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).
Best Answer
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, $$ \int_0^t f(B_k)\,d B_t=F(B_t)-\frac12\int_0^t f'(B_k)\,dt $$