Brownian Increments and Left-Endpoint Sums

Why the Left Endpoint Matters

A stochastic integral is built from many small products: an integrand value multiplied by a Brownian increment. The subtlety is when the integrand value is chosen. If it is determined by information available at the start of the interval, it cannot use the Brownian movement that is about to occur. This nonanticipating convention is the defining feature of the Itô construction.

In a pharmacokinetic interpretation, an integrand might depend on a concentration state that fluctuates over time. At time $t_i$, the current state can be used to weight the next Brownian increment; the future state at $t_{i+1}$ cannot be used as if it were already known at $t_i$. The distinction is not cosmetic: different choices of endpoint can produce different limiting sums.

Brownian Increments and Available Information

Let $W_t$, $0\le t\le T$, be standard Brownian motion on a probability space with a filtration $(\mathcal F_t)$. A filtration is an increasing family of sigma-algebras describing information available over time. Here, assume $W$ is Brownian motion with respect to this filtration: $W_t$ is $\mathcal F_t$-measurable, and for $s<t$, $W_t-W_s$ is independent of $\mathcal F_s$ and has distribution $N(0,t-s)$. In particular, \[ E[W_t-W_s\mid\mathcal F_s]=0,\qquad E[(W_t-W_s)^2\mid\mathcal F_s]=t-s. \] These conditional statements are exactly what the left-endpoint construction uses.

Definition: A process $H_t$ is adapted to $(\mathcal F_t)$ if $H_t$ is $\mathcal F_t$-measurable for each $t$. For a partition $0=t_0<t_1<\cdots<t_n=T$, an elementary left-endpoint process has the form $H_t=H_i$ on $[t_i,t_{i+1})$, where each $H_i$ is $\mathcal F_{t_i}$-measurable and $E[H_i^2]<\infty$. Its left-endpoint Brownian sum is $$ \sum_{i=0}^{n-1}H_i\bigl(W_{t_{i+1}}-W_{t_i}\bigr). $$

The coefficient $H_i$ may be random; the requirement is that it be determined by information at $t_i$. Thus a coefficient can depend on earlier Brownian increments. It cannot depend on the increment $W_{t_{i+1}}-W_{t_i}$ that it is multiplying, or on information arriving after $t_i$.

Worked Example: A Deterministic Coefficient on One Interval

On the interval $[0,0.25]$, take $H_t=2$ and suppose a particular realized Brownian increment is $W_{0.25}-W_0=0.3$. The left-endpoint sum has one term:

2(W_{0.25}-W_0)=2(0.3)=0.6.

The coefficient is known at time 0. More generally, this increment has distribution $N(0,0.25)$, so the sum $2(W_{0.25}-W_0)$ has distribution $N(0,1)$: multiplying by 2 multiplies the variance by $2^2$. The displayed value 0.6 is one possible realized sum, not its expectation.

Worked Example: A Coefficient Based on Past Noise

Divide $[0,2]$ into intervals $[0,1]$ and $[1,2]$. Set $H_0=1$, and set $H_1=1+0.5(W_1-W_0)$. The first coefficient is known at time 0; the second is known at time 1. Write $\Delta W_0=W_1-W_0$ and $\Delta W_1=W_2-W_1$. The sum is

S=1\Delta W_0+(1+0.5\Delta W_0)\Delta W_1.

For a realized pair $(\Delta W_0,\Delta W_1)=(0.2,-0.1)$, the first coefficient is 1 and the second is $1+0.5(0.2)=1.1$. Therefore

$$ S=1(0.2)+1.1(-0.1)=0.2-0.11=0.09. $$

The coefficient $1+0.5\Delta W_0$ uses the first increment to weight the second, independent increment. Conditional on $\mathcal F_1$, that coefficient is fixed and $E[\Delta W_1\mid\mathcal F_1]=0$. This is a valid nonanticipating construction.

Why Elementary Left Sums Behave Well

The Brownian property gives a useful identity for elementary left-endpoint sums. It says that their second moment is controlled exactly by the squared coefficients and interval lengths. This identity will justify convergence when finer partitions are used.

Theorem: Let $W$ be Brownian motion with respect to $(\mathcal F_t)$, and let $H_t=H_i$ on each $[t_i,t_{i+1})$, with $H_i$ square-integrable and $\mathcal F_{t_i}$-measurable. Then $$ E\left[\left(\sum_{i=0}^{n-1}H_i(W_{t_{i+1}}-W_{t_i})\right)^2\right] = \sum_{i=0}^{n-1}E[H_i^2](t_{i+1}-t_i). $$

Proof. Write $\Delta W_i=W_{t_{i+1}}-W_{t_i}$ and expand the square. For each diagonal term, independence of $\Delta W_i$ from $\mathcal F_{t_i}$, together with $E[(\Delta W_i)^2]=t_{i+1}-t_i$, gives

E[H_i^2(\Delta W_i)^2] =E[H_i^2]E[(\Delta W_i)^2] =E[H_i^2](t_{i+1}-t_i).

For a cross term with $i<j$, the random variable $H_i\Delta W_iH_j$ is $\mathcal F_{t_j}$-measurable: the earlier increment is already known by $t_j$, and $H_j$ is measurable at $t_j$. The later increment has conditional mean zero given that information. Hence

E[H_i\Delta W_iH_j\Delta W_j] =E[H_i\Delta W_iH_j E[\Delta W_j\mid\mathcal F_{t_j}]] =0.

All cross terms vanish, while the diagonal terms sum to the stated expression. Square-integrability ensures the expectations used in this expansion exist. $\square$

From Sums to the Itô Integral

For a continuous adapted process $H_t$, use a partition of $[0,T]$ and replace $H_t$ on each interval by its value at the left endpoint. The resulting sum is

S_n=\sum_{i=0}^{n-1}H_{t_i}(W_{t_{i+1}}-W_{t_i}).

Under suitable conditions, these sums converge as the largest interval length tends to zero. The next result gives one direct set of sufficient conditions and proves the convergence. Boundedness is a convenient assumption here; broader integrability conditions are treated separately.

Theorem: Suppose $H$ is adapted, has continuous paths almost surely on $[0,T]$, and there is a finite constant $K$ such that $|H_t|\le K$ for all $t\in[0,T]$, almost surely. For any sequence of deterministic partitions whose mesh tends to zero, the left-endpoint sums $S_n$ converge in $L^2$. Their limit defines the Itô integral $\int_0^T H_t\,dW_t$ for this class of processes.

Proof. For each partition, define $H^{(n)}_t=H_{t_i}$ on $[t_i,t_{i+1})$. For almost every sample path, continuity of $H$ on the compact interval implies uniform continuity. Since the partition mesh tends to zero, $H^{(n)}_t\to H_t$ for every $t<T$ on that path. The endpoint $T$ does not affect a time integral. Also, $|H^{(n)}_t-H_t|^2\le4K^2$. Dominated convergence on the product of time and probability therefore gives

E\left[\int_0^T|H^{(n)}_t-H_t|^2\,dt\right]\longrightarrow0.

For two partitions $n,m$, refine them together. The difference $H^{(n)}-H^{(m)}$ is an elementary adapted process on this common refinement. The identity just proved applies to its sum, so

E[|S_n-S_m|^2] =E\left[\int_0^T|H^{(n)}_t-H^{(m)}_t|^2\,dt\right].

The right-hand side tends to zero as $n,m\to\infty$, by the convergence of each approximation to $H$ in the displayed time-and-probability norm and the triangle inequality. Thus $(S_n)$ is Cauchy in $L^2$, which is complete, so it has an $L^2$ limit. This limit is the Itô integral for the stated class. $\square$

Worked Example: Left and Right Endpoints Give Different Limits

Take $H_t=W_t$ on $[0,1]$, and use a partition $0=t_0<\cdots<t_n=1$. Since $W_0=0$, the left sum is $S_n=\sum_i W_{t_i}\Delta W_i$. The identity

W_{t_{i+1}}^2-W_{t_i}^2 =2W_{t_i}\Delta W_i+(\Delta W_i)^2

holds for each interval. Summing and rearranging gives

S_n=\frac12\left(W_1^2-\sum_i(\Delta W_i)^2\right).

Brownian quadratic variation along partitions with mesh tending to zero satisfies $\sum_i(\Delta W_i)^2\to1$ in probability (and almost surely along standard nested partitions). Consequently, the left sums tend to $\frac12(W_1^2-1)$ in probability.

The right sum uses $W_{t_{i+1}}$ instead. Since $W_{t_{i+1}}=W_{t_i}+\Delta W_i$, it equals $S_n+\sum_i(\Delta W_i)^2$, and therefore tends to $\frac12(W_1^2+1)$, not the left-sum limit. For example, if a realized path has $W_1=0.4$ and its squared increments sum to approximately 1.05 on a fine partition, the left sum is approximately $(0.16-1.05)/2=-0.445$, while the right sum is approximately $(0.16+1.05)/2=0.605$. The difference is due to the accumulated squared increments, not rounding error.

Reading the Convention in a PK Setting

Suppose a concentration process $C_t$ is used as the integrand for a Brownian-driven term. On a grid, the corresponding sum is $\sum_i C_{t_i}\Delta W_i$. Each concentration value is the state available at the start of its interval; later concentration values are not substituted into that interval's coefficient.

Worked Example: A Concentration-Weighted Increment Sum

Suppose two successive half-time-unit intervals have starting concentrations $C_0=5.0$ and $C_{0.5}=4.2$ mg/L. For one realized path, let the Brownian increments be $\Delta W_0=0.3$ and $\Delta W_1=-0.2$. The left-endpoint sum is

C_0\Delta W_0+C_{0.5}\Delta W_1 =5.0(0.3)+4.2(-0.2) =1.50-0.84=0.66.

The value $4.2$ is used for the second increment because it is the concentration at the beginning of that interval. Replacing it with the concentration at time 1 would use information from the end of the interval and would define a different sum. The units of the full stochastic term also depend on the diffusion coefficient and the time units used for $W$; the arithmetic here isolates the endpoint convention.

Diagnostics, Alternatives, and Limits

When constructing or checking a stochastic sum, ask whether each coefficient is measurable with respect to the information available at its left endpoint. Also check that the filtration really makes $W$ Brownian: merely having Brownian marginal distributions is not enough to guarantee that future increments are independent of current information. The earlier material on Brownian motion distinguishes pathwise and distributional claims; here, the conditional independence relative to the filtration is essential.

A tempting alternative is a right-endpoint sum. It is a legitimate algebraic construction, but it is not the nonanticipating construction of the Itô integral. Midpoint sums are another possible convention, but they also do not define the Itô integral by the argument given here. Comparing or converting those conventions is a separate topic. The practical diagnostic is simple: if a purported left sum uses a coefficient that depends on the increment it multiplies, the nonanticipating condition has failed.

The convergence result above assumes bounded, continuous, adapted paths and deterministic partitions with vanishing mesh. It does not claim that every irregular process or every arbitrary sequence of sums converges. The elementary-sum identity and the convergence proof provide a controlled construction under stated hypotheses, not a license to interchange limits or use future observations. In applications, a numerical scheme that substitutes a future simulated state into an earlier coefficient should be identified as a different scheme rather than silently described as an Itô left sum.

Key takeaway: An Itô left-endpoint sum weights each Brownian increment with a coefficient already measurable at the beginning of that interval. This nonanticipating choice makes the cross terms vanish conditionally and provides the basis for the integral's construction.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

For $H_i$ measurable with respect to $\mathcal F_{t_i}$, what information is unavailable when $H_i$ is chosen?
On $[0,1]$, with one interval, $H_0=3$. If the realized increment is $-0.2$, calculate the left sum.
Why does the cross term involving $\Delta W_i$ and $\Delta W_j$, for $i<j$, have expectation zero in the elementary-sum identity?
For $H_t=W_t$, state the limits of the left and right sums over $[0,1]$.
A proposed coefficient on $[t_i,t_{i+1})$ is $H_i=W_{t_{i+1}}$. Is this an admissible left-endpoint coefficient? Explain.

Solutions. (1) The increment $W_{t_{i+1}}-W_{t_i}$ and all later information are unavailable at $t_i$. (2) The sum is $3(-0.2)=-0.6$. (3) The earlier coefficient and increment, together with $H_j$, are measurable at $t_j$, while $E[\Delta W_j\mid\mathcal F_{t_j}]=0$; conditioning therefore makes the cross-term expectation zero. (4) The left sums tend to $\frac12(W_1^2-1)$, while the right sums tend to $\frac12(W_1^2+1)$, using quadratic variation on $[0,1]$. (5) No. $W_{t_{i+1}}$ depends on the increment over the interval, so it is not generally measurable with respect to $\mathcal F_{t_i}$ and anticipates the movement being weighted.