# Large deviation bound for O-U process

Assume $X_t$ is an Ornstein-Uhlenbeck process in the form of $$d X_t = -\alpha X_t dt + \sigma dB_t$$ Is there an exponential bound (large-deviation bound) for $$P\left( \max_{t\le T} |X_t| \ge z \right) \le ?$$

P.S. when $\alpha = 0$ above $X_t$ is a Brownian motion, and hence $$P\left( \sup_{t\le T} |\sqrt{\varepsilon} B_t | \ge c \right) \le 4d \exp\left( - \frac{c^2 }{2dT \varepsilon} \right)$$ for Brownian motion (Nonasymptotics result wanted). See e.g. the last display in

https://en.wikipedia.org/wiki/Schilder%27s_theorem

Is there an analogous version for the OU process?

• What is the relation between $T$ and $z$ in the asymptotic you are after? – ofer zeitouni Jul 12 '17 at 4:26
• There is no relations, we want nonasymptotic result analogous to the Brownian motion version in the (new) P.S. part in the question – Koltchinskii Jul 13 '17 at 17:56
• The asymptotics you wrote is meaningful when $T/z^2\to 0$. To me this looks like a relation. Anyway, you already got two answers in different asymptotic regimes, unless you clarify what you are after let me guess that you won't get a third. – ofer zeitouni Jul 13 '17 at 22:25
• I actually wondered if we can resolve this problem using Girsanov transformation type of argument ... – Koltchinskii Jul 18 '17 at 1:24
• @Koltchinskii My answer is already along these lines; one derives the LDP for the scaled Wiener process and then uses a Girsanov type argument to reduce white noise perturbations of a dynamical system to the Wiener process case. – Ian Jul 18 '17 at 3:13

Fernique's theorem, valid for all continuous Gaussian processes, implies that there are constants $C,c$ (depending on $T$) such that $P(\max_{0 \le t \le T} |X_t| \ge z) \le C e^{-c z^2}$.
Yes: for white noise perturbations of the 1D dynamical system $$\dot{x}=b(x)$$, the action functional is $$S(\phi)=\frac{1}{2} \int_0^T |b(\phi(s))-\dot{\phi}(s)|^2 ds$$ for $$\phi \in H^1$$ and otherwise infinity. The LDP says that $$\sigma^2 \log(P(X_\cdot \in A))$$ is asymptotically between $$-\inf_{f \in \mathrm{Int}(A)} S(f)$$ and $$-\inf_{f \in \mathrm{Cl}(A)} S(f)$$ as $$\sigma \to 0$$.
In your case $$A$$ is the exterior of a ball in $$C([0,T])$$ centered at zero, so these two are the same, and so the problem is (presumably) to compute $$I=\inf_{\phi \in C([0,T]) : \| \phi \|_\infty \geq z,\phi(0)=0} \frac{1}{2} \int_0^T |\dot{\phi}(s)+\alpha \phi(s)|^2 ds$$. Then the logarithmic asymptotic for your quantity is $$e^{-\sigma^{-2} I}$$. I suspect the minimizer is attained by putting $$\phi(T)=z$$ (giving yourself the maximum possible amount of time to get to a magnitude of $$z$$) so that by the Euler-Lagrange equation $$\phi(t)=z \frac{\sinh(\alpha t)}{\sinh(\alpha T)}$$ is a minimizer; you should check this guess though. If I'm right then that means the minimum of $$I$$ is $$\int_0^T \left ( z \alpha \left ( \frac{\cosh(\alpha s) + \sinh(\alpha s)}{\sinh(\alpha T)} \right ) \right )^2 ds = \alpha z^2(\coth(\alpha T)+1)$$. This is large compared to $$z^2$$ for $$T \ll \alpha^{-1}$$ but quickly settles toward $$2\alpha z^2$$; unsurprisingly this means that in the long run you should expect to see maximum deviations from zero on the order of $$\frac{\sigma}{\sqrt{\alpha}}$$.