Let us consider $X$ to be an Ornstein–Uhlenbeck process, i.e. the solution of $\text{d}X\_t = \text{d}W\_t - X\_t \text{d}t$. We define $Y$ by


$$\mathbb{P}(Y\in \cdot):= \lim\_{t\to+\infty}\mathbb{P}(X\in \cdot|\forall\_{s\leq t}X\_s \geq 0).$$ 


I expect that random variable $Y\_t$ is well-concentrated, for example


$$\mathbb{P}(Y\_t\geq x) \leq C \exp\{-c x^2\},$$
for some $C,c>0$ (independent of $t$). 


Is there an "easy proof" of this fact? I can think about a proof using the first eigen-value of the OU operator on the half-line and then applying the $h$-transform but this seems a big gun for an easy problem.



Ornstein-Uhlenbeck conditioned to be positive
One way to do it in a computational way is to use a representation as time changed Brownian motion $\{W\_t\}$ and known results concerning Brownian motion.


Write $X\_t=xe^{-t}+e^{-t}W\_{u(t)}$ where $u(t)=e^{2t}-1$ and $x$ is your starting point, which of course you need to assume $>0$ (the result is not true for stationary OU so I assume that 
you meant to start from a fixed point). With $A\_t=\{W\_{u(t)}>ye^t-xe^{-t}\}$ and 
$B\_t=\{W\_{u(s)}>-xe^{-s}, s\leq t\}$, what you are trying to compute is 
$P(A\_t\cap B\_t)/P(B\_t)$.


Now, $P(B\_t)$ is larger than
$$ P(W\_s>-xe^{-1}, s\in [0,1]; W\_1>1; W\_s-W\_1\geq -1, s\in (1,u(t)))\geq -C(x)e^{-t}$$ by the reflection principle. Similarly, by decomposing at time $1$, you get for large $t$ that, for $y>2$ say,
$P(A\_t\cap B\_t)\leq P(W\_s\geq -x, s\leq e^{2t}-1, W\_{e^{2t}-1}\geq (y-1)e^t)\leq C(x)e^{-t} 
ye^{-y^2/2}$. This gives what you wanted. 


