Conditional Expectations and the Martingale Property

From the Ito Integral to Conditional Mean Zero

For an Ito integral, zero unconditional mean is not the whole story. The stronger fact is that an integral accumulated after time $s$ has conditional mean zero given the information available at time $s$. This is what makes the integral process a martingale. The extension from step integrands to all square-integrable integrands requires care: approximations to an integral over $(s,t]$ must themselves have no contribution before $s$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, and suppose $B_t$ is a Brownian motion with respect to this filtration. In particular, for $u<v$, $B_v-B_u$ is independent of $\mathcal{F}_u$, has mean zero, and has variance $v-u$. Fix a finite horizon $T$. The admissible integrands on $[0,T]$ are progressively measurable processes $H$ satisfying $\mathbb{E}\int_0^T H_u^2\,du<\infty$; this class is denoted by $\mathcal{L}^2$. The integral $\int_0^t H_u\,dB_u$ below is the Ito integral.

Definition: An integrable process $M_t$ is a martingale with respect to $(\mathcal{F}_t)$ if it is $\mathcal{F}_t$-measurable at each time $t$ and $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely whenever $0\le s\le t$. It is square-integrable if $\mathbb{E}[M_t^2]<\infty$ at every time in its domain.

For an adapted step process $G_u=\sum_{i=0}^{n-1}G_i\mathbf{1}_{(u_i,u_{i+1}]}(u)$, with $G_i$ square-integrable and $\mathcal{F}_{u_i}$-measurable, its Ito integral is the finite sum $\sum_iG_i(B_{u_{i+1}}-B_{u_i})$. Independence of Brownian increments from the filtration gives the conditional mean-zero property for each such increment. The Ito isometry extends these integrals by $L^2$ limits to $\mathcal{L}^2$; the density of adapted step processes in this space is part of the construction of the extended integral.

The Conditional-Expectation Result

Theorem: Let $H\in\mathcal{L}^2$ on $[0,T]$, and define $M_t=\int_0^t H_u\,dB_u$ for $0\le t\le T$. Then $M_t$ is square-integrable and adapted, and for every $0\le s\le t\le T$, $$ \mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\quad\text{and}\quad \mathbb{E}[M_t\mid\mathcal{F}_s]=M_s $$ almost surely. Thus $M$ is a square-integrable martingale.

Proof. Fix $s\le t$. By time-interval additivity of the Ito integral, $M_t-M_s=\int_s^t H_u\,dB_u$. Choose adapted step processes $G^n$ supported on $(s,t]$ such that $\mathbb{E}\int_s^t|G^n_u-H_u|^2\,du\to0$. Such approximations are available by applying the density of adapted step processes on this interval; equivalently, one may approximate on $[0,T]$ and multiply by $\mathbf{1}_{(s,t]}$. The latter multiplication preserves adapted step form and removes every contribution outside the interval. Write each $G^n$ on a partition $s=u_0<\cdots<u_m=t$ as $G^n_u=\sum_{i=0}^{m-1}G_i^n\mathbf{1}_{(u_i,u_{i+1}]}(u)$, where $G_i^n$ is $\mathcal{F}_{u_i}$-measurable. Each product $G_i^n(B_{u_{i+1}}-B_{u_i})$ is integrable by Cauchy–Schwarz. Its conditional expectation given $\mathcal{F}_{u_i}$ is zero, since the Brownian increment is independent of $\mathcal{F}_{u_i}$ and has mean zero. As $\mathcal{F}_s\subseteq\mathcal{F}_{u_i}$, the tower property gives conditional expectation zero given $\mathcal{F}_s$ as well. Adding the finitely many terms yields $\mathbb{E}[\int_s^tG^n_u\,dB_u\mid\mathcal{F}_s]=0$. By the Ito isometry, $\int_s^tG^n_u\,dB_u\to\int_s^tH_u\,dB_u$ in $L^2(\mathbb{P})$. Conditional expectation is a contraction in $L^2$, so the conditional expectations converge in $L^2$ too. Their limit is therefore zero, proving the conditional mean-zero assertion.

For each $t$, the Ito isometry gives $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^tH_u^2\,du\le\mathbb{E}\int_0^TH_u^2\,du<\infty$. Thus $M_t$ is integrable and square-integrable. The step-integral approximations to $\int_0^tH_u\,dB_u$ are $\mathcal{F}_t$-measurable; their $L^2$ limit is also $\mathcal{F}_t$-measurable. Finally, $M_t=M_s+(M_t-M_s)$, so linearity of conditional expectation and the conditional mean-zero result imply $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$. This verifies adaptation, integrability, and the martingale identity. $\square$

The proof establishes a true square-integrable martingale, not merely a local martingale: its second moment is finite at every time because the integrand has finite expected squared energy on the full horizon. No stopping or localization is needed under this global $\mathcal{L}^2$ assumption.

Worked Calculations

Worked Example: A Deterministic Two-Interval Integrand

Let $T=2$, and set $H_u=1$ on $(0,1]$ and $H_u=3$ on $(1,2]$. Compute $M_2$, its conditional expectation given $\mathcal{F}_1$, and its second moment.

The integrand is adapted and square-integrable because $\mathbb{E}\int_0^2H_u^2\,du=1^2\cdot1+3^2\cdot1=10$. The integral is the step sum $M_2=(B_1-B_0)+3(B_2-B_1)=B_1+3(B_2-B_1)$, using $B_0=0$. The increment $B_2-B_1$ is independent of $\mathcal{F}_1$ and has mean zero. Therefore $\mathbb{E}[M_2\mid\mathcal{F}_1]=B_1+3\mathbb{E}[B_2-B_1\mid\mathcal{F}_1]=B_1=M_1$.

For the second moment, the two Brownian increments are independent and centered, with variances $1$ and $1$. The cross term has expectation zero, so $\mathbb{E}[M_2^2]=\mathbb{E}[B_1^2]+9\mathbb{E}[(B_2-B_1)^2]=1+9=10$. Check: this equals $\mathbb{E}\int_0^2H_u^2\,du=10$, as required by the Ito isometry.

Worked Example: A Random Amplitude Known at the Interval Start

Let $T=2$, $H_u=0$ on $(0,1]$, and $H_u=B_1$ on $(1,2]$. Find the integral at time $2$, its conditional expectation given $\mathcal{F}_1$, and its second moment.

The coefficient $B_1$ is $\mathcal{F}_1$-measurable, so this is an adapted step integrand. It belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^2H_u^2\,du=\mathbb{E}[B_1^2]\cdot1=1$. The integral is $M_2=B_1(B_2-B_1)$, while $M_1=0$. Conditioning on $\mathcal{F}_1$ gives $\mathbb{E}[M_2\mid\mathcal{F}_1]=B_1\mathbb{E}[B_2-B_1\mid\mathcal{F}_1]=0=M_1$.

Independence of $B_2-B_1$ from $\mathcal{F}_1$ also gives $\mathbb{E}[M_2^2]=\mathbb{E}[B_1^2]\mathbb{E}[(B_2-B_1)^2]=1\cdot1=1$. Check: the Ito isometry gives the same value, $\mathbb{E}\int_0^2H_u^2\,du=\mathbb{E}[B_1^2]=1$. The random coefficient need not be independent of the past; it must be known at the left endpoint of the interval whose Brownian increment it multiplies.

Worked Example: Conditioning at an Earlier Time

Use $H_u=1$ on $(0,1]$ and $H_u=B_1$ on $(1,2]$. Compute $\mathbb{E}[M_2\mid\mathcal{F}_{1/2}]$ and compare it with $M_{1/2}$.

As in the step-integral definition, $M_2=B_1+B_1(B_2-B_1)$. Since $B_1=B_{1/2}+(B_1-B_{1/2})$, independent increments give $\mathbb{E}[B_1\mid\mathcal{F}_{1/2}]=B_{1/2}$. For the other term, condition first on $\mathcal{F}_1$: the increment $B_2-B_1$ is independent of $\mathcal{F}_1$, so $\mathbb{E}[B_1(B_2-B_1)\mid\mathcal{F}_1]=0$. Applying the tower property then gives $\mathbb{E}[B_1(B_2-B_1)\mid\mathcal{F}_{1/2}]=0$. Hence $\mathbb{E}[M_2\mid\mathcal{F}_{1/2}]=B_{1/2}$.

On $[0,\tfrac12]$, the integrand is $1$, so $M_{1/2}=B_{1/2}$, confirming the martingale identity. The integrability condition is $\mathbb{E}\int_0^2H_u^2\,du=1+\mathbb{E}[B_1^2]=2<\infty$. Check: conditioning the two terms of $M_2$ separately gives $B_{1/2}+0$, exactly the value obtained by the theorem.

Why the Interval Support Matters

When proving a conditional statement about $\int_s^tH_u\,dB_u$, an approximation must represent that interval integral, rather than an integral with an extra contribution from earlier times. For example, take the target integrand $H=0$ and $s>0$. The step process $G^n_u=n^{-1}\mathbf{1}_{(0,s]}(u)$ converges to zero in $L^2$, but its integral over $[0,T]$ is $B_s/n$. Given $\mathcal{F}_s$, its conditional expectation is $B_s/n$, not zero in general. Thus convergence of arbitrary global approximants does not justify saying that each approximating integral has conditional mean zero.

The correction is to approximate on $(s,t]$, or to multiply each global approximant by $\mathbf{1}_{(s,t]}$ before taking its integral. The resulting step integrals have no pre-$s$ term. Their conditional expectations given $\mathcal{F}_s$ are zero, and the Ito isometry carries that property to the $L^2$ limit.

A second common error is to infer the martingale property from $\mathbb{E}[M_t]=0$ alone. That only checks conditioning on the trivial initial information; the martingale property requires $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for every $s\le t$. The increment argument proves this stronger identity. Under the present finite-energy assumption it also proves square-integrability, while an integrand with only local square-integrability need not yield a true square-integrable martingale on the full interval.

Check Your Understanding

For each question, distinguish what follows from a step-integral computation and what requires passage to the $L^2$ limit.

For an adapted step integrand on $(s,t]$, explain why each coefficient-times-increment term has conditional expectation zero given $\mathcal{F}_s$.
Why must approximants used for $\int_s^tH_u\,dB_u$ be supported on $(s,t]$? Give the extra term produced by $n^{-1}\mathbf{1}_{(0,s]}$.
If $H_u=B_2$ on $(1,2]$, is this an admissible adapted step integrand? Explain using the left endpoint information condition.
For $H_u=2$ on $(0,1]$, compute $\mathbb{E}[M_1^2]$ and verify it using the Ito isometry.
State the conditional expectation identity that turns the zero-mean increment result into the martingale property.

Conditional Expectations and the Martingale Property

What You'll Learn