Ornstein-Uhlenbeck conditioned to be positive 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.
 A: 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. 
A: Started from a non-zero point, because the OU process started from 0 is symmetric about 0, I think it must enjoy a reflection principle about 0 only which should allow easy estimation of the probabilities.
