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

  • $\begingroup$ Hi I think that this Question should be also tagged with the "stochastic-calculus" item Regards $\endgroup$
    – The Bridge
    Sep 8, 2010 at 9:52

2 Answers 2


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.

  • $\begingroup$ Thanks Has2 for your comments. I will be thinking about how you get these equations :D $\endgroup$
    – Nameless
    Sep 9, 2010 at 1:47
  • $\begingroup$ You are welcome. I tried to explain how you get the equations in the answer. Now I will add some further details, hope they are useful. $\endgroup$
    – has2
    Sep 13, 2010 at 12:08

Hi Nameless,

I think that your question is related to killed diffusions for which infinitesimal generators are available I think.

Here is a reference but I recommand that you "google" killed diffusions


  • $\begingroup$ The link to springerlink.com is broken. I'm also unable to find any copy saved on the Wayback Machine. $\endgroup$ Aug 11, 2022 at 12:02

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