I have the transition distribution $p(X_{t+1}|X_t=x_t) = \text{N}(\phi x_t,1)$ where $−1<\phi<1$.

Can we calculate the stationary distribution and its mean and variance? I know I can do that if the Markov chain is discrete, by multiplying the transition probability matrix $Q$ many times. How can I do that if the transition probability is continuous?

I am talking in general not just for this example. Can we calculate the theoretical stationary distribution of a continuous Markov chain if we have the transition distribution?


1 Answer 1


Your Markov chain is a Gaussian $\text{AR}(1)$ process with unit error variance:

$$X_{t+1} = \phi X_t + \varepsilon_{t+1} \quad \quad \quad \varepsilon_t \sim \text{IID N}(0,1).$$

The elements in this process can be written as weighted sums of normal error terms (in $\text{MA}(\infty)$ form), so the stationary distribution (if it exists) will be a normal distribution, $X_t \sim \text{N}(\mu, \sigma^2)$.

It remains only to find a mean and variance parameter that yield a stationary distribution. To do this, we apply the recursive equation to give equations for the mean and variance parameters. Under the condition of stationarity we have a fixed mean, which means gives the mean equation:

$$\mu = \mathbb{E}(X_{t+1}) = \mathbb{E}(\phi X_t + \varepsilon_{t+1}) = \phi \mathbb{E}(X_t) + \mathbb{E}(\varepsilon_{t+1}) = \phi \mu.$$

Since $-1 < \phi < 1$ this yields the unique solution $\mu = 0$. Under the condition of stationarity we also have a fixed variance, which means gives the variance equation:

$$\sigma^2 = \mathbb{V}(X_{t+1}) = \mathbb{V}(\phi X_t + \varepsilon_{t+1}) = \phi^2 \mathbb{V}(X_t) + \mathbb{V}(\varepsilon_{t+1}) = 1 + \phi^2 \sigma^2.$$

This equation yields the solution $\sigma^2 = 1/(1-\phi^2)$. Thus, the stationary distribution exists, and it is:

$$X_t \sim \text{N} \Bigg( 0, \frac{1}{1-\phi^2} \Bigg).$$

Note that this form is well-known as the stationary distribution of an $\text{AR}(1)$ process with unit variance on the noise terms. It can easily be generalised to broader $\text{AR}(1)$ processes with a non-zero mean, or non-unit variance for the error term.

  • $\begingroup$ Thank you so much for the answer but when you multiply the stationary distribution from your answer with the transition distribution from my question, It won't equal the posterior. I know that In Markov chain, $\pi = Q* \pi$ where $\pi$ is the posterior distribution and $Q$ is the transition distribution. Am I right? $\endgroup$
    – floyd
    Commented Aug 23, 2019 at 17:33
  • $\begingroup$ Sorry, just noticed this comment - I've answered it at your other question. $\endgroup$
    – Ben
    Commented Sep 1, 2019 at 3:43
  • $\begingroup$ Could you take a look at my related question on MSE: math.stackexchange.com/q/4728844/47771? $\endgroup$
    – 0xbadf00d
    Commented Jul 2, 2023 at 13:25

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