Let $X_t$ be a continuous-time simple random on $\mathbb{Z}$ starting at $0$.
My question is whether $$\lim_{t \to \infty}\mathbb{P}(X_t>k)$$ exists, and if it does, calculate it.
I know that in the discrete case, if $t$ counts steps, then the probability of not returning to the origin in the first $t$ steps is bounded by: $$ \frac{c}{\sqrt{t}}$$ for some constant $c$. But I'm not sure this has something to do with my question. I suppose the limit in my question should be equal to $1$, otherwise the walk would be very tight.
Thanks for your help!
Just elaborating on Sangchul Lee's answer.
Consider the discrete version of this random walk. Denote by $S_n$ the current location in the random walk, then $$S_n = \sum_{i=1}^nX_i$$ where $X_i$ are independent $Ber(\frac{1}{2}$) random variables.
Then, according to the central limit theorem, $\frac{S_n}{n}$ converges in distribution to the normal distribution $N(0,1)$.
Let $k$ be some positive integer, and let $\epsilon > 0$.
For $n$ large enough, $\frac{k}{\sqrt{n}} < \epsilon$.
From this $n$ onwards, $\mathbb{P}(S_n>k)=\mathbb{P}(\frac{S_n}{\sqrt{n}}>\frac{k}{\sqrt{n}})\geq \mathbb{P}(\frac{S_n}{\sqrt{n}}>\epsilon).$
By the definition of convergence in distribution, $\mathbb{P}(\frac{S_n}{\sqrt{n}}>\epsilon)\xrightarrow{n \to \infty} 1 -\phi(\epsilon)$,
where $\phi$ is the cumulative distribution function of the normal distribution.
$\epsilon$ was an arbitrary positive number, so
$\lim_{n\to \infty}\mathbb{P}(S_n>k)\geq \sup_{\epsilon>0}[1-\phi(\epsilon)]=\frac{1}{2}.$
But $\lim_{n\to \infty}\mathbb{P}(S_n>k) \leq \lim_{n\to \infty}\mathbb{P}(S_n>0) = \lim_{n\to \infty}\frac{1}{2} = \frac{1}{2}.$
Thus, $\lim_{n\to \infty}\mathbb{P}(S_n>k) = \frac{1}{2}.$ In particular, $\lim_{n\to \infty}\mathbb{P}(S_n>|k|) = 1.$
P.S. This holds for the continuous version of the random walk because the continuous markov chain is just slowed down / accelerated version of the discrete. So the limiting behaviour, as time approaches infinity, is the same.