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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.

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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.

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