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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.

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2 Answers 2

up vote 2 down vote accepted

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.

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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.

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The reflection principle does not work in the case of the OU process since it is not space-homogenous. –  Piotr Miłoś Apr 28 at 13:30

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