exactly, then $\xi$ follows the target distribution in the long run.

However, in general the solution to and has no closed form, so some discretization methods have to be adopted. Consider the following discretized chain for $W_{k}=(\xi_{k}^{\mathrm{T}},\rho_{k}^{\mathrm{T}})^{\mathrm{T}}$, $k\ge0$: $$\begin{aligned}
\xi_{k+1} & =\xi_{k}+\delta\rho_{k},\label{eq:langevin_xi}\\
\rho_{k+1} & =(1-\gamma\delta)\rho_{k}-\delta\cdot\nabla V(\xi_{k})+\sqrt{2\gamma\delta}\eta_{k},\label{eq:langevin_rho}\end{aligned}$$ where $\delta$ is the step size, $\{\eta_{k}\}_{k=0}^{\infty}\overset{iid}{\sim}N(0,I_{r})$, and $\eta_{k}$ is independent of $\{W_{k}\}_{i=0}^{k-1}$. The iterations and are typically referred to as the *underdamped Langevin algorithm*.

The importance and usefulness of the $\{W_{k}\}$ sample will be illustrated in Theorem \[thm:control\_bias\_mse\]. Before that we need to first guarantee that $\{W_{k}\}$ is well defined and does not explode as time goes on. Formally, we show that under some mild conditions, $\{W_{k}\}$ is stable in the sense that it has finite moments of any order, uniformly in the step count $k$. The result is summarized in Theorem \[thm:finite\_moments\], along with the assumptions we need to impose.

\[assu:v\_boundedness\](a) $V(x)$ is bounded from below, i.e., $V(x)\ge\nu_{0}$ for some constant $\nu_{0}\in\mathbb{R}$ and all $x\in\mathbb{R}^{r}$. (b) The operator norm of the second derivative of $V$ is bounded, i.e., $\Vert\nabla^{2}V(x)\Vert\le\nu$ for some constant $\nu>0$ and all $x\in\mathbb{R}^{r}$. (c) $V(x)\in\mathscr{C}_{poly}^{\infty}$.

For Assumption \[assu:v\_boundedness\](a), we can assume $\nu_{0}=0$ without loss of generality. This is because 