Four Checks Before Trusting an Integral Calculation
A stochastic-integral computation can look plausible while failing at a basic step: the integrand may use future information, a product may be treated as independent without justification, or a second moment may be mistaken for a variance. These errors are especially easy to make when an Ito integral is first represented by sums. The checks below use the definition of the Ito integral and examples where the relevant limits and moments can be computed explicitly.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual completeness and right-continuity conditions. Assume \(B_t\) is a Brownian motion with respect to this filtration: it is adapted, has continuous paths and \(B_0=0\), and, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). The independence from \(\mathcal{F}_s\), not adaptedness alone, is what permits the usual Ito-integral construction.
For an adapted step process \(H_s=\sum_{i=0}^{n-1}H_i\mathbf{1}_{(t_i,t_{i+1}]}(s)\), where \(H_i\) is square-integrable and \(\mathcal{F}_{t_i}\)-measurable, the corresponding integral is \(\sum_i H_i(B_{t_{i+1}}-B_{t_i})\). Brownian increments are independent of the information available at the left endpoint. That is why these sums, and their \(L^2\) limits, have the properties used below.
The Step-Process Test: What Independence Does and Does Not Give
Proof. Put \(\Delta B_i=B_{t_{i+1}}-B_{t_i}\) and \(\Delta t_i=t_{i+1}-t_i\). Since \(H_i\) is \(\mathcal{F}_{t_i}\)-measurable and \(\mathbb{E}[\Delta B_i\mid\mathcal{F}_{t_i}]=0\), we have \(\mathbb{E}[H_i\Delta B_i]=0\). For \(i<j\), the product \(H_i\Delta B_iH_j\) is \(\mathcal{F}_{t_j}\)-measurable, while \(\Delta B_j\) is independent of \(\mathcal{F}_{t_j}\) and has mean zero. Therefore \(\mathbb{E}[H_i\Delta B_iH_j\Delta B_j]=0\). Also, independence of \(\Delta B_i\) from \(\mathcal{F}_{t_i}\) gives \(\mathbb{E}[H_i^2(\Delta B_i)^2]=\mathbb{E}[H_i^2]\Delta t_i\). Expanding \(I^2\) and summing these diagonal and cross terms proves the second-moment identity; the first identity follows by summing the zero means. The same conditional-mean argument, applied to sums between two times, gives the martingale property for step-process integrals. For \(H\in\mathcal{L}^2\), step-process approximations converge in \(L^2\) by the Ito isometry. Conditional expectation is an \(L^2\)-contraction, so the martingale conditional-expectation identity passes to the limit. The isometry and integrability pass to the limit as well. \(\square\)
This result does not say that all terms in the sum are independent. A later integrand value can depend on earlier increments. What the proof uses is the conditional mean zero of the next increment, together with its independence from the information available before that increment.
Worked Example: Adapted Terms Can Be Dependent
On the interval \([0,1]\), let \(X=B_{1/2}-B_0\), \(Y=B_1-B_{1/2}\), and set \(H_s=1\) on \((0,1/2]\) and \(H_s=\mathbf{1}_{\{X>0\}}\) on \((1/2,1]\). This is an adapted step process: its second coefficient is measurable at time \(1/2\). Both intervals have length \(1/2\), and \(X,Y\) are independent \(N(0,1/2)\) variables.
The integral is \(I=X+\mathbf{1}_{\{X>0\}}Y\). The two displayed summands are not independent. Indeed, the second summand is zero whenever \(X\le0\), an event of probability \(1/2\); but it is nonzero almost surely whenever \(X>0\). If the summands were independent, the probability that the second is zero conditional on \(X\le0\) would equal its unconditional probability. Instead those probabilities are \(1\) and \(1/2\).
Its mean and second moment can still be computed. The mean is \(\mathbb{E}[X]+\mathbb{E}[\mathbf{1}_{\{X>0\}}]\mathbb{E}[Y]=0\). For the second moment, the cross term has expectation \(2\mathbb{E}[X\mathbf{1}_{\{X>0\}}]\mathbb{E}[Y]=0\). The remaining terms give
Check: the Ito isometry gives the same value: \(\mathbb{E}\int_0^1H_s^2\,ds=\frac12+\frac12\mathbb{P}(X>0)=\frac12+\frac14=\frac34\). The terms’ dependence does not invalidate the integral or its isometry; it invalidates a calculation that assumes those terms are independent.
Right Endpoints Are Not Ito Left Endpoints
The Ito integral is built from left-endpoint values of adapted integrands. A right-endpoint sum can use information from the end of its own interval, so it is not the defining approximation. For the particular integrand \(H_s=B_s\), the difference between the two choices has a direct calculation. Here and below, \([B]_T=T\) denotes Brownian quadratic variation, the limit in probability of sums of squared increments along partitions whose mesh tends to zero.
Worked Example: The Brownian Self-Integral with Right Endpoints
For a partition \(0=t_0<\cdots<t_n=T\), write \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\). The Ito left-endpoint sum and the right-endpoint sum are
For each interval, \(B_{t_i}=B_{t_{i-1}}+\Delta B_i\). Thus \(B_{t_i}\Delta B_i=B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2\). Also, telescoping squares gives \(B_T^2=\sum_i(2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2)\). Therefore the sums satisfy the exact identities
The left sums converge in \(L^2\) to the Ito integral \(\int_0^T B_s\,dB_s\); \(B_s\) is progressively measurable and \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty\). The quadratic-variation sums converge in probability to \(T\). Hence the right sums converge in probability to \(\frac12(B_T^2+T)\), not to the Ito integral \(\frac12(B_T^2-T)\). Their limiting difference is \(T\).
Check: for \(T=1\), the difference of the limits is \(\frac12(B_1^2+1)-\frac12(B_1^2-1)=1\), as also follows from \(R_\Pi-L_\Pi=\sum_i(\Delta B_i)^2\to1\). This is an endpoint-choice error, not an algebraic discrepancy. A right-endpoint sum is also not, in general, the Stratonovich integral: for this integrand the symmetric (midpoint) sums have limit \(\frac12(B_T^2)\), which differs from the right-endpoint limit by \(T/2\).
Variance, Second Moment, and Gaussianity
For a square-integrable random variable \(I\), \(\operatorname{Var}(I)=\mathbb{E}[I^2]-(\mathbb{E}[I])^2\). The proposition shows that an Ito integral with an \(\mathcal{L}^2\) integrand has mean zero. Consequently its unconditional variance equals its second moment, and the Ito isometry computes both. This equality does not make the integral Gaussian. Gaussianity holds for deterministic square-integrable integrands, but it need not hold for random integrands.
Worked Example: A Random-Integrand Integral That Is Not Gaussian
Take \(H_s=B_s\) on \([0,1]\), with \(B\) Brownian relative to the stated filtration. The process is progressively measurable and belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=1/2\). The preceding left-sum computation gives \(I=\int_0^1B_s\,dB_s=\frac12(B_1^2-1)\).
The Ito isometry gives \(\mathbb{E}[I^2]=1/2\), and the martingale mean-zero property gives \(\mathbb{E}[I]=0\). Directly, using \(B_1\sim N(0,1)\), \(\mathbb{E}[B_1^2]=1\) and \(\mathbb{E}[B_1^4]=3\), so
This integral is not Gaussian: \(B_1^2\ge0\), so \(I\ge-1/2\) almost surely, whereas any nondegenerate Gaussian variable has positive probability below every real threshold. The integral is nondegenerate because its variance is \(1/2>0\). Check: its lower bound does not contradict mean zero; the expression \(\frac12(B_1^2-1)\) has expectation \(\frac12(1-1)=0\). Its second moment and variance are both \(1/2\) precisely because its mean is zero.
A Diagnostic Checklist and a Common Mistake
Before accepting a computation, check the following points.
- Endpoint: Is the sum using an adapted left-endpoint value, as in the Ito construction? If not, identify and compute the limiting correction rather than calling it an Ito sum.
- Filtration: Is Brownian motion Brownian with respect to the filtration? Adaptedness alone does not give independence of future increments from current information.
- Independence: Are the terms genuinely independent, or is only a conditional-mean-zero argument available? Later coefficients may depend on earlier increments.
- Distribution: Is the integrand deterministic, or is there a separate argument for Gaussianity? The Ito isometry gives a second moment, not a distribution.
- Moment: Has the mean been established before identifying variance with second moment? For a square-integrable Ito integral under these assumptions, the mean is zero, so the two agree unconditionally.
Common mistake. For the example \(H_s=B_s\) on \([0,1]\), it is wrong to write \(\int_0^1 B_s\,dB_s=\lim R_\Pi\) and then report \(\frac12(B_1^2+1)\) as the Ito integral. The exact identity above shows that this is the right-sum limit. The correction is to use the left sums: \(\int_0^1B_s\,dB_s=\lim L_\Pi=\frac12(B_1^2-1)\). The difference is \(\lim(R_\Pi-L_\Pi)=\lim\sum_i(\Delta B_i)^2=1\), so the two candidate answers cannot be interchangeable.
The most reliable computations keep three questions separate: which integral is being approximated, what conditional information makes each increment calculation valid, and which moment or distribution is actually being claimed. The Ito isometry answers a second-moment question for \(\mathcal{L}^2\) integrands; it does not by itself establish independence of summands or Gaussianity of the integral.
Check Your Understanding
For each question, specify the assumptions or calculation that justify your conclusion.
- Why does adaptedness of \(B\) alone not ensure that \(B_t-B_s\) is independent of \(\mathcal{F}_s\)?
- For the two-interval step process in the example, which fact makes the cross term in \(\mathbb{E}[I^2]\) vanish?
- For the Brownian self-integral on \([0,T]\), what is the limiting difference between right- and left-endpoint sums?
- When does the Ito isometry give the variance directly, rather than requiring subtraction of a nonzero squared mean?
- What property of \(\frac12(B_1^2-1)\) rules out a nondegenerate Gaussian distribution?