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Suppose one has the an Ito process of the form:

$$dX_t = b(X_t)dt + \sigma(X_t)dW_t$$

The infinitesimal generator $LV(x)$ is defined by:

$$\lim_{t\rightarrow 0} \frac{E^x\left[V(X_t) \right]-V(x)}{t}$$

One can show that $$LV(x) = \sum_{i}b_i(x)\frac{\partial V}{\partial x_i}(x) + \frac{1}{2}\sum_{i,j}(\sigma(x)\sigma(x)^T)_{ij}\frac{\partial^2V}{\partial x_i \partial x_j}(x)$$

I'm wondering if there's an equivalent one infinitesimal generator for $$dX_t = b(X_t)dt + \sigma(X_t)d\eta(t)$$

$$d\eta(t) = \lambda\eta(t) dt + \alpha dW(t)$$

This is a stochastic process where the perturbations are from an Ornstein-Uhlenbeck process instead of a Brownian Motion. Wiki gives the infinitesimal generator of an Ornstein-Uhlenbeck process to be:

$$LV(x) = -\lambda x V'(x) + \frac{\alpha^2}{2}V''(x)$$

But I don't know if there's a way to use that fact to combine it with $LV(x)$ for the Ito process to get the infinitesimal generator for the stochastic process perturbed by an Ornstein-Uhlenbeck process

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You can look at your process $X_{t}$ as a two dimensional stochastic process

$$Y_{t}=\left[\begin{array}{cc}X_{t}\\ \eta_{t}\end{array}\right]$$

Then

$$dY_{t}=\left[\begin{array}{cc}dX_{t}\\ d\eta_{t}\end{array}\right]=\left[\begin{array}{cc}b(X_t)+\lambda\eta_{t}\sigma(X_{t})\\ \lambda\eta_{t}\end{array}\right]dt+\left[\begin{array}{cc}\alpha\sigma(X_t)&0\\ 0&\alpha\end{array}\right]\left[\begin{array}{cc}dW_{t}\\ dW_{t}\end{array}\right]$$

and the infinitesimal generator is of the form

$$LV(y)=LV(x,\eta)=\left(b(x)+\lambda\eta \sigma(x)\right)V'_{x}(x,\eta)+\lambda\eta V'_{\eta}(\eta,x)$$ $$+\frac{1}{2}\alpha^{2}\sigma^{2}(x)V''_{xx}(x,\eta)+\alpha^{2}\sigma(x)V''_{x\eta}(x,\eta)+\frac{1}{2}\alpha^{2}V''_{\eta\eta}(x,\eta)$$

By the way, the infinitesimal generator of an Ornstein-Uhlenbeck process of the form

$$d\eta_{t} = \lambda\eta_{t} dt + \alpha dW_{t}$$

is

$$LV(\eta)=\lambda \eta V'(\eta) + \frac{\alpha^2}{2}V''(\eta)$$

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  • $\begingroup$ Can please give proof of the expression of the generator $LV(y)=LV(x,\eta)=....$ $\endgroup$
    – Zbigniew
    Jan 5, 2018 at 19:17

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