# How to find the transition distribution functions of these two processes?

What are the transition distribution (or density) functions of processes defined by

$dX_t=\mu dt +\sigma dW_t$

and

$dX_t= \theta(\mu-X_t) dt +\sigma dW_t,$

where $\theta>0$, $\mu$ is a real number, $\sigma>0$, and $W_t$ is a standard Brownian motion.

I know it is a solved problem, but I cannot find a reference that presents the detailed steps of the derivations. Could you please provide some good references? Or, could you please come with the derivations?

• @Gordon the defintion of transition distribution is: $f(X=x,T=t_0+s|X=x_0, T=t_0)$ that is, the probability distribution function for $s$ period of times ahead, given that know the pocess is in $X=x_0$ – John Feb 13 '17 at 18:31
• @Gordon it is the limit of the previous function with $s$ going to infinity. But if you provide me the trasition distribution derivation for both (or at least one) process that would be already really helpfull. – John Feb 13 '17 at 19:30
• @gordon the lim(theta->0) of the second one ought to look like the lim(mu->0) of the first one , shouldn't it? – dm63 Feb 14 '17 at 12:26
• Thanks @dm63. I mixed $\mu$ and $\theta$ for the second one, which is now corrected. – Gordon Feb 14 '17 at 13:53
• @John please don't self delete posts once they receive an answer. – Bob Jansen Mar 13 '17 at 19:07

We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$, for $t>s$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-\mu(t-s)-X_s \le y-\mu(t-s)-X_s \mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu(t-s)-X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.
For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\theta(t-s)}X_s + \mu\left(1-e^{-\theta(t-s)} \right)+\sigma\int_s^te^{-\theta(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \le y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\theta(t-v)}dW_v \le y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\theta}\big(1-e^{-2\theta(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\theta(t-s)}x - \mu\big(1-e^{-\theta(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\theta}\big(1-e^{-2\theta(t-s)} \big)}} \right). \end{align*}
• Could you please give some little detail about the fourth equality of the second part, that is: $= \phi ()....$ @Gordon – John Feb 14 '17 at 12:07