Conditional Second Moments of Simple Integral Increments

Conditioning on the Information at the Start

The Ito isometry for simple predictable processes gives an unconditional second moment. Often the information available at the beginning of an interval is itself important: an adapted coefficient may already be known then, while future Brownian increments remain random. Conditioning on that starting information preserves this distinction. It gives a second-moment identity that is useful for tracking how the variability of an integral increment depends on the current state of knowledge.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ with respect to the filtration. In particular, for $u<v$, $B_v-B_u$ is independent of $\mathcal{F}_u$, has mean zero, and has second moment $v-u$. All integrals below are Ito integrals. An admissible integrand is progressively measurable and belongs to $\mathcal{L}^2$ on the time interval in question, meaning $\mathbb{E}\int_0^T H_u^2\,du<\infty$. The simple processes considered below satisfy these conditions, so their integrals are defined by finite sums. No Stratonovich integral or Stratonovich correction is involved.

Definition: On a deterministic partition $s=t_0<t_1<\cdots<t_n=t$, a simple predictable integrand has the form $H_u=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(u)$, where each $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and $\mathbb{E}[\xi_i^2]<\infty$. Its Ito integral increment is $J_{s,t}=\int_s^t H_u\,dB_u=\sum_{i=0}^{n-1}\xi_i(B_{t_{i+1}}-B_{t_i})$. Values at the finitely many partition endpoints do not affect the time integral.

The coefficient $\xi_i$ is chosen using information available at the left endpoint $t_i$; it is not required to be deterministic. For conditioning on $\mathcal{F}_s$, all coefficients after $s$ may still be random. The identity below says to take the conditional expectation of their squared contributions, rather than to treat those coefficients as already known at time $s$.

The Conditional Second-Moment Identity

Theorem: For the simple predictable integrand above, let $J_{s,t}=\int_s^t H_u\,dB_u$. Then, almost surely, $$ \mathbb{E}[J_{s,t}\mid\mathcal{F}_s]=0,\qquad \mathbb{E}[J_{s,t}^2\mid\mathcal{F}_s] =\mathbb{E}\!\left[\int_s^t H_u^2\,du\middle|\mathcal{F}_s\right] =\mathbb{E}\!\left[\sum_{i=0}^{n-1}\xi_i^2(t_{i+1}-t_i)\middle|\mathcal{F}_s\right]. $$ Consequently, the conditional variance of $J_{s,t}$, defined as $\mathbb{E}[(J_{s,t}-\mathbb{E}[J_{s,t}\mid\mathcal{F}_s])^2\mid\mathcal{F}_s]$, equals this same conditional second moment.

Proof. Set $\Delta B_i=B_{t_{i+1}}-B_{t_i}$, $\Delta t_i=t_{i+1}-t_i$, and $Y_i=\xi_i\Delta B_i$. Each $Y_i$ is square-integrable because $\xi_i$ is square-integrable and $\Delta B_i$ is independent of $\mathcal{F}_{t_i}$, so $\mathbb{E}[Y_i^2]=\Delta t_i\mathbb{E}[\xi_i^2]<\infty$. Thus all the finite sums and conditional expectations used below are well-defined. Since $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and $\mathbb{E}[\Delta B_i\mid\mathcal{F}_{t_i}]=0$, the tower property gives $\mathbb{E}[Y_i\mid\mathcal{F}_s]=0$. Indeed, $\mathcal{F}_s\subseteq\mathcal{F}_{t_i}$, and conditioning first on $\mathcal{F}_{t_i}$ gives zero. Summing proves the conditional-mean identity.

For a diagonal term, independence of $\Delta B_i$ from $\mathcal{F}_{t_i}$ and $\mathbb{E}[(\Delta B_i)^2]=\Delta t_i$ give $\mathbb{E}[Y_i^2\mid\mathcal{F}_{t_i}]=\xi_i^2\Delta t_i$. Conditioning again on $\mathcal{F}_s$ yields $\mathbb{E}[Y_i^2\mid\mathcal{F}_s]=\mathbb{E}[\xi_i^2\Delta t_i\mid\mathcal{F}_s]$.

For distinct terms, take $i<j$. The product $Y_i\xi_j$ is $\mathcal{F}_{t_j}$-measurable, and $\Delta B_j$ is independent of $\mathcal{F}_{t_j}$ with mean zero. The product is integrable by the Cauchy–Schwarz inequality, since $Y_i$ and $Y_j$ are square-integrable. Therefore $\mathbb{E}[Y_iY_j\mid\mathcal{F}_{t_j}]=Y_i\xi_j\mathbb{E}[\Delta B_j\mid\mathcal{F}_{t_j}]=0$. The tower property gives $\mathbb{E}[Y_iY_j\mid\mathcal{F}_s]=0$. Expanding $J_{s,t}^2=(\sum_iY_i)^2$, all cross terms thus vanish after conditioning on $\mathcal{F}_s$, leaving the sum of the diagonal conditional expectations. Finally, $\int_s^tH_u^2\,du=\sum_i\xi_i^2\Delta t_i$ pathwise. The conditional variance equals the second moment because the conditional mean is zero. $\square$

This conditional identity is not the assertion that each random coefficient in the sum is known at time $s$. Only a coefficient measurable with respect to $\mathcal{F}_s$ can be taken outside $\mathbb{E}[\cdot\mid\mathcal{F}_s]$. Later coefficients must remain inside the conditional expectation.

Worked Calculations with Adapted Coefficients

Worked Example: A Single Coefficient Known at the Start

Let $s=1$, $t=3$, and $H_u=B_1$ on $(1,3]$. Compute the conditional second moment of $J=\int_1^3B_1\,dB_u$, then its unconditional second moment.

The coefficient $B_1$ is $\mathcal{F}_1$-measurable and square-integrable, since $\mathbb{E}[B_1^2]=1$. The defining simple sum has one term: $J=B_1(B_3-B_1)$. The conditional identity, or independence and the variance of $B_3-B_1$, gives

\mathbb{E}[J^2\mid\mathcal{F}_1] =B_1^2(3-1)=2B_1^2.

Taking expectations gives $\mathbb{E}[J^2]=2\mathbb{E}[B_1^2]=2$. Directly, $B_3-B_1$ is independent of $B_1$, has mean zero and second moment $2$; hence $\mathbb{E}[B_1^2(B_3-B_1)^2]=\mathbb{E}[B_1^2]\mathbb{E}[(B_3-B_1)^2]=1\cdot2=2$. This check also shows why the conditional answer is the random variable $2B_1^2$, not the constant $2$.

Worked Example: Two Coefficients on a Two-Step Partition

On $[0,2]$, take $H_u=1$ on $(0,1]$ and $H_u=1+B_1$ on $(1,2]$. Find the second moment of the full increment $J=\int_0^2H_u\,dB_u$, conditioning first on $\mathcal{F}_0$.

The coefficient on the second interval is $\mathcal{F}_1$-measurable and square-integrable, since $\mathbb{E}[(1+B_1)^2]=1+2\mathbb{E}[B_1]+\mathbb{E}[B_1^2]=2$. The finite-sum definition gives $J=B_1+(1+B_1)(B_2-B_1)$. Since $\mathcal{F}_0$ is the starting sigma-field, the theorem gives

\mathbb{E}[J^2\mid\mathcal{F}_0] =\mathbb{E}\!\left[1^2(1-0)+(1+B_1)^2(2-1)\middle|\mathcal{F}_0\right] =1+2=3.

Here the result is constant because $B_1$ has mean zero and second moment one. Check it directly by writing $X=B_1$ and $Y=B_2-B_1$. These are independent, each with mean zero and second moment one. Expanding $J=X+(1+X)Y$ gives $\mathbb{E}[J^2]=\mathbb{E}[X^2]+2\mathbb{E}[X(1+X)Y]+\mathbb{E}[(1+X)^2Y^2]$. The middle term is zero because $Y$ is independent of $X$ and has mean zero. The first is $1$, and the last is $\mathbb{E}[(1+X)^2]\mathbb{E}[Y^2]=2\cdot1=2$. The total is $1+0+2=3$, in agreement with the conditional identity.

Worked Example: Conditioning Before a Later Random Coefficient

On $[1,3]$, use the partition $1<2<3$. Set $H_u=1+B_1$ on $(1,2]$, and $H_u=B_2-B_1$ on $(2,3]$. Compute $\mathbb{E}[J^2\mid\mathcal{F}_1]$, where $J=\int_1^3H_u\,dB_u$.

The first coefficient is $\mathcal{F}_1$-measurable; the second is $\mathcal{F}_2$-measurable. Both are square-integrable. The theorem gives $\mathbb{E}[J^2\mid\mathcal{F}_1]=(1+B_1)^2+\mathbb{E}[(B_2-B_1)^2\mid\mathcal{F}_1]=(1+B_1)^2+1$. To check this with the finite sum, let $U=B_2-B_1$ and $V=B_3-B_2$. Then $J=(1+B_1)U+UV=U(1+B_1+V)$. Given $\mathcal{F}_1$, $U$ and $V$ are independent standard normal increments, independent of $\mathcal{F}_1$. Thus

\mathbb{E}[J^2\mid\mathcal{F}_1] =\mathbb{E}[U^2]\mathbb{E}[(1+B_1+V)^2\mid\mathcal{F}_1] =1\cdot\big((1+B_1)^2+1\big).

In the last step, the cross term has conditional expectation $2(1+B_1)\mathbb{E}[V]=0$, and $\mathbb{E}[V^2]=1$. This direct computation agrees with the formula, including the contribution from the coefficient that is random at time $1$.

Why the Conditional Form Matters

Because the conditional mean of the increment is zero, its conditional second moment is also its conditional variance. The theorem therefore describes the increment's conditional variability using the squared integrand accumulated over the interval. If the integrand is already known at the starting time, this quantity can remain random even though it is measurable with respect to the conditioning sigma-field. If it depends on information arriving later, its squared value must itself be averaged conditionally.

The result depends on both predictability and Brownian independence relative to the filtration. A coefficient chosen at the left endpoint can depend on the past; the next Brownian increment is nevertheless independent of that past. A coefficient chosen using the right endpoint's Brownian increment does not satisfy the stated hypothesis. In that case the conditional mean-zero and cross-term arguments in the proof need not hold, so this identity cannot be applied without further justification.

Common mistake: For the single-interval example with coefficient $B_1$, replacing the coefficient by its unconditional second moment before conditioning gives the incorrect claim $\mathbb{E}[J^2\mid\mathcal{F}_1]=\mathbb{E}[B_1^2](3-1)=2$. This drops the information in $\mathcal{F}_1$. The correct calculation keeps the coefficient inside the conditional expression until it is clear that it is known at the conditioning time: $\mathbb{E}[B_1^2(B_3-B_1)^2\mid\mathcal{F}_1]=B_1^2\cdot2=2B_1^2$. The unconditional expectation of this answer is $2$, but the conditional second moment itself is generally random.

Check Your Understanding

In each question, assume Brownian motion is Brownian with respect to the stated filtration and that each coefficient is measurable at the left endpoint of its interval.

For $J=B_2(B_4-B_2)$, compute $\mathbb{E}[J^2\mid\mathcal{F}_2]$ and then $\mathbb{E}[J^2]$.
Let $H_u=2$ on $(0,1]$ and $H_u=B_1$ on $(1,3]$. Find $\mathbb{E}[(\int_0^3H_u\,dB_u)^2\mid\mathcal{F}_0]$.
For a simple integral increment, why do the cross terms vanish after conditioning on the starting sigma-field?
In the theorem, which coefficient values can be taken outside $\mathbb{E}[\cdot\mid\mathcal{F}_s]$, and why?
For $J=(1+B_1)(B_3-B_1)$, state the conditional second moment given $\mathcal{F}_1$ and explain why it need not be a constant.

Conditional Second Moments of Simple Integral Increments

What You'll Learn