Proof $Z_t = e^{-(a+\frac{b^2}{2})t +X_t}$ is a martingale I want to prove that $E[Z_t|Z_s]=Z_s, 0<s<t$ and $Z_t, X_t$ defined as
$$
Z_t = e^{-(a+\frac{b^2}{2})t + X_t}\\
X_t = at + bB_t
$$
where $B_t \sim N(0, t)$ is Brownian motion, $a=\mu(x), b = \sigma(x)$, i.e. drift and volatility are constant.
First, I rewrote
\begin{align}
Z_t &= Z_s + (Z_t - Z_s) \Rightarrow\\
E[Z_t|Z_s] &= E[Z_s + (Z_t - Z_s)|Z_s] = E[Z_s|Z_s] + E[(Z_t-Z_s)|Z_s] \\
&= Z_s + E[(Z_t-Z_s)|Z_s]
\end{align}
The first term is $Z_s$ because conditional expectation of a random variable on itself is equal to the random variable. For the second term, I noticed that without the condition on $Z_s$, ( I set $e^{-(a+\frac{b^2}{2})t} = h(a,b,t)$)
$$
\frac{Z_t - Z_s}{h(a,b,s)} = h(a,b,t-s)e^{X_{t}} - e^{X_s}
$$
Without the $h(\cdot)$ term, I would expand the functions of $X_t$ using Taylor series up to the first term:
$$
e^{X_t} - e^{X_s} \approx (X_{t}-X_s)e^{X_s}
$$
so that $X_t -X_s \sim N(a(t-s), b^2(t-s))$ and $e^{X_{s}} \sim LogNormal(as, b^2)$, so if they're independent, then
$$
E[e^{X_t} - e^{X_s}] \approx a(t-s) \cdot as = a^2s(t-s)
$$
which is not equal to $0$, so $Z_t$ is not a martingale. But, as I said earlier, I'm not sure how to account for conditioning on $Z_s$ and how to add the $h(\cdot)$ term.
 A: I think your approach of setting $Z_t = Z_s + (Z_t - Z_s)$ may work for the Brownian motion, but fails in this case due to the multiplicative nature of the exponential, which leads to complications if you try to linearize.
Instead, go for the multiplicative kind of equality you would like : $Z_t = \frac{Z_t}{Z_s} Z_s$ (where these quotients are well defined a.s. since $Z_t$ is positive almost surely).
Then $E[Z_t|Z_s] = Z_sE[\frac{Z_t}{Z_s} |Z_s]$.
Write the expression for $\frac{Z_t}{Z_s}$, and see that it somehow only depends upon $B_t - B_s$. Now, $Z_s$ belongs in the sigma algebra of $B_s$, as it is a Borel function of $B_s$. Hence, by the independent increment property of Brownian motion, this variable is independent of $Z_s$, and therefore the above conditional expectation, is just the usual expectation, which should be $1$ once you get everything together and evaluate. (has to do with moment generating function of the normal random variable)

EDIT : $$\frac{Z_s}{Z_t} = e^{X_t - X_s - (a+\frac{b^2}{2})(t-s)} = e^{b(B_t - B_s) + a(t-s) - (a-\frac{b^2}{2})(t-s)} =e^{b(B_t-B_s) - \frac{b^2}{2}(t-s)} $$,
which is independent of $Z_s$, hence the conditional expecation is just the expectation of this. Now, $(B_t - B_s) \sim N(0,(t-s))$. Looking up the MGF of $N(\mu,\sigma)$, we have $E[e^{tN}] = e^{t \mu + \frac{t^2\sigma^2}{2}}$. For our case, $N(0,(t-s))$ and $t=b$, we have $E[e^{b(B_t - B_s)}] = e^{\frac{b^2(t-s)}{2}}$. That cancels out with whatever is left to give $1$.
A: We have that $X_t$ is normally distributed with mean $\mu=at$ and variance $\sigma^2=b^2t$. Therefore, $e^{X_t}$ is log-normally distributed, with mean $e^{at+0.5\sigma^2}$. Therefore:
$$ Z_t:=e^{-at-0.5\sigma^2+X_t} $$
Must be a martingale, since $\forall t$:
$$\mathbb{E}[Z_t]=e^{0}=1=\mathbb{E}[Z_t|\mathcal{F}_0]=\mathbb{E}[Z_t|Z_0] $$
If you wan't to prove the martingale condition explicitely for some $s\leq t$, you can write:
$$\mathbb{E}[Z_t|Z_s]=\mathbb{E}\left[exp\left(-as-0.5b^2s+as+bW_s-a(t-s)-0.5b^2(t-s)+a(t-s)+bW(t-s) \right)|Z_s \right]=\\=Z_s\mathbb{E}[exp(-a(t-s)-0.5b^2(t-s)+a(t-s)+bW(t-s)]=\\=Z_s\mathbb{E}[exp(-a(h)-0.5b^2(h)+a(h)+bW(h)]=Z_s $$
The first line of the equality above uses the independence of increments property of the Brownian motion (i.e $W_t=W_s + W(t-s)$ in distribution). The last line evaluates to $Z_s$ with the expectation equal to 1, since again, $E[Z_t]=1$ $\forall t$.
