Finding $X_t$ of an Itô Diffusion

Question

Someone please help me with this:

I have that $X_t$ is the Ito diffusion with genertator $A(f)(x)=\alpha xf'(x)+f''(x).$

Then, if $X_0=x \in \mathbb{R}^+$, how do I find $X_t$?

Answer 1

In this case, the dynamics is given by \begin{align*} dX_t = \alpha X_t dt + \sqrt{2}\,dW_t. \end{align*} Then, \begin{align*} d\left(e^{-\alpha t} X_t\right) &=-\alpha e^{-\alpha t} X_tdt + e^{-\alpha t} dX_t\\ &=\sqrt{2}e^{-\alpha t} dW_t. \end{align*} Consequently, \begin{align*} e^{-\alpha t} X_t = X_0 + \sqrt{2}\int_0^te^{-\alpha s} dW_s. \end{align*} That is, \begin{align*} X_t = e^{\alpha t} X_0 + \sqrt{2}\int_0^te^{-\alpha (s-t)} dW_s.\tag{1} \end{align*}

EDIT:

From $(1)$ and Ito isometry, \begin{align*} E(X_t) &= E\left(e^{\alpha t} X_0 + \sqrt{2}\int_0^te^{-\alpha (s-t)} dW_s\right)\\ &=e^{\alpha t} X_0 + \sqrt{2}e^{\alpha t} E\left( \int_0^t e^{-\alpha s} dW_s\right)\\ &= e^{\alpha t} X_0,\\ Var(X_t) &= E\left(\left( \sqrt{2}\int_0^te^{-\alpha (s-t)} dW_s\right)^2 \right) \\ &=2\int_0^t e^{-2\alpha (s-t)} dt \\ &= \frac{1}{\alpha}\left(e^{2\alpha t}-1 \right). \end{align*}