# Deriving a closed form expression for stochastic integral (to show its a martingale)

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?

• what is this notation and what are you trying to say? – elbarto Jun 4 '15 at 20:18
• "is there a way to actually derive the closed form expression, without prior knowledge, and working from the integral itself?" Well yes, the $dB$-term in Itô for $G(B)$ is $G'(B)dB$ since $d(G(B))=G'(B)dB+\frac12G''(B)dt$, and you want this to be $BdB$, hence $G'(B)=B$, etc. – Did Jun 4 '15 at 20:28
• If my understanding is correct, then we proceed: $$f'(B_t)dB_t = B_t dB_t$$ $$f'(x) = x, f''(x) = 1$$ $$f(x) = \frac{x^2}{2} + c \implies f(B_t) = \frac{B_t ^2}{2}$$ 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. – elbarto Jun 4 '15 at 21:00

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.