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Let $X=(X_t)_{t\ge 0}$ be a stochastic process (Ornstein-Uhlenbeck process) determined by

$$dX_t=-aX_tdt+\sigma dW_t,$$

where $X_0=0$, $a>0$ and $\sigma>0$ are constants, and $W=(W_t)_{t\ge 0}$ is a standard Brownian motion. Consider the following integral

$$I(t)=\int_0^tX_s^2ds,$$

where $t>0$ is fixed. Now we aim to estimate the probability

$$P\left[I(t)>\alpha\right],$$

where $\alpha>0$ is some given number. Does someone have an idea? I was considering to apply Ito's formula to $X^2$ and then to use integration by parts, but didn't obtain the wanted result. Thanks a lot for the reply!

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  • $\begingroup$ you might look up square root diffusions of the type that go into the CIR model, the square of a ou process is one, and I think you can find the laplace transform of the distribution you are interested in,. $\endgroup$ – user83457 Aug 4 '16 at 13:31
  • $\begingroup$ Revus & Yor have a nice section on them unde squared Bessel processes, I think $\endgroup$ – user83457 Aug 4 '16 at 13:39
  • $\begingroup$ Thank you very much for your reply and I'll read it immediately. $\endgroup$ – CodeGolf Aug 4 '16 at 21:51
  • $\begingroup$ @michael Indeed, setting $Y_t=X^2_t$, we obtain $dY_t=(\sigma^2-2aY_t)dt+\sqrt{|Y_t|}dW_t$, which is similar to a Bessel process. Do you have any idea about tranforming $Y_t$ to $f(Y_t)$ for some function $f$, such that $f(Y)$ is a Bessel process? $\endgroup$ – CodeGolf Aug 5 '16 at 7:21
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    $\begingroup$ Some of the bounds here should be useful. $\endgroup$ – Nate Eldredge Aug 12 '16 at 8:44
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Using the notation of the OP, let $I(t)=\int_0^t X_s^2 ds$ where $X$ solves the above SDE. By Chebyshev's inequality, we have that $$ P(I(t) > \alpha \mid X_0 = x) \le \frac{E\left\{ I(t) \mid X_0 = x \right\}}{\alpha} \;. $$ Fortunately, thanks to a Feynmann-Kac formula, the function $$ u(t,x)=E\left\{ I(t) \mid X_0 = x \right\} $$ appearing in the upper bound of this inequality is a local solution to an inhomogeneous, linear PDE: $$ \begin{cases} \partial_t u = L u + x^2 &\forall x, t\ge 0 \\ u(0,x) = 0 &\forall x \end{cases} \tag{$\star$} $$ where $L = - a x \partial_x + \frac{\sigma^2}{2} \partial_{xx}$ is the infinitesimal generator of the above SDE. Recall, that the linear operator $L$ has eigenvalues and eigenvectors: $$ L e_k(x) = - k \cdot a \cdot e_k(x) $$ where $e_k(x)$ is the $k$th Hermite polynomial and $k$ ranges over all natural numbers including zero. Expand the solution to $(\star)$ using these eigenvectors: $$ u(t,x) = \sum_{k \ge 0} s_k(t) e_k(x) $$ and similarly, expand the inhomogeneity: $$ x^2 = (e_2(x) + 2 e_0(x) ) \frac{\sigma^2}{4 a} $$ Substitute these expansions back into $(\star)$ and invoke orthogonality of these eigenvectors (in a weighted inner product space) to obtain the following system of ODEs for the spectral coefficients of $u(t,x)$: $$ \begin{cases} \dot s_0 = \frac{\sigma^2}{2 a} \;, & s_0(0) = 0 \;, \\ \dot s_1 = - a \cdot s_1 \;, &s_1(0) = 0 \;, \\ \dot s_2 = -2 \cdot a \cdot s_2 + \frac{\sigma^2}{4 a} \;, &s_2(0) = 0 \;, \\ \dot s_k = - k \cdot a \cdot s_k \;, &s_k(0) = 0 \;, \quad k \ge 3 \end{cases} $$ The solutions to these ODES are all zero with the exception of: $$ \begin{cases} s_0(t) = \frac{\sigma^2}{2 a} t \\ s_2(t) = \sigma^2 \left( \frac{1 - e^{-2 a t}}{8 a^2} \right) \end{cases} $$ Thus, the solution to ($\star$) is: $$ u(t,x) = \frac{\sigma^2}{2 a} t + \sigma^2 \left( \frac{1 - e^{-2 a t}}{8 a^2} \right) \left( \frac{4 a}{\sigma^2} x^2 - 2 \right) \;. $$ To answer your question, take $x=0$ in this function to obtain: $$ P(I(t) >\alpha \mid X_0 = 0) \le \frac{1}{\alpha} \left( \frac{\sigma^2}{2 a} t - \sigma^2 \left( \frac{1 - e^{-2 a t}}{4 a^2} \right) \right) $$ You can adapt this technique to estimate all sorts of path-dependent expected values.

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  • $\begingroup$ @ Nawaf Bou-Rabee Thank you very much for the reply $\endgroup$ – CodeGolf Aug 12 '16 at 4:21
  • $\begingroup$ Is this estimate not sharp enough? $\endgroup$ – Nawaf Bou-Rabee Aug 20 '16 at 1:12
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    $\begingroup$ Well, this question appears when I solve another problem. For the purpose of applications, it is sufficient for me. But I don't know whether it is the best estimation or not... $\endgroup$ – CodeGolf Aug 20 '16 at 9:05
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    $\begingroup$ It's true that it's recommended that answers that are helpful at least be upvoted. $\endgroup$ – Todd Trimble Aug 20 '16 at 14:23

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