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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

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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.

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    $\begingroup$ Haha wow how do I even get stuck with these easy problems, thanks $\endgroup$ – John Fernley Oct 24 '16 at 13:57
  • $\begingroup$ @JohnFernley You are welcome. $\endgroup$ – saz Oct 24 '16 at 13:57
  • $\begingroup$ 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? $\endgroup$ – John Fernley Oct 24 '16 at 14:00
  • $\begingroup$ @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 $\int_0^t f(s,\color{red}{t}) \, dW_t$. $\endgroup$ – saz Oct 24 '16 at 14:18
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    $\begingroup$ @JohnFernley No, it's not a martingale. $\endgroup$ – saz Oct 24 '16 at 14:43

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