I have $B_s = $ brownian motion at time $s$.
$$ \int_0 ^t B_s \, dB_s$$ $$0 \leq t \leq T$$
And want to check if it is a martingale, first from its closed form expression, and then via conditions on the Ito integral.
From exercises immediately before this one, it is known to me that the closed form expression is $$ \int _0 ^t B_s \, dB_s = \frac{1}{2}(B_t^2 - t) $$
But - is there a way to actually derive the closed form expression, without prior knowledge, and working from the integral itself?
By ito: $$d(f(B_t)) = f'(B_t)dB_t + \cfrac{1}{2}f''(B_t)dt$$
equate:
$$f'(B_t)dB_t = B_t dB_t$$
$$f'(x) = x, f''(x) = 1$$
$$f(x) = \frac{x^2}{2} + c \implies f(B_t) = \frac{B_t ^2}{2}$$
so we have $$d (f(B_t)) = B_t dB_t + \frac{1}{2} dt$$
$$\int_0^T d (f(B_t)) = \int_0 ^T B_t dB_t + \frac{1}{2} \int_0 ^T dt$$
$$\int_0 ^T d(\frac{B_t ^2}{2}) = \int_0 ^T B_t dB_t + \frac{T}{2} $$
$$ \frac{B_T^2}{2} = \int_0 ^T B_t dB_t + \frac{T}{2}$$
$$ \int_0 ^T B_t dB_t = \frac{B_T^2}{2} - \frac{T}{2}$$
ofcourse, it could be any constant, but this wouldnt change whether or not it is a martingale.