The Martingale Property of the Integral Process

From Mean-Zero Sums to a Martingale Process

A simple stochastic integral is first defined by multiplying Brownian increments by coefficients known at the start of their time intervals. The zero-mean property of such sums gives a useful fact at a fixed terminal time. The stronger process-level statement is that, at every time $s$, the conditional expectation of the integral at a later time $t$ equals its value at $s$. The key is to condition each future increment on the filtration at the time its coefficient is chosen.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative 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 $T>0$ and a partition $0=t_0<t_1<\cdots<t_n=T$. A simple predictable integrand on this partition has the form $H_t=\sum_{k=0}^{n-1}\xi_k\mathbf{1}_{(t_k,t_{k+1}]}(t)$, where each $\xi_k$ is $\mathcal{F}_{t_k}$-measurable.

Definition: The simple Ito integral process associated with $H$ is $$ I_t=\int_0^t H_u\,dB_u =\sum_{k=0}^{n-1}\xi_k\big(B_{t\wedge t_{k+1}}-B_{t\wedge t_k}\big),\qquad 0\le t\le T. $$ We assume $\mathbb{E}\int_0^T H_u^2\,du=\sum_{k=0}^{n-1}(t_{k+1}-t_k)\mathbb{E}[\xi_k^2]<\infty$. Thus $H$ is in the class $\mathcal{L}^2$ of progressively measurable integrands with finite expected integrated square. Every integral below is an Ito integral for such an integrand.

The endpoint convention makes the coefficient on $(t_k,t_{k+1}]$ known at its left endpoint. At an individual endpoint the value of $H$ does not affect the integral, but the left-endpoint measurability condition does matter: it is what allows the Brownian increment to have conditional mean zero after the coefficient is fixed.

Adaptedness and Integrability

Proposition: Under the assumptions above, $I_t$ is $\mathcal{F}_t$-measurable and integrable for every $t\in[0,T]$. In fact, $I_t$ is square-integrable.

Proof. Fix $t\in[0,T]$. Each nonzero summand in the definition of $I_t$ has $t_k<t$; its coefficient $\xi_k$ is therefore $\mathcal{F}_{t_k}$-measurable and hence $\mathcal{F}_t$-measurable. Its Brownian increment is also $\mathcal{F}_t$-measurable. The finite sum $I_t$ is consequently $\mathcal{F}_t$-measurable. For a summand with $t_k<t$, put $v=t\wedge t_{k+1}$. The increment $B_v-B_{t_k}$ is independent of $\mathcal{F}_{t_k}$ and has second moment $v-t_k$. Conditioning on $\mathcal{F}_{t_k}$ gives $$ \mathbb{E}\!\left[\xi_k^2(B_v-B_{t_k})^2\right] =(v-t_k)\mathbb{E}[\xi_k^2] \le (t_{k+1}-t_k)\mathbb{E}[\xi_k^2]<\infty. $$ A finite sum of square-integrable random variables is square-integrable, so $I_t$ is square-integrable and, in particular, integrable. If $t\le t_k$, the corresponding summand is zero. $\square$

This argument checks integrability without using the Ito isometry: it bounds each of the finitely many terms separately. It does not calculate the second moment of their sum, which would require handling cross terms.

The Martingale Property

Theorem: For $H$ as defined above, the integral process $I_t=\int_0^t H_u\,dB_u$ is a true martingale with respect to $(\mathcal{F}_t)$ on $[0,T]$. That is, $I_t$ is integrable and adapted, and for every $0\le s\le t\le T$, $$ \mathbb{E}[I_t\mid\mathcal{F}_s]=I_s\quad\text{almost surely}. $$

The conditional expectation is meaningful because the variables are integrable and the filtration is increasing, so $\mathcal{F}_s\subseteq\mathcal{F}_t$. To prove the identity, separate an interval containing $s$, if there is one, from intervals beginning at or after $s$. On the first kind, the coefficient is already known at time $s$. On the second kind, condition first at the interval's left endpoint, where the coefficient is known.

Proof. Adaptedness and integrability were established in the Proposition. Fix $s\le t$. For each partition interval, the contribution to $I_t-I_s$ is zero if the interval does not intersect $(s,t]$. If $s\le t_k<t$, its contribution is $$ \xi_k\big(B_{v_k}-B_{t_k}\big),\qquad v_k=t\wedge t_{k+1}. $$ Since $t_k\ge s$, $\mathcal{F}_s\subseteq\mathcal{F}_{t_k}$. The coefficient $\xi_k$ is $\mathcal{F}_{t_k}$-measurable, and the increment $B_{v_k}-B_{t_k}$ is independent of $\mathcal{F}_{t_k}$ with mean zero. The product is integrable by the square-integrability calculation in the Proposition and the Cauchy–Schwarz inequality. Thus $$ \mathbb{E}\!\left[\xi_k(B_{v_k}-B_{t_k})\mid\mathcal{F}_{t_k}\right] =\xi_k\,\mathbb{E}[B_{v_k}-B_{t_k}\mid\mathcal{F}_{t_k}]=0. $$ The tower property then gives $\mathbb{E}[\xi_k(B_{v_k}-B_{t_k})\mid\mathcal{F}_s]=0$.

There can also be at most one interval with $t_k<s<t_{k+1}$. If it contributes to $I_t-I_s$, its contribution is $\xi_k(B_{v_k}-B_s)$, where $v_k=t\wedge t_{k+1}$. Here $\xi_k$ is $\mathcal{F}_{t_k}$-measurable and therefore $\mathcal{F}_s$-measurable. The increment $B_{v_k}-B_s$ has conditional mean zero given $\mathcal{F}_s$, so $$ \mathbb{E}\!\left[\xi_k(B_{v_k}-B_s)\mid\mathcal{F}_s\right] =\xi_k\,\mathbb{E}[B_{v_k}-B_s\mid\mathcal{F}_s]=0. $$ If $s$ is a partition endpoint, there is no interval crossing it; if $s=t$, the entire difference is zero. These cases are therefore covered as well. Summing the finitely many contributions and using linearity of conditional expectation yields $\mathbb{E}[I_t-I_s\mid\mathcal{F}_s]=0$. Since $I_s$ is $\mathcal{F}_s$-measurable, $\mathbb{E}[I_t\mid\mathcal{F}_s]=I_s$ almost surely. Hence $I$ is a true martingale, not merely a local martingale. $\square$

Worked Conditional Calculations

Worked Example: A Deterministic Two-Step Integrand

Let $T=2$, $t_0=0,t_1=1,t_2=2$, and choose $H_t=1$ on $(0,1]$ and $H_t=2$ on $(1,2]$. Find $I_t$ and verify the martingale identity for $s=\tfrac12$, $t=\tfrac32$.

For $0\le t\le1$, only the first interval contributes, so $I_t=B_t$. For $1\le t\le2$, the first interval contributes $B_1$, and the second contributes $2(B_t-B_1)$. In particular, $$ I_{1/2}=B_{1/2},\qquad I_{3/2}=B_1+2(B_{3/2}-B_1). $$

Rewrite the later value relative to time $\tfrac12$: $$ I_{3/2}=B_{1/2}+(B_1-B_{1/2})+2(B_{3/2}-B_1). $$ Both increments after time $\tfrac12$ have conditional mean zero given $\mathcal{F}_{1/2}$. Therefore $\mathbb{E}[I_{3/2}\mid\mathcal{F}_{1/2}]=B_{1/2}=I_{1/2}$. Check: taking expectations gives $\mathbb{E}[I_{3/2}]=0$, since each Brownian increment in its defining expression has mean zero. This agrees with the conditional identity, because $\mathbb{E}[I_{1/2}]=\mathbb{E}[B_{1/2}]=0$.

Worked Example: A Random Coefficient Known at the Left Endpoint

On the partition $0<1<2$, let $H_t=1$ on $(0,1]$ and $H_t=B_1$ on $(1,2]$. The second coefficient is $\mathcal{F}_1$-measurable, and $\mathbb{E}[B_1^2]=1$, so the square-integrability requirement is satisfied. Verify the conditional identity at $s=\tfrac32$, $t=2$.

For times in the second interval, $$ I_t=B_1+B_1(B_t-B_1). $$ Consequently, $$ I_2=I_{3/2}+B_1(B_2-B_{3/2}). $$ The factor $B_1$ is $\mathcal{F}_{3/2}$-measurable. The increment $B_2-B_{3/2}$ is independent of $\mathcal{F}_{3/2}$ and has mean zero. It follows that $$ \mathbb{E}[I_2\mid\mathcal{F}_{3/2}] =I_{3/2}+B_1\mathbb{E}[B_2-B_{3/2}\mid\mathcal{F}_{3/2}] =I_{3/2}. $$ Check: the terminal expectation is $\mathbb{E}[B_1]+\mathbb{E}[B_1(B_2-B_1)]=0+0=0$. For the product term, condition on $\mathcal{F}_1$: its conditional expectation is $B_1\mathbb{E}[B_2-B_1\mid\mathcal{F}_1]=0$. Thus the direct expectation agrees with the martingale's zero initial value.

Worked Example: A Bounded Random Coefficient and an Earlier Conditioning Time

Let $H_t=1$ on $(0,1]$ and $H_t=\mathbf{1}_{\{B_1>0\}}$ on $(1,2]$. This coefficient is $\mathcal{F}_1$-measurable and bounded. Verify $\mathbb{E}[I_2\mid\mathcal{F}_{1/2}]=I_{1/2}$, including the contribution whose coefficient is not yet known at time $\tfrac12$.

The terminal value is $I_2=B_1+\mathbf{1}_{\{B_1>0\}}(B_2-B_1)$, while $I_{1/2}=B_{1/2}$. First condition the second term at time $1$. The indicator is then measurable, and $$ \mathbb{E}\!\left[\mathbf{1}_{\{B_1>0\}}(B_2-B_1)\mid\mathcal{F}_1\right] =\mathbf{1}_{\{B_1>0\}}\mathbb{E}[B_2-B_1\mid\mathcal{F}_1]=0. $$ The tower property gives conditional mean zero for this term given $\mathcal{F}_{1/2}$. Brownian independent increments also give $\mathbb{E}[B_1\mid\mathcal{F}_{1/2}]=B_{1/2}$, because $B_1=B_{1/2}+(B_1-B_{1/2})$ and the latter increment has conditional mean zero. Hence $\mathbb{E}[I_2\mid\mathcal{F}_{1/2}]=B_{1/2}$. Check: taking expectations gives $\mathbb{E}[I_2]=\mathbb{E}[B_1]+0=0$, consistent with $I_0=0$.

Why Predictability Matters

The proof relies on the coefficient being known before the increment is observed. A coefficient chosen using the end of its own interval does not satisfy this assumption, and the conditional mean-zero step can fail.

Common mistake. On $(0,1]$, suppose one informally uses the coefficient $B_1$. It is not $\mathcal{F}_0$-measurable, so this is not an admissible simple predictable Ito integrand on that interval. Treating it as though it were admissible would give the formal product $B_1(B_1-B_0)=B_1^2$, whose expectation is $\mathbb{E}[B_1^2]=1$, not zero. The invalid calculation is exactly the error of taking the coefficient outside a conditional expectation at time $0$ when it is not known then. By contrast, the admissible coefficient $1$ gives the integral $B_1-B_0=B_1$, with expectation zero.

The result proved here is a true martingale statement: integrability, adaptedness, and the conditional expectation identity have all been checked directly. The argument is specific to simple predictable integrands and finite sums. Extending the construction and the martingale property to general $\mathcal{L}^2$ integrands requires the later $L^2$-limit construction; no such extension is needed for the finite partition argument above.

Check Your Understanding

In each question, assume Brownian increments are independent of the stated filtration and equalities of conditional expectations are understood almost surely.

For a simple integrand on a three-interval partition, identify which coefficient is used on an interval containing the conditioning time $s$, and why it is measurable with respect to $\mathcal{F}_s$.
Explain why a full interval $(t_k,t_{k+1}]$ lying after $s$ can be handled by conditioning first on $\mathcal{F}_{t_k}$.
For $H_t=B_1$ on $(1,2]$, write $I_2-I_{3/2}$ and compute its conditional expectation given $\mathcal{F}_{3/2}$.
Which parts of the martingale definition are verified by square-integrability, and which part requires the conditional-increment argument?
Give the precise measurability condition that fails when $B_1$ is used as a coefficient on $(0,1]$, and compute the expectation of the resulting formal product.