How to prove this using Ito's lemma?

Question

I am stuck at the following formula which I need to prove using Ito's calculus. $$\frac{1}{T}\int_0^T W_t^2 dt=\frac{T}{2}+\int_0^T 2W_t \left(1-\frac{t}{T}\right) dW_t$$

How to prove this?

Answer 1

Using the hint provided by @Surb set $f(x,t)=x^2(1-\frac{t}{T})$. By Ito's lemma:

$$df(x,t)=2W_t(1-\frac{t}{T})dW_t+(1-\frac{t}{T})dt-W_t^2/Tdt$$

or

$$W_T^2\times 0-W_0=\int_0^{T}2W_t(1-\frac{t}{T})dW_t+\int_0^{T}(1-\frac{t}{T})dt-\int_0^{T}W_t^2/Tdt$$

or

$$0=\int_0^{T}2W_t(1-\frac{t}{T})dW_t+(T-0.5T)-\int_0^{T}W_t^2/Tdt$$

Hence

$$\frac{1}{T}\int_0^{T}W_t^2dt=\frac{T}{2}+\int_0^{T}2W_t(1-\frac{t}{T})dW_t$$