Let $X\_t\in\mathbb{R}$ be an Ito diffusion process given by $$ dX\_t=a(b-X\_t)dt+\sigma dW\_t$$, then the characteristic operator of $X\_t$ is given by $$L=a(b-x)\frac{\partial}{\partial x}+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}$$


(more details about the characteristic operator can be found here <http://en.wikipedia.org/wiki/It%C5%8D_diffusion>). Where in this case $a>0$, $b\in\mathbb{R}$, and $\sigma>0$. 


Now I assume that $\tau$ is a random variable with density function $f(t)=\lambda e^{-\lambda t}\chi(t)\_{[0,\infty)}$, $\alpha\_t$ is a random variable taking only two values $1$ and $2$.
 $$\alpha\_{t}=1\chi\_{(0,\tau)}(t)+2\chi\_{[\tau,\infty)}(t) $$


Then I change $b=b(\alpha\_{t}$), i.e, b is random and taking two values $b(1)\in\mathbb{R}$ and $b(2)\in\mathbb{R}$


My question is : how we find the characteristic of Ito diffusion $X\_t$ given by $$dX\_t=a[b(\alpha\_t)-X\_t]dt+\sigma dW\_t $$ ?


Thanks for your time and consideration 



Characteristic operator  
The wikipedia page cited in the question provides most of the answer: to get your operator compute
\begin{equation}
\lim\_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X\_\delta)] -f(x)}{\delta}
\end{equation}
The difference between your problem and the case covered in the wikipedia article is that $f$ in the above display is a function of $x$ only. However, your problem has an additional state variable (the binary variable that takes one of the values $1$ or $2$ depending on $\alpha$). So, the correct limit to study is:
\begin{equation}
\lim\_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X\_\delta,\alpha\_t)] -f(x,1)}{\delta}.
\end{equation}
Thus, you don't have one function, but two functions $f(x,1)$ and $f(x,2)$ and two PDEs that these functions satisfy.


It is implicitly assumed that $\tau$ is independent of the dynamics of $X$ before $\tau$. Furthermore, before $\tau$ the dynamics of $X$ are governed by the first SDE given in the question. One can use these to write the above expectation in two pieces: one piece over the set $\{\delta < \tau\}$ the other over $\{\delta < \tau\}^c$. Once this is done, the usual use of Ito's formula gives:
$$
L\_1 f(x,1)=-\lambda f(x,2)~~~ (\*)
$$
and
$$
L\_2 f(x,2) = 0.
$$
where
$$
L\_i = a(b\_i - x) \frac{\partial}{\partial x} + \frac{1}{2} \sigma^2\frac{\partial^2}{\partial x^2}
$$


Further details:
\begin{align\*}
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) ]&=
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) 1\_{\{ \tau > \delta\}} ] +
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) 1\_{\{ \tau \le \delta\} }]\\\\
&\approx (1-\lambda \delta){\mathbb E}[ f(X^1\_\delta,\alpha\_\delta)] + 
\lambda \delta f(x,2),
\end{align\*}
where $X^1$ is a process that is independent of $\tau$ with dynamics determined by $L\_1$. 


Here you use several things: 1) $P( \tau < \delta) \approx \delta \lambda$ 2) if a jump occurs before $\delta$, you can ignore what happens between $\tau$ and $\delta$ (the contribution of this part is in the order of $\delta^2$ and when divided by $\delta$ and
$\delta$ is let go to $0$, it disappears).


To get (\*) from the previous display: use Ito's formula on the first expectation, subtract $f(x,1)$, divide by $\delta$ and let $\delta \rightarrow 0$. $f(x,2)$ is a function of what happens after $\tau$; after $\tau$ the stochastic process is a simple diffusion with generator $L\_2$: this is why (\*\*) holds.


