# Why does no drift factor in a stochastic differential equation imply that the process is a Martingale?

I was working on this problem:

Use Itô's formula to prove that the following stochastic processes are martingales:

$$$$\bullet \quad X_t = e^\frac{t}{2}\cos B_t \\ \bullet \quad X_t = e^\frac{t}{2}\sin B_t \\ \bullet \quad X_t = (B_t + t)e^{-B_t - \frac{t}{2}} \\$$$$

I used Ito's lemma to prove that the resulting stochastic differential equation is independent of any drift factor. How does it imply that $X_t$ is a martingale, mathematically. How can you prove it?

$$$$\bullet \quad dX_t = -e^\frac{t}{2}\sin B_t dB_t \\ \bullet \quad dX_t = e^\frac{t}{2}\cos B_t dB_t \\ \bullet \quad dX_t = e^{-B_t - \frac{t}{2}}(1 - B_t -t)dB_t \\$$$$

These are the resulting differential equations, all independent of drift. How can you prove, in general, that an SDE independent of drift, is a Martingale?

Because $dB_t$ is independent of $B_t$ and thus $$\Bbb E_s(f(t,B_t)dB_t)=\Bbb E_s(f(t,B_t))\Bbb E_s(dB_t)=0$$ for $s\le t$ which implies $$\Bbb E_s(X_t-X_0)=\Bbb E_s\left(\int_0^t f(u,B_u)dB_u\right)=\int_0^s f(u,B_u)dB_u=X_s-X_0$$ which is the main property for a matringale.