Switching order of stochastic integration

Question

I am looking at the following relation and I am having a hard time convincing myself that this is indeed true:

Standing at time $t$, for $t < T_{i-1} < T_i $ we observe that $$ \int _{T_{i-1}}^{T_i} \left(\int_t^s dW_u \right) ds = \int_t^{T_{i-1}} \left( \int_{T_{i-1}}^{T_i} ds \right) dW_u + \int_{T_{i-1}}^{T_i} \left( \int_u^{T_i} ds \right) dW_u $$

Can the relation be shown using additional steps or does it require a particular theorem?

thanks

Answer 1

Hints:

1. Recall the integration by parts formula for Brownian motion: $$\int_s^t f(u) \, dW_u = \bigg[f(u) W_u \bigg]_{u=s}^t - \int_s^t W_s f'(s) \, ds, \tag{1}$$ here $f$ is a nice (deterministic) differentiable function.
2. Use $(1)$ to show that $$\int_{T_{i-1}}^{T_i} u \, dW_u = (T_i W_{T_i}-W_{T_{i-1}} T_{i-1}) - \int_{T_{i-1}}^{T_i} W_u \, du. \tag{2}$$
3. Using elementary calculations show that the left-hand side of the identity which you want to prove equals $$\int_{T_{i-1}}^{T_i} W_s \, ds - W_t (T_{i-1}-T_{i-1}) \tag{3}$$ and the right-hand side $$\begin{align*} &(T_i-T_{i-1})(W_{T_{i-1}}-W_t) + T_i (W_{T_i}-W_{T_{i-1}}) - \int_{T_{i-1}}^{T_i} u \, dW_u \\ &= - (T_i-T_{i-1}) W_t+ (T_i W_{T_i}-T_{i-1} W_{T_{i-1}})- \int_{T_{i-1}}^{T_i} u \, dW_u. \tag{4} \end{align*}$$
4. Use $(2)$ to conclude that $(3)=(4)$.