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).

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    $\begingroup$ 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 $\xi_n$. $\endgroup$
    – Did
    Commented May 27, 2017 at 12:01

1 Answer 1


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.


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