ProbabilityTheoryStochasticProcessesStochasticCalculus Suppose one has the SDE Misplaced &&= \mu X_t \space dt + \sigma X_t \space dW_t \end{aligned}$$ where $\mu \in \mathbb{R}$ and $\sigma > 0$ are deterministic. Note that the presence of $X_t$ in both terms hints that a solution would also solve the following integral equation $$\int_0^t \frac1{X_s} \space dX_s = \mu \int_0^t \frac1{X_s} X_s \space ds + \sigma \int_0^t \frac1{X_s} X_s \space dW_s = \mu t + \sigma W_s$$ So that $Y_t := \log{X_t}$ would solve the SDE, however as we know from Itô's formula, there needs to be a correction term for this to be rigorously shown. Therefore, take $Y_t$ and compute using Itô's $$\begin{aligned} \log{X_t} - \log{X_0} &= \int_0^t \frac1{X_s} \space dX_s - \frac12 \int_0^t \frac1{X_s^2} \space d\left\langle X \right\rangle_s \\ &= \int_0^t \frac1{X_s} (\mu X_s \space ds + \sigma X_s \space dW_s) - \frac12 \int_0^1 \frac1{X_s^2} \sigma^2 X_s^2 \space ds \\ &= \int_0^t \mu \space ds + \int_0^t \sigma \space dW_s - \frac12 \int_0^1 \sigma^2 \space ds \\ &= (\mu - \frac12 \sigma^2) t + \sigma W_t \end{aligned}$$ Where I have substituted $dX_t$ into the integral and computed $\left\langle X_t \right\rangle$ using the fact that it is a semi-martingale and the [[Quadratic Variation]] is defined to be the quadratic variation of the [[Continuous Local Martingale]] term, $\sigma X_t \space dW_t$. As this SDE has time-homogenous and Lipschitz coefficients, $b(x) = \mu x$ and $\sigma(x) = \sigma x$, it is time-homogeneous and it's solution is a [[Markov Process]]. The generator can be computed explicitly $$\begin{aligned} Lf(x) &= \frac12 \sigma(x)^2 \frac{\partial^2}{\partial x^2} f(x) + b(x) \frac{\partial}{\partial x} f(x)\\ &= \frac12 \sigma^2 x^2 f''(x) + \mu x f'(x) \end{aligned}$$