Suppose we have a Markov process $X=(X_t)_{t\geq 0}$, taking values on $\mathcal{N}$, with this Q-matrix (Example 17.27 of Probability Theory, 3rd version, by A. Klenke): $$ q(x,y) = \begin{cases} x &\text{if } y=x+1\\ -x &\text{if } y=x\\ 0 &\text{otherwise} \end{cases} $$
Is $W_t=e^{-t}X_t$ a martingale?
I know that $P_1[X_t>e^tx]=(1-e^{-t})^{\lfloor e^tx\rfloor}$ from a previous result in the same book above. To prove if it is a martingale, I am thinking about the trick about the expected value of a positive random variable but with no success.
I would also like to know if the time growth was linear, how could I spot it?
Let me know if more context is needed. Thank you.
Using the $Q$-matrix you can show that $X_t-\int_0^t X_s\,ds$ is a martingale. Then by Ito's formula you see that $e^{-t}X_t$ is indeed a martingale.
Added at OP's request: Given a function $f$ on the state space, define $g(i):=\sum_{j\in\mathcal N} q(i,j)f(j)$. One way to think about the $Q$-matrix is this: $f(X_t)-\int_0^t g(X_s)\,ds$ is a martingale. For your $Q$-matrix, if $f(i) = i$ is the identity function, then $g(i)=i$ as well, by direct computation.
Another way to get at the martingale property of $e^{-t}X_t$ is as follows. For $t\ge 0$ and small $h>0$, we have $$ \eqalign{E[e^{-(t+h)}X_{t+h}\mid \mathcal F_t] &=e^{-(t+h)}\left[(X_t+1)\cdot(hX_t+o(h))+X_t\cdot(1-hX_t+o(h))\right]\cr &=e^{-t}\cdot(1-h+o(h))\left[X_t\cdot(1+h)+o(h)\right]\cr &=e^{-t}[X_t+o(h)].\cr } $$ Now to compute $E[e^{-t}X_t\mid \mathcal F_s]$, cut the interval $[s,t]$ into $n$ subintervals of length $\delta:=(t-s)/n$ and "telescope", using the tower property of conditional expectations: $$ \eqalign{ E[e^{-t}X_t-e^{-s}X_s\mid\mathcal F_s] &=\sum_{k=0}^{n-1}E[e^{-(s+(k+1)\delta}X_{s+(k+1)\delta}-e^{-(s+k\delta)}X_{s+k\delta}\mid\mathcal F_s]\cr &=\sum_{k=0}^{n-1}E\left[E[e^{-(s+(k+1)\delta}X_{s+(k+1)\delta}-e^{-(s+k\delta)}X_{s+k\delta}\mid\mathcal F_{s+k\delta}]\,\mid \mathcal F_s\right]\cr } $$ Each inner conditional expectation in this final sum is $o(1/n)$ by what has gone before; therefore the sum is $n\cdot o(1/n) = o(1)$. Send $n\to\infty$ to see that $E[e^{-t}X_t-e^{-s}X_s\mid\mathcal F_s]=0$, which is the martingale property.
A good place to learn more about the martingale connection to the $Q$-matrix might be the book Continuous Time Markov Processes by T. Liggett.