Assume we generate a sample path $X_t$ from an Ornstein-Uhlenbeck distribution (i.e. a mean-reverting random walk), where $dX_t = −\rho(X_t − \mu)dt + \sigma dW_t$. For concreteness, take $\mu = 0$, $\rho = 1$ and $\sigma^2 = 2$ so that both the characteristic time scale and width are $1$, i.e. after $t=1$ the expected value of $X_t \rightarrow N(0,1)$, the unit normal distribution.
I'm interested in the distribution of $Z_t = \min_{s<t} X_s$, i.e. the cumulative minimum of a sample path. Is there a closed form (exact or approximate) for the asymptotic distribution ($t >> 1/\rho$) for the inverse cdf of $Z_t$, say $\Psi_t^{-1}(p)$, so for example I could find the median of $Z_t$, namely $\Psi_t^{-1}(0.5)$?
I'm hoping this is well-known and I'm just using the wrong terminology to search for an answer.
I made some progress starting from the intuition that every time step ($\Delta t = 1/\rho$), we're effectively sampling an uncorrelated value from $N(0,1)$, so after $t$ time steps the likelihood that $Z_t > z$ is equivalent to the probability that $t$ samples from $N(0,1)$ are greater than $z$.
If $\Phi(x)$ is the cumulative distribution function for $N(0,1)$, then we have $P\{ Z_t <= z \} = \Psi_t(z) = 1 - (1-\Phi(z))^t$, so that $(1-\Psi_t(z))^{1/t} = 1-\Phi(z)$ or $z = \Phi^{-1}(1 - (1-\Psi_t(z))^{1/t})$. For example the median line (50-th percentile) of $Z_t$, where $\Psi_t(z) = 0.5$, would be given by $\Phi^{-1}(1 - 0.5^{1/t})$.
By simulation, this seems to have the right shape for a range of percentiles $p = \Psi_t(z)$, but is shifted upwards by ~ 0.5 (independent of $p$), i.e. the expression $\Phi^{-1}(1- (1-p)^{1/t}) - 1/2$ seems a much better approximation. Again intuitively it makes sense that the estimate above is too high, since the samples at each time step aren't really independent: if we observe a new minimum value $Z_t$ at time $t$, then we're quite likely to see an even lower value at a nearby time when the the new value is still highly correlated with the old one. But I don't immediately see how to incorporate that into the estimate.