefore $$ \mathbb{E}|I_1(H)-I_1(K)|^2 =\tfrac14\mathbb{E}(B_1^2-1)^2 =\tfrac14(3-2+1)=\tfrac12. $$
Check: the isometry gives \(\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=\tfrac12\), agreeing with the direct computation. The uniform estimate gives \(\mathbb{E}[\sup_{0\le t\le1}|I_t(H)|^2]\le4\cdot\tfrac12=2\). The latter is a bound on the pathwise-in-time error, not a claim that its expectation equals \(2\).
Worked Example: A Left-Endpoint Approximation
On \([0,1]\), divide time into four intervals of length \(h=\tfrac14\). Let \(t_j=j/4\), and on \(t_j\le s<t_{j+1}\), \(j=0,1,2,3\), set \(H_s=B_s\) and \(K_s=B_{t_j}\). Both processes are progressively measurable and square-integrable. On each interval \(B_s-B_{t_j}\) has second moment \(s-t_j\), so the integrated squared error is $$ \mathbb{E}\int_0^1|H_s-K_s|^2\,ds =\sum_{j=0}^3\int_{t_j}^{t_{j+1}}(s-t_j)\,ds =4\int_0^{1/4}u\,du =4\cdot\tfrac1{32} =\tfrac18. $$
Consequently the terminal mean-square error is \(\tfrac18\), while the expected squared supremum of the error is at most \(4\cdot\tfrac18=\tfrac12\). For a direct terminal check, on each interval Itô’s formula for \(f(x)=x^2\), with \(f'(x)=2x\), \(f''(x)=2\), and correction \(\tfrac12f''\,ds=ds\), yields $$ \int_{t_j}^{t_{j+1}}(B_s-B_{t_j})\,dB_s =\tfrac12\big((B_{t_{j+1}}-B_{t_j})^2-h\big). $$
The four Brownian increments are independent, each with distribution \(N(0,h)\). For \(Z\sim N(0,h)\), \(\operatorname{Var}(Z^2)=3h^2-h^2=2h^2\). Hence the variance of the sum of the four displayed terms is \(4\cdot\tfrac14\cdot2h^2=2h^2=2/16=\tfrac18\); its mean is zero. This verifies the terminal mean-square error independently. Check: for the strict threshold \(\varepsilon=\tfrac12\), the corollary bounds the probability that the uniform error exceeds \(\tfrac12\) by \((4/\tfrac14)\cdot\tfrac18=2\). Since probabilities are at most \(1\), this particular bound is valid but uninformative.
Worked Example: A Short-Interval Approximation Error
On \([0,1]\), take \(H_s=1\) and \(K_s=\mathbf{1}_{[0,3/4]}(s)\). Both are progressively measurable and have finite expected integrated square. Their difference is zero up to time \(3/4\) and equals \(1\) afterwards, apart from the single endpoint, which does not affect the integral. Thus $$ \mathbb{E}\int_0^1|H_s-K_s|^2\,ds =\int_{3/4}^1 1\,ds =\tfrac14. $$
The integral error is zero for \(t\le3/4\) and equals \(B_t-B_{3/4}\) for \(3/4\le t\le1\). At the terminal time its mean is zero and its mean square is \(\mathbb{E}(B_1-B_{3/4})^2=1-\tfrac34=\tfrac14\), as also follows from the isometry. The uniform estimate gives \(\mathbb{E}[\sup_{0\le t\le1}|I_t(H)-I_t(K)|^2]\le4\cdot\tfrac14=1\). Therefore, for the strict threshold \(2\), the probability that the supremum exceeds \(2\) is at most \(1/2^2=1/4\). Check: the terminal second moment \(\tfrac14\) is exactly the integrated squared integrand error, while the probability bound controls the larger event involving the maximum over the interval.
How to Use the Estimates and a Common Pitfall
Suppose an approximation procedure supplies \(K\) for a target integrand \(H\). First calculate or bound \(\delta=\mathbb{E}\int_0^T|H_s-K_s|^2\,ds\). Then \(\delta\) is exactly the terminal mean-square error, while \(4\delta\) bounds the expected squared uniform error. For any chosen tolerance \(\varepsilon>0\), the chance of a uniform error larger than \(\varepsilon\) is at most \(4\delta/\varepsilon^2\). If this last upper bound exceeds \(1\), it remains correct but gives no useful probability guarantee.
Common mistake: using the terminal estimate to control the whole path. From \(\mathbb{E}|I_T(H)-I_T(K)|^2=\delta\), one cannot conclude that the expected squared maximum error is at most \(\delta\). The terminal value observes only one time; the maximum includes every time in \([0,T]\). The valid conclusion from Doob’s inequality is the bound \(4\delta\). For example, in the left-endpoint approximation above, the terminal mean-square error is \(\tfrac18\), whereas the justified uniform bound is \(\tfrac12\), not \(\tfrac18\). The factor \(4\) is the cost of this general maximal control.
The comparison concerns Itô integrals driven by the same Brownian motion and measured under the same probability measure \(\mathbb{P}\). The assumptions ensure both integrals exist as square-integrable martingales; no claim about a local martingale outside the square-integrable setting is needed. The central practical message is that closeness of integrands in the \(L^2\) norm on time and probability controls both terminal accuracy and uniform-in-time accuracy, with explicit constants.
Check Your Understanding
Use the stated estimates, and distinguish the terminal error from the uniform error in each answer.
- If \(\mathbb{E}\int_0^T|H_s-K_s|^2\,ds=0.01\), what bounds follow for the terminal mean-square error and for the expected squared supremum error?
- Why must the Brownian increments be independent of \(\mathcal{F}_s\), not merely adapted, for the martingale argument in the proof?
- For the four-interval approximation, calculate the uniform-error probability bound at the strict threshold \(\varepsilon=1\).
- If \(K^{(n)}\) approximates \(H\) with integrated squared error tending to zero, what mode of convergence of the integral processes follows from the uniform estimate?
- In the short-interval example, explain why the single endpoint \(s=3/4\) does not change the integrated squared error.