For $X_{n+1}=(1-\Delta t)X_n + \sqrt{2\Delta t}\xi_n$, why is $X_n$ centered normal with variance $1/(1-\delta t/2)$ for $n\gg1$?

We have that $\xi_n\sim\mathcal{N}(0,1)$, and these $\xi_n$'s a independent.

The above equation is the Euler-Maruyama discretisation of the stochastic differential equation $$\mathrm{d}X_t = -X_t\mathrm{d}t + \sqrt{2}\mathrm{d}W_t,$$ where $X_t$ is a continuous time stochastic process and $W_t$ denotes a Wiener process (standard Brownian motion).

• Assume that $X_n$ and $X_{n+1}$ have the same variance and deduce this variance from the recursion giving $X_{n+1}$ in terms of $X_n$ and $\xi_n$. – Did May 27 '17 at 12:01

A discrete process of the form $$x_{n}=a x_{n-1} + e_n$$ where $e_n$ is white noise (independent and stationary) is known as an autoregresive (AR) process. To get its variance, assuming $x_n$ is stationary and zero mean, multiply the equation by itself, take expectations, recall that $E(x_n e_n) = E(x_n)E(e_n)=0$ by independence, and obtain:
$$E(X_n^2) = a^2 E(X_{n-1}^2) + E(e_n^2)$$ Or
$$\sigma^2_x = \frac{\sigma^2_e}{1-a^2}$$
In your case, you get $$\sigma^2_x =\frac{2 \, \Delta t}{2 \Delta t-(\Delta t)^2}=\frac{1}{1-\Delta t/2}$$
The assumption of stationarity is justified if $n$ is large. That $x_n$ is Normal depends on the initial condition.