Martingale Representation Theorem Applied to Integral of Squared Brownian Motion wrt time

I am trying to find the martingale representation of $$F = \int_0^TW_t^2dt$$.

I wrote that $$\int_0^TW_t^2dt = TW_T^2 - \frac{T^2}{2} - \int_0^T 2tW_t dW_t$$ and I'm stuck.

Any ideas on how to proceed?

• There's a typo: $2$ should be in front of the integral. Dec 30, 2020 at 0:33

As $$\frac12 W_{T}^2 - \frac{T}2= \int_{0}^T W_t dW_t$$ we may proceed:
$$TW_T^2 - \frac{T^2}{2} - \int_0^T tW_t dW_t =2T(\frac12 W_{T}^2 - \frac{T}2) + \frac{T^2}2 - \int_0^T tW_t dW_t =$$ $$= \frac{T^2}2 + 2T \int_{0}^T W_t dW_t - \int_0^T tW_t dW_t=$$ $$=\frac{T^2}2 +\int_0^T (2TW_t - tW_t) dW_t .$$
Indeed, by Ito formula $$dW_t^2 = 2W_t d W_t + dt$$ and thus
$$\int_0^TW_t^2dt = TW_T^2 - 0W_0^2 - \int_0^T t dW_t^2 = TW_T^2 - 0W_0^2 - \int_0^T t (2W_t d W_t + dt) =$$ $$= TW_T^2 - \frac{T^2}{2} - 2\int_0^T tW_t dW_t$$
Thus $$\int_0^TW_t^2dt = TW_T^2 - \frac{T^2}{2} - 2\int_0^T tW_t dW_t = \frac{T^2}2 + 2\int_0^T (T-t)W_tdW_t .$$