How to find a function $f$ to transform a stochastic process $X$ to a martingale $f(X)$ Let's think about a property we want to show for a process $X=(X_t)_{t\geq0}$, which is very hard to show. But maybe we can show another property for a martingale $H$ very easy and the hope is that the desiered property can be derived from this. So the Idea is to find a function $f:\Bbb R\to\Bbb R$, so that $H_t:=f(X_t)$ is a martingale with respect to the original filtration.
An exemplary property for this could be e.g. the expected value is bounded in any kind. Let's assume we found such a function $f$ which is concave and strictly monotonically increasing. As $f(X)$ is a martingale we get for a constant $c=\Bbb Ef(X_t)\leq f(\Bbb EX_t)$ (by jensen's inequality) and thus ($f^{-1}$ is increasing as well) $f^{-1}(c)\leq \Bbb EX_t$.
As this question is only about the general procedure how to find such a function, we could do this e.g. with the process $X_t:=B_t+t$, whereas $B$ is a Brownian motion. I know the property 'bounded below' could be shown ways easier in this case, but it's only about the general procedure.
 A: The easiest solution of $f(X)$ to be a martingale is to choose $f$ as a constant function. But this is not what we want. So let's look for a better solution:
By the Itô-formula we get
$$f(B_t+t)=\int_0^t f'(B_s+s)\mathrm dB_s+\int_0^t f'(B_s+s)\mathrm ds+\frac{1}{2}\int_0^t f''(B_s+s)\mathrm ds.
$$
Since the first summand is already a local martingale (a stochastic integral w.r.t a local martingale is a local martingale again), we could try to find a function $g=f'$ so that $g(x)+\frac{1}{2}g'(x)=0$ for all $x$. This is satisfied for example by $g(x)=e^{-2x}$. Since
$$f(B_t+t)=-\frac{1}{2}e^{-2(B_t+t)}=\int_0^t e^{-2(B_s+s)}\mathrm dB_s
$$ is a local martingale we just have to check some conditions anymore that it is even a (true) martingale. One of them is to proof $\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t<\infty$ for all $t\geq0$ ($[.]_t$ denotes the quadratic variation). Note that
$$\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t=\mathbb E\int_0^t e^{-4(B_s+s)}\mathrm ds=\int_0^t e^{-4s} \mathbb Ee^{-4B_s}\mathrm ds.
$$
As $\mathbb Ee^{-4B_s}=e^{8s}$ is the moment-generating function of a $(0,s)$-normally distributed random variable evaluated at the point $-4$, we get $$\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t=\int_0^t e^{4s} \mathrm ds=(e^{4t}-1)/4<\infty.$$
