# Ito Representation for Integral of Brownian Motion

By Ito's representation theorem because the integral of Brownian Motion is mean zero we should have some random function $\phi$ such that

$$\int_0^T W_t dt = \int_0^T\phi(\omega,t)dW_t$$

I'm a little uncomfortable about the left hand side being finite variation and the right a local martingale even for fixed $T$, I just can't get any closer to the form above than the obvious Ito formula result

$$\int_0^T W_t dt = TW_T-\int_0^TtdW_t$$

Thanks

The identity

$$\int_0^T W_t \, dt = T W_T - \int_0^T t \, dW_t$$

is a good start. Now note that

$$T W_T = \int_0^T T \, dW_t$$

which implies that

$$\int_0^T W_t \, dt = \int_0^T (T-t) \, dW_t.$$

This is exactly the representation you are looking for.

• Haha wow how do I even get stuck with these easy problems, thanks – John Fernley Oct 24 '16 at 13:57
• @JohnFernley You are welcome. – saz Oct 24 '16 at 13:57
• Actually also wondering in $\int_0^T W_t \, dt = \int_0^T (T-t) \, dW_t$, am I right that here the right hand side is a local martingale in $T$ and the left hand side is not? – John Fernley Oct 24 '16 at 14:00
• @JohnFernley Why do you think so? Since $\int_0^T W_t \, dt = \int_0^T (T-t) \, dW_t$ the left-hand side is a (local) martingale if and only if the right-hand side is a (local) martingale. The stochastic integral $\int_0^t f(s) \, dW_s$ is a (local) martingale, but in our setting we have a (more general) expression of the form $\int_0^t f(s,\color{red}{t}) \, dW_t$. – saz Oct 24 '16 at 14:18
• @JohnFernley No, it's not a martingale. – saz Oct 24 '16 at 14:43