A Worked Calculation With a Multi-Interval Integrand

From Interval Coefficients to an Integral

A simple Ito integral is a finite sum, but its terms need not be deterministic: on each time interval, the coefficient can be a random variable determined by the information available at the interval's start. This makes a multi-interval calculation a useful test of what the construction requires. One must keep track of which increment belongs to which coefficient, and use the fact that each Brownian increment is independent of the information already available.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to the filtration: for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Expectations and conditional expectations below are under $\mathbb{P}$.

Definition: Let $0=t_0<t_1<\cdots<t_n=T$ be a finite partition. An adapted partition-step integrand has the form $H_t=H_i$ on each interval $(t_i,t_{i+1}]$, where $H_i$ is $\mathcal{F}_{t_i}$-measurable and $\mathbb{E}[H_i^2]<\infty$. Its value at the finitely many partition endpoints does not affect its time integral. It is progressively measurable and satisfies $\mathbb{E}\int_0^T H_t^2\,dt=\sum_{i=0}^{n-1}\mathbb{E}[H_i^2](t_{i+1}-t_i)<\infty$; thus it belongs to $\mathcal{L}^2$. Its Ito integral process is $M_t=\int_0^t H_s\,dB_s$, defined here by the finite sum of coefficient-times-increment terms.

More explicitly, if $t\in[t_k,t_{k+1}]$, then the construction gives

M_t=\sum_{i=0}^{k-1}H_i(B_{t_{i+1}}-B_{t_i})+H_k(B_t-B_{t_k}).

This formula specifies the value at every time, not only at the terminal time. The coefficient $H_k$ multiplies the Brownian increment after $t_k$; it is not replaced by a value observed at the end of that increment. The integrability assumption ensures each term is square-integrable, since $\mathbb{E}[H_i^2(B_{t_{i+1}}-B_{t_i})^2]=\mathbb{E}[H_i^2](t_{i+1}-t_i)$, using independence of the increment from $\mathcal{F}_{t_i}$.

Mean, Variance, and Martingale Increments

The next result collects the direct checks that are useful in a calculation. A martingale here means an integrable adapted process satisfying $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $s\le t$. The conclusion is a true square-integrable martingale, not merely a local martingale: its second moment is finite at each time.

Theorem: For an adapted partition-step integrand with square-integrable coefficients, its integral process $M_t$ is a square-integrable martingale. In particular, for $0\le s\le t\le T$, $\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0$, $\mathbb{E}[M_t]=0$, and $$ \mathbb{E}[M_T^2]=\sum_{i=0}^{n-1}\mathbb{E}[H_i^2](t_{i+1}-t_i). $$

Proof. First consider $M_T=\sum_i H_i\Delta B_i$, where $\Delta B_i=B_{t_{i+1}}-B_{t_i}$. Since $H_i$ is $\mathcal{F}_{t_i}$-measurable and $\Delta B_i$ is independent of $\mathcal{F}_{t_i}$ with mean zero, $\mathbb{E}[H_i\Delta B_i]=0$. For $i<j$, the product $H_i\Delta B_iH_j$ is $\mathcal{F}_{t_j}$-measurable; the later increment $\Delta B_j$ is independent of $\mathcal{F}_{t_j}$ and has mean zero. Thus $\mathbb{E}[H_i\Delta B_iH_j\Delta B_j]=0$. These expectations exist by the square-integrability of the terms and the Cauchy–Schwarz inequality. For each diagonal term, independence gives $\mathbb{E}[(H_i\Delta B_i)^2]=\mathbb{E}[H_i^2](t_{i+1}-t_i)$. Expanding the square of the finite sum proves the stated second-moment formula, which is finite. The same argument applies to the sum defining $M_t$, with the last interval shortened to $t-t_k$, so $M_t$ is square-integrable for every $t$.

To verify the conditional martingale assertion, split the finite sum for $M_t-M_s$ at $s$ and at the original partition points. Each resulting term has the form $H_i(B_b-B_a)$, where $a\ge s$, $a\ge t_i$, and $b\le t_{i+1}$. The coefficient is $\mathcal{F}_{t_i}$-measurable, hence $\mathcal{F}_a$-measurable. Independence and the zero mean of the Brownian increment give $\mathbb{E}[H_i(B_b-B_a)\mid\mathcal{F}_a]=0$. Taking conditional expectations again with respect to $\mathcal{F}_s\subseteq\mathcal{F}_a$ gives zero. Summing the finitely many terms proves $\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0$. Taking $s=0$, and using $M_0=0$, gives $\mathbb{E}[M_t]=0$. $\square$

Worked Calculations

Worked Example: Three Random Coefficients on Three Unit Intervals

On the partition $0<1<2<3$, take $H_0=2$, $H_1=B_1$, and $H_2=1+B_2-B_1$. Each coefficient is measurable at the left endpoint of its interval. The coefficients are square-integrable: their second moments are $4$, $1$, and $2$, respectively. Compute the terminal integral, its mean, and its variance.

Write the independent unit-time increments as $X_1=B_1$, $X_2=B_2-B_1$, and $X_3=B_3-B_2$. Each has mean zero and second moment $1$. The integral is

$$ M_3=2X_1+X_1X_2+(1+X_2)X_3. $$

The first term has mean $2\mathbb{E}[X_1]=0$. Independence gives $\mathbb{E}[X_1X_2]=\mathbb{E}[X_1]\mathbb{E}[X_2]=0$. Conditioning on $X_1,X_2$ gives $\mathbb{E}[(1+X_2)X_3]=0$. Hence $\mathbb{E}[M_3]=0$.

For the variance, the mean is zero, so compute $\mathbb{E}[M_3^2]$. The three squared terms have expectations $\mathbb{E}[(2X_1)^2]=4\mathbb{E}[X_1^2]=4$, $\mathbb{E}[(X_1X_2)^2]=\mathbb{E}[X_1^2]\mathbb{E}[X_2^2]=1$, and $\mathbb{E}[((1+X_2)X_3)^2]=\mathbb{E}[(1+X_2)^2]\mathbb{E}[X_3^2]=(1+1)\cdot1=2$. Every cross term has expectation zero: for the first and second terms, it is $2\mathbb{E}[X_1^2X_2]=0$; for terms containing $X_3$, condition on $X_1,X_2$ and use $\mathbb{E}[X_3]=0$. Therefore $\operatorname{Var}(M_3)=4+1+2=7$.

Check: the theorem gives $4\cdot1+1\cdot1+2\cdot1=7$, the same variance. The direct expansion also shows why the random coefficients do not create extra cross terms: each later Brownian increment has conditional mean zero given the earlier information.

Worked Example: Unequal Interval Lengths

Use the partition $0<\tfrac12<\tfrac32$, with $H_0=1$ and $H_1=B_{1/2}$. The first interval has length $\tfrac12$, the second length $1$. Both coefficients are square-integrable, and $H_1$ is known at time $\tfrac12$. Evaluate the terminal integral and its variance.

Let $X=B_{1/2}$ and $Y=B_{3/2}-B_{1/2}$. These are independent, with $\mathbb{E}[X]=\mathbb{E}[Y]=0$, $\mathbb{E}[X^2]=\tfrac12$, and $\mathbb{E}[Y^2]=1$. The integral is $M_{3/2}=X+XY$. Thus $\mathbb{E}[M_{3/2}]=0$, since both $X$ and $XY$ have mean zero. Its second moment is

\mathbb{E}[(X+XY)^2] =\mathbb{E}[X^2]+2\mathbb{E}[X^2Y]+\mathbb{E}[X^2Y^2] =\frac12+0+\frac12\cdot1=1.

The middle term vanishes by independence and $\mathbb{E}[Y]=0$; the last factors into the second moments of $X$ and $Y$. Since the mean is zero, the variance is $1$. Check: the coefficient formula gives $\mathbb{E}[H_0^2]\cdot\tfrac12+\mathbb{E}[H_1^2]\cdot1=1\cdot\tfrac12+\tfrac12\cdot1=1$. This also confirms that the length of each interval, not just the number of intervals, enters the calculation.

Worked Example: A Conditional Increment Across an Interval Boundary

On $0<1<2<3$, take coefficients $H_0=1$, $H_1=B_1$, and $H_2=1+B_2$. Consider the increment $M_{5/2}-M_{3/2}$, conditional on $\mathcal{F}_{3/2}$. All coefficients are square-integrable and known at the start of their respective intervals.

Set $U=B_2-B_{3/2}$ and $V=B_{5/2}-B_2$. Both are independent of $\mathcal{F}_{3/2}$, independent of one another, and have mean zero and variance $\tfrac12$. Since $B_2=B_{3/2}+U$, the increment is

M_{5/2}-M_{3/2}=B_1U+(1+B_{3/2}+U)V.

For the first term, $B_1$ is $\mathcal{F}_{3/2}$-measurable and $\mathbb{E}[U\mid\mathcal{F}_{3/2}]=0$. For the second, condition first on $\mathcal{F}_2$: $1+B_2$ is then known and $V$ has conditional mean zero. Consequently $\mathbb{E}[M_{5/2}-M_{3/2}\mid\mathcal{F}_{3/2}]=0$.

The conditional second moment can also be checked explicitly. The square of the first term has conditional expectation $\tfrac12 B_1^2$. For the second term, $V$ is independent of $U$, so its conditional second moment is $\tfrac12\mathbb{E}[(1+B_{3/2}+U)^2\mid\mathcal{F}_{3/2}] =\tfrac12((1+B_{3/2})^2+\tfrac12)$. The cross term has conditional expectation zero because it contains the factor $V$, whose conditional mean given $\mathcal{F}_2$ is zero. Thus

\mathbb{E}[(M_{5/2}-M_{3/2})^2\mid\mathcal{F}_{3/2}] =\frac12 B_1^2+\frac12\left((1+B_{3/2})^2+\frac12\right).

Check: taking expectations gives $\tfrac12\mathbb{E}[B_1^2]+\tfrac12(\mathbb{E}[(1+B_{3/2})^2]+\tfrac12)=\tfrac12+\tfrac12(1+\tfrac32+\tfrac12)=2$. The interval-by-interval second-moment calculation gives the same answer: the portion from $3/2$ to $2$ contributes $\tfrac12\mathbb{E}[B_1^2]=\tfrac12$, and the portion from $2$ to $5/2$ contributes $\tfrac12\mathbb{E}[(1+B_2)^2]=\tfrac12(1+2)=\tfrac32$. Their sum is $2$.

A Common Mistake: Using an End-of-Interval Coefficient

Do not infer that every coefficient-times-increment product has mean zero. For example, replacing the left-endpoint coefficient on $(0,1]$ by $B_1$ gives $B_1(B_1-B_0)=B_1^2$, whose expectation is $1$, not zero. The zero-mean calculation for a simple Ito integral requires the coefficient to be known at the interval's start, so that it is independent of the later Brownian increment. Here $B_1$ is not $\mathcal{F}_0$-measurable; this expression is not the simple Ito integral with an adapted left-endpoint coefficient. With the valid coefficient $H_0=1$, the integral is $B_1-B_0=B_1$, and its expectation is $0$.

For a multi-interval calculation, a reliable order is to write down each left-endpoint coefficient, attach it to the increment over that interval, and only then compute moments. Check square-integrability of each coefficient before using second moments. To verify a claimed martingale increment, condition on the information at the starting time and apply the zero-mean independent-increment property to the subsequent Brownian increments. These checks establish the result directly for the finite-sum construction; no limiting argument or local-martingale conclusion is needed.

Check Your Understanding

Use the finite-sum definition and condition on the filtration at the relevant interval starts.

For $H_0=3$ on $(0,1]$ and $H_1=B_1$ on $(1,2]$, write $M_2$ using the two Brownian increments and compute its mean and variance.
On $0<1<3$, take $H_0=1$ and $H_1=B_1$. Compute the terminal variance and identify the contribution from each interval.
For an adapted partition-step integrand, explain why two terms from distinct intervals have zero expected product.
In the third worked example, explain why the coefficient $1+B_2$ is allowed on $(2,3]$ but not on $(1,2]$.
State the integrability and measurability conditions on the coefficients that ensure the finite sum defines a square-integrable Ito integral.

A Worked Calculation With a Multi-Interval Integrand

What You'll Learn