The Ito Integral with Respect to Brownian Motion — Examples

Substituting the deterministic coefficients into the definition gives the displayed finite sum for every $t$, and at $t=T$ it is ${\sum }_k{a}_k(B_{t_{k+1}}-B_{t_k})$, a linear combination of the independent increments over the partition intervals.

The squared $L^2[0,T]$ norm of $h$ is ${\int }_0^T{h}^{2}ds={\sum }_k{a}_k^2(t_{k+1}-t_k)$, so by [F2] the law of the terminal integral is $N(0,{\sum }_k{a}_k^2(t_{k+1}-t_k))$, with mean $0$ and that variance.

The cases are covered: a single-interval step ($m=1$, $a_0=a$) gives $a(B_T-B_0)=a{B}_T$ of law $N(0,a^{2}T)$; the degenerate case $a_k=0$ for some $k$ contributes zero variance on that block; and $h=0$ gives the Dirac law $N(0,0)$ at $0$. AC enters only through [F3].

The predictable finite-energy process $H\equiv 1$ is represented in $L^2(\mathrm{d}t\otimes P)$ on each finite horizon by the elementary process $1_{(0,T]}$ of [F1]. Hence its integral is the elementary sum $B_t-B_0$, and the stopped quantity is $(1\cdot B{)}_{t\wedge \tau }=B_{t\wedge \tau }-B_0$.

By [F2] the stopping identity applies with $H\equiv 1$: $(1\cdot B{)}_{t\wedge \tau }={\int }_0^t{1}_{[0,\tau ]}(s)d{B}_s$ up to indistinguishability, and substituting step 1.1 for the left-hand side gives the displayed identity; both sides are continuous in $t$ because $B$ is and $t\mapsto t\wedge \tau$ is.

The cases $\tau \equiv \mathrm{\infty }$ (integral equals $B_t-B_0$), $\tau \equiv t_0$ (integral equals $B_{t\wedge t_0}-B_0$) and $\tau \equiv 0$ (integral vanishes) are all instances; the identity is a statement about the localized integral of a bounded integrand, so no integrability of $\tau$ is required. AC enters only through [F3].

Since both integrals have mean $0$ by [F1], the covariance is the expectation of the product, and [F2] evaluates it as ${\int }_0^{T}hkds$.

By [F3] the pair is jointly Gaussian with covariance matrix $\mathrm{\Sigma }$ whose off-diagonal entry is ${\int }_0^{T}hkds$; if that entry vanishes, $\mathrm{\Sigma }$ is diagonal, and the multivariate normal law with diagonal covariance is the law of a pair with independent coordinates (realization $m+{\mathrm{\Sigma }}^{1\mathrm{/}2}Z$ with independent standard normals), so the pair is independent.

Disjoint supports give $h(s)k(s)=0$ for every $s$, hence ${\int }_0^{T}hkds=0$ and independence; the degenerate cases $h=0$ or $k=0$ are included (a Dirac factor is independent of every variable), and AC enters only through [F4].

For each $n\ge 1$, set $L_0^n=B_0$ and $L_s^n=B_{k{2}^{-n}}$ on $(k{2}^{-n},(k+1)2^{-n}]$, $k=0,1,\dots$. Each $L^n$ is predictable by the countable interval generators in [F1]; the coefficients need not be bounded to give measurability. Define ${\beta }_s={\lim }_n{L}_s^n$ wherever this limit exists as a finite real number, and zero elsewhere. The convergence set is measurable by the countable Cauchy criterion, hence $\beta$ is predictable by [F1]. On the single full event of continuous Brownian paths the left grid points tend to $s$ for every $s>0$, so ${\beta }_s=B_s$ simultaneously for all $s\ge 0$. No membership of that full event in ${\mathcal{F}}_0$ is needed, and the limiting map is never defined by multiplying B by that event.

In particular for every fixed $s$, $E{\beta }_s^2=E{B}_s^2=s$. Tonelli in [F4] applies to the measurable nonnegative map ${\beta }^2$ and gives $E{\int }_0^t{\beta }_s^{2}ds=t^2\mathrm{/}2$. For the dyadic grid $t_k=kt\mathrm{/}{2}^n$ put $H_s^n={\sum }_{k<2^n}{B}_{t_k}{1}_{(t_k,t_{k+1}]}(s)$. This is predictable, and its finite energy follows from $E{B}_{t_k}^2=t_k$. By deterministic-time equality of ${\beta }_s$ and $B_s$, [F2] and Tonelli give $E{\int }_0^t\mathrm{\mid }{H}_s^n-{\beta }_s{\mathrm{\mid }}^{2}ds={\sum }_k{\int }_{t_k}^{t_{k+1}}(s-t_k)ds=\frac{t^2}{2^{n+1}}\mathrm{.}$ Thus the integrals of $H^n$ converge in $L^2(P)$ to the integral of $\beta$ by [F3].

Fix $n$ and truncate the coefficient $B_{t_k}$ to $c_r(B_{t_k})$, where $c_r(x)=\max (-r,\min (x,r))$, to obtain bounded elementary $H^{n,r}$. Dominated convergence applies to each coefficient error squared, bounded by $B_{t_k}^2$ and tending to zero. Hence $H^{n,r}\to H^n$ in predictable $L^2$. For each increment ${\mathrm{\Delta }}_{k}B=B_{t_{k+1}}-B_{t_k}$, independence in [F2] gives $E\mathrm{\mid }(c_r(B_{t_k})-B_{t_k}){\mathrm{\Delta }}_{k}B{\mathrm{\mid }}^2=(t_{k+1}-t_k)E\mathrm{\mid }{c}_r(B_{t_k})-B_{t_k}{\mathrm{\mid }}^2⟶0.$ The finite sum therefore converges in $L^2$ by the triangle inequality. Comparing this with the isometric convergence of the elementary integrals proves ${\int }_0^t{H}_s^{n}d{B}_s=S_n:={\sum }_k{B}_{t_k}{\mathrm{\Delta }}_{k}B$ in $L^2(P)$. This explicitly licenses unbounded step coefficients without calling them elementary.

Finite telescoping gives $2S_n=B_t^2-B_0^2-{\sum }_k({\mathrm{\Delta }}_{k}B{)}^2$. Since $B_0=0$ almost surely, [F5] implies $S_n\to Y_t=(B_t^2-t)\mathrm{/}2$ almost surely. Write $I_t$ for the integral class of $\beta$. Steps 2.1 and 3.1 give $E\mathrm{\mid }{I}_t-S_n{\mathrm{\mid }}^2\to 0$, whereas Fatou in [F4] gives $E\mathrm{\mid }{I}_t-Y_t{\mathrm{\mid }}^2\le {\mathrm{lim\hspace{0.17em}inf}}_{n}E\mathrm{\mid }{I}_t-S_n{\mathrm{\mid }}^2=0$. Thus $I_t=Y_t$ almost surely.

Choose the continuous adapted integral version supplied by [F3]. Intersect its continuity event, the common Brownian continuity and zero-start event, and the equality events of step 4.1 for all positive rational t. This is a measurable full event; continuity of both sides extends the equality from rational to all nonnegative real times. At time zero the integral is zero and $B_0=0$ on this event. No claim is made that the identity holds on every exceptional constant Brownian path, or that the entire all-time equality set must itself be measurable in an incomplete space.

By [F3] the mean is zero. By [F2], $E{Y}_t^2=\frac{1}{4}(E{B}_t^4-2tE{B}_t^2+t^2)=t^2\mathrm{/}2,$ in agreement with the isometry and step 2.1. For $t>0$ this variance is positive, while $Y_t\ge -t\mathrm{/}2$. Every centered normal with positive variance gives positive probability to an interval below $-t\mathrm{/}2$, because its density there is positive; a zero-variance normal has zero variance. Thus [F6] rules out a deterministic-integrand law for $t>0$. At $t=0$ both sides vanish and there is no such non-Gaussian claim. Finite grids include both endpoints, and n can start at 1 without changing any limit. Full AC covers [F6]; the predictable representative and grids are explicit and no additional choice of paths is made. No later Ito formula is used.

The integrand $H_s=s$ is deterministic and continuous, hence predictable with finite energy $A_t=t^3\mathrm{/}3$ at every $t$; under the given usual conditions its localized integral $M=H\cdot B$ is defined and is a continuous square-integrable martingale.

Applying [F2] to $H_s=s$ and to the given partition sequence gives $[M{]}_t={\int }_0^t{s}^{2}ds=t^3\mathrm{/}3$ for every $t\in [0,T]$, with the convergence uniform in probability; the value does not depend on the chosen deterministic partition sequence because the theorem holds for every such sequence.

Sanity cases: at $t=0$ the value is $0$; the example's integrand grows with time, so the accumulated quadratic variation $t^3\mathrm{/}3$ is not linear, in contrast with the Brownian case $[B{]}_t=t$; and a constant integrand $H\equiv c$ would give $[M{]}_t=c^{2}t$, of which this is the $c=s$ analogue. AC enters only through [F3].

On the event $A$ one has $B_T^2>T$ by definition of $A$, while off $A$ the integrand $H=1_A$ vanishes; hence $H^2{B}_T^2=1_A{B}_T^2\ge 1_{A}T$, and the inequality is strict on the nonempty event $A$ (where $B_T^2>T$ strictly), so $E[H^2{B}_T^2]>TP(A)$.

The integrand $H=1_A$ is not admissible: $A\in \sigma (B_T)\subseteq {\mathcal{F}}_T$ and $P(A)\in (0,1)$ by [F1]; an event in ${\mathcal{F}}_0$ differs from a germ event in ${\mathcal{F}}_{0+}^0$ only by a null set, so by [F2] every set in ${\mathcal{F}}_0$ has probability $0$ or $1$; hence $A\mathrm{\notin }{\mathcal{F}}_0$, so a block coefficient equal to $1_A$ on $(0,T]$ violates the left-endpoint measurability requirement at time $0$.

Since $H^2=1_A$, one has $E[H^2]=P(A)$ and ${\int }_0^T{H}^{2}ds=H^{2}T=1_{A}T$, so $TE[H^2]=TP(A)<E[H^2{B}_T^2]$ by step 1.1; this is the asserted violation of the isometry-type equality, and $P(A)\in (0,1)$ by [F1], so the strict inequality is between finite positive numbers.

The witness therefore separates the two hypotheses: with the left-endpoint measurability clause enforced, the elementary isometry of item 7 computes $E[H^2{B}_T^2]=TE[H^2]$ for the admissible case; dropping that clause and using the future information contained in $B_T$ admits the strict violation above. Note that the symmetry of $B_T$ does not rescue the mean: it is the second moment, not the first, that the isometry governs, and the failure exhibited is a second-moment failure. AC enters only through [F3].

By [F1] there is an event of probability one on which the Brownian path is of unbounded variation on every nondegenerate compact interval; on that event the hypothesis "$g$ has bounded variation" of [F2] fails for $g=B$ and every interval $[0,t]$ with $t>0$, so the Riemann--Stieltjes existence theorem for a general continuous integrand is not available pathwise.

For the specific integrand $f(x)=x$ and the dyadic partitions, the two evaluation rules give the sums $L_n:={\sum }_k{B}_{t_k}{\mathrm{\Delta }}_k$ and $R_n:={\sum }_k{B}_{t_{k+1}}{\mathrm{\Delta }}_k$, whose telescoping identities [F3] express them as $L_n=\frac{1}{2}(B_t^2-{\sum }_k{\mathrm{\Delta }}_k^2)$ and $R_n=\frac{1}{2}(B_t^2+{\sum }_k{\mathrm{\Delta }}_k^2)$.

By the quadratic-variation limit of [F3], $L_n\to \frac{1}{2}(B_t^2-t)$ and $R_n\to \frac{1}{2}(B_t^2+t)$ almost surely; the two limits differ by $t>0$, so the Riemann--Stieltjes sums of $fdg$ with $g=B$ have no common limit along dyadic partitions and the pathwise Riemann--Stieltjes integral ${\int }_0^{t}BdB$ does not exist in that sense, even though the Ito integral does.

Consequently the bounded-variation construction cannot serve as the definition of the Brownian Ito integral: its central hypothesis fails almost surely on every nondegenerate interval [F1], and its conclusion fails explicitly for the witness $f(x)=x$, $g=B$ by the limit mismatch of step 2.1. The Ito integral of the same integrand exists and equals $\frac{1}{2}(B_t^2-t)$ by the companion example page, so the failure is a failure of the pathwise construction, not of the stochastic integral. AC enters only through [F4].

The two raw processes differ exactly on the time section $\{1\mathrm{/}2\}$: for $t=1\mathrm{/}2$ one has $H=1\mathrm{\ne }0=K$ at every $\omega$, while for $t\mathrm{\ne }1\mathrm{/}2$ both vanish; by [F2] the exceptional set has product measure zero, so the processes are equal in the $L^2(\mathrm{d}t\otimes P)$ sense while failing pointwise equality for every $\omega$.

Both processes are predictable by [F1] and have finite energy: $H^2=H$ integrates to $0$ and $K^2=0$, so both integrals over $[0,T]$ are defined as classes.

By [F3] the integrals agree, ${\int }_0^{T}HdB={\int }_0^{T}KdB=0$ almost surely, because the integrands differ on a product-null set; the equality of integrals is therefore not evidence of pointwise equality of the integrands.

The example isolates the convention in force throughout this development: representatives of predictable $L^2$ classes are interchangeable, deterministic singleton sections are invisible to the product measure, and every statement about an Ito integral is a statement about an almost-everywhere class. The degenerate variant $H=1_{\{0\}}$ agrees with $K=0$ even as a raw process on $(0,T]$, since the elementary convention makes the value at time zero irrelevant; the singleton $\{1\mathrm{/}2\}$ exhibits the genuine pointwise failure. AC enters only through [F4].