We know the following is a martingale, and is commonly used to represent Ito's Integral. If $W$ is a brownian motion.
$ \int_0^T W(u) dW(u) = \frac{1}{2} W^2 (t) - \frac{1}{2}t, \ for \ \ T \ge t \ge 0 $
how ever if we drop the $ - \frac{1}{2}t $ term it is not a martingale. Is there a proof showing when we drop the $ - \frac{1}{2}t$ term why the above is not a martingale?
Every martingale has constant expectation, but only one of the processes $(\frac12 W^2(t)-\frac12t)_t$ and $(\frac12 W^2(t))_t$ can have this property.