# Find the constant $a$ such that $Y_t$ is a martingale.

Let $$X_t$$ be the solution of SDE $$\text{d}X_t=3X_t\text dt+2X_t\text dB_t$$ and $$X_0=1$$ which $$B_t$$ denotes the Brownian motion with $$B_0=0$$. Let $$Y_t=e^{at}X_t$$, Find the constant $$a$$ such that $$Y_t$$ is a martingale.

$$dY_t=ae^{at}X_t+e^{at}dX_t=ae^{at}X_t+e^{at}(3X_t\text dt+2X_t\text dB_t)=ae^{at}X_t+3e^{at}X_t\text dt+2e^{at}X_t\text dB_t$$

That's all I can get, I fail to apply this to prove $$E[Y_{t+1}|\mathcal{F}_s]=Y_t$$ Any hints would be helpful.

You've made a mistake when applying Ito's lemma. You should have $$dY_t = ae^{at}X_t dt + e^{at} dX_t = (3+a)e^{at}X_t dt + 2e^{at} X_t dB_t = (3+a) Y_t dt + 2 Y_t dB_t$$ (notice the presence of $$dt$$!). For this to be a martingale, it must at least have no $$dt$$ component and hence (since the initial condition rules out $$Y = 0$$) we have $$a=-3$$. In this case, $$Y$$ is (up to a constant) the stochastic exponential of Brownian motion and hence is a martingale.