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


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


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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ Some of the bounds here should be useful. $\endgroup$ – Nate Eldredge Aug 12 '16 at 8:44

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.

  • $\begingroup$ @ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.