$\bf 3.9.$ Suppose $f\in\mathcal V(0,T)$ and that $t\to f(t,\omega)$ is continuous for a.a. $\omega$. Then we have shown that $$\int\limits_0^T f(t,\omega)dB_t(\omega)=\lim_{\Delta t_j\to0}\sum_jf(t_j,\omega)\Delta B_j\qquad\text{ in }\, L^2(P).$$ Similarly we define the Stratonovich integral of $f$ by $$\int\limits_0^Tf(t,\omega)\circ dB_t(\omega)=\lim_{\Delta t_j\to0}\sum_j f(t_j^*,\omega)\Delta B_j,\quad\text{ where }\; t_j^*=\tfrac12(t_j+t_{j+1}),$$ whenever the limit exists in $L^2(P)$. In general these integrals are different. For example, compute $$\int\limits_0^T B_t\circ dB_t$$ and compare with example $3.1.9.$
I am struggling to evaluate the integral $\displaystyle \int^{T}_{0} B_t \circ dB_t $ from definition.
So far I have that
$\begin{align} \displaystyle \sum B_{\frac{t_j+t_{j+1}}{2}}(B_{t_{j+1}}-B_{t_{j}}) &= \\ \sum [ B_{\frac{t_j+t_{j+1}}{2}}(B_{t_{j+1}}-B_{\frac{t_j+t_{j+1}}{2}})+B_{t_{j}}(B_{\frac{t_j+t_{j+1}}{2}}-B_{t_j})]+ \sum (B_{\frac{t_j+t_{j+1}}{2}}-B_{t_j})^2 \end{align}$
I can see that the first term goes to $\int^{T}_{0} B_t dB_t$ in $L^2(P)$. Not sure where to go next nor with the general direction of where this solution is going.
Best Answer
The key point is to write $B_T^2$ in a clever way:
$$\begin{align*} B_T^2 &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2) \\ &= \sum_{j=1}^n \big( (B_{t_j}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 - \big( (B_{t_{j-1}}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 \\ &= 2 \sum_{j=1}^n B_{t_j^{\ast}} (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2- \sum_{j=1}^n (B_{t_{j-1}}-B_{t_j^{\ast}})^2 \tag{1} \end{align*}$$
where $t_j^{\ast}:= \frac{1}{2}(t_j+t_{j-1})$. Now, by definition, the first term on the right-hand side converges to the Stratonovich integral
$$\int_0^T B_t \circ dB_t.$$
Consequently, it remains to show that the remaining terms converge to $0$ as the mesh size $|\Pi|$ converges to $0$. To this end, we set
$$S_n := \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2$$
and note that by the stationarity of the increments
$$\mathbb{E}(S_n) = \mathbb{E} \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 \right) = \sum_{j=1}^n (t_j-t_j^{\ast}) = \frac{T}{2}.$$
Consequently, by the independence of the increments,
$$\begin{align*} \mathbb{E} \left[ \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 - \frac{T}{2} \right)^2 \right] &= \text{Var} \, (S_n) \\ &= \sum_{j=1}^n \text{Var} \, \left[ (B_{t_j}-B_{t_j^{\ast}})^2 - (t_j-t_j^{\ast}) \right]. \end{align*}$$
Using the scaling property, i.e. $B_t \sim \sqrt{t} B_1$ for any $t \geq 0$, it is therefore not difficult to see that this expression converges to $0$ as $|\Pi| \to 0$. In fact, we find that
$$\mathbb{E} \left[ \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 - \frac{T}{2} \right)^2 \right] \leq C|\Pi| T$$
for $C:= \mathbb{E}((B_1^2-1)^2)<\infty$. Hence,
$$\sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 \stackrel{L^2}{\to} \frac{T}{2}.$$
Exactly the same calculation goes through for the third addend in $(1)$. Finally, we conclude
$$B_T^2 = 2\int_0^T B_t \circ dB_t + \frac{T}{2} - \frac{T}{2} = 2\int_0^T B_t \circ dB_t.$$