Let $B_t$ be a brownian motion adapted to $\mathcal F_t$. For general $\mathcal F_t$-adapted processes $X_t$ the Ito-integral could be defined as $$ \int_0^t X_s dB_s = \lim_{n\to \infty} \int_0^t X_s^{(n)} dB_s $$ where the $X_t^{(n)}$ are simple processes approximating $X_t$ such that for each $0 = t_0 < t_1 < \ldots < t_k = t$ as the constant value on the interval $[t_i, t_{i+1})$ the left end point is choosen, i.e. $X_t^{(n)}(t) = X_{t_i}$ for $t_i \le t < t_{i+1}$ and such that $$ \lim_{n\to \infty} E\int_0^T | X_t^{(n)} - X_t |^2 dt = 0. $$ Now I am looking for examples where this definition is used to compute the integral (and not Ito's Lemma). The only one I find in the literature are $$ \int_0^t B_s dB_s \quad \mbox{ and } \quad \int_0^t B_s^s dB_s $$ For example, for the first we can choose $$ \int_0^1 B_s dB_s = \lim_n \sum_{i=0}^{n-1} B_{i/n} (B_{(i+1)/n} - B_{i/n}) $$ and by $\sum_{i=0}^{n-1} B_{i/n} (B_{(i+1)/n} - B_{i/n}) = \frac{1}{2} B_1^2 - \frac{1}{2} \sum_{i=0}^{n-1} (B_{(i+1)/n} - B_{i/n})^2$ we have $$ \int_0^t B_s dB_s. = \frac{1}{2} B_1^2 - \frac{1}{2} $$ where the last term approaches $1$ as this is the quadratic variation of the Brownian motion. Another computation using the definition with the law of large numbers could be found on the german wikipedia.

Now I am looking for other examples where the definition is used, do you know any? For example for $\exp(B_t)$, by using Ito's Lemme we find $$ \int_0^t \exp(B_s) dB_s = \exp(B_t) - \frac{1}{2}\int_0^t B_s ds $$ but I do not know how to derive this result using just the definition (i.e. finding an approximating series of simple processes).

  • $\begingroup$ This is not really an answer to your question but it is relevant. There is a general result that says the convergence of integrals occurs in probability and uniformly on compact intervals if you take any sequence of partitions such that the mesh of the partitions along this sequence goes to $0$. $\endgroup$
    – Calculon
    Jul 9, 2015 at 14:43
  • $\begingroup$ You can calculate the integral $\int f(B_s) \, dB_s$ this way for any function $f$ which is differentiable - that's actually one possibility to prove Itô's lemma. $\endgroup$
    – saz
    Jul 9, 2015 at 17:02
  • 1
    $\begingroup$ Did you really mean $\int_0^t B_s^s dB_s$ where the power of the Brownian integrand increases with $s$? $\endgroup$
    – user940
    Oct 8, 2015 at 21:05

1 Answer 1


Instead of using exp, I'll show how to derive $\int_0^t f(B_t)\,dB_t$, from the definition, assuming $\,f''$ exists and is continuous.

Let $F(x)=\int_0^xf(s)\,ds$. We of course start with a partition $0=t_0<t_1<\dots<t_m=t$, and setting $B_k=B_{t_k}$. Let $\mu=\sup_k t_{k+1}-t_k$ be the mesh of the partition.

Write $$ F(B_{k+1})=F(B_k)+f(B_k)(B_{k+1}-B_k)+\frac12f'(B_k)(B_{k+1}-B_k)^2+R_k(B_{k+1}) $$ where $R_k(x)$ is the remainder term for the Taylor series centered at $B_k$. Rearranging, and letting $\Delta B_k=B_{k+1}-B_k$, $$ f(B_k)\Delta B_k=F(B_{k+1})-F(B_{k})-\frac12 f'(B_k)(\Delta B_k)^2-R_k(B_{k+1}) $$ We then sum the above expression over $0\le k\le m-1$, and take the limit in probability as $\mu\to 0$ (meaning we are really using a sequence of partitions). The LHS of above becomes $\int f(B_t)dB_t$, by definition.

On the right, the $F(B_{k+1})-F(B_{k})$ part is telescoping, adding up to $F(t_n)-F(t_0)=F(t)$.

To evaluate the $f'(B_k)(\Delta B_k)^2$ part, we let $\Delta t_k=t_{k+1}-t_k$, and replace this with $$ f'(B_k)\Delta t_k + f'(B_k)((\Delta B_k)^2 - \Delta t_k) $$ Then $\sum_{k=0}^{m-1}f'(B_k)\Delta t_k \to \int_0^t f'(B_k)dt$, and $\sum_{k=0}^{m-1}f'(B_k)(\Delta B_k - \Delta t_k)\to 0$. You show the last limit approaches zero by computing its variance: $$ Var\left(\sum_{k=0}^{m-1}f'(B_k)((\Delta B_k)^2 - \Delta t_k)\right)=\sum_{k=0}^{m-1}Var(f'(B_{k}))(\Delta t_k)^2 Var\left(\frac{\Delta (B_k)^2}{\Delta t_k}-1\right)$$ $$\qquad \qquad \qquad\le Ct\cdot \mu\cdot \kappa\cdot \sum \Delta t_k=Ct\mu \kappa t\stackrel{\mu\to 0}\to 0$$ where $C=\sup_{s\in [0,t]} f'(s)$, so that $Var(f'(B_k))\le Var(CB_k)\le C^2 t_k\le C^2t$, and $\kappa$ is the variance of a $\chi^2$ random variable (note that $\frac{\Delta (B_k)^2}{\Delta t_k}$ are i.i.d. $\chi^2$). Since the variance $\to0$, and its mean is zero, the LHS $\to 0$ in $L_2$.

Finally, letting $K=\sup_{0\le s\le t}f''(t)$, the remainder term $R(B_{k+1})$ is uniformly bounded by $K(\Delta B_k)^3$, and a similar variance calculation shows that $\sum_{k=0}^{m-1} R_k(B_{k+1})\to 0$.

Putting this all together, $$ \int_0^t f(B_k)\,d B_t=F(B_t)-\frac12\int_0^t f'(B_k)\,dt $$

  • $\begingroup$ Actually, there are some kinks in this argument, in particular when it comes to bounding $Var(f'(B_k))$. $\endgroup$ Oct 8, 2015 at 21:32

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