Showing ergodicity for the mean of an AR(1) I am asked to show ergodicity of the mean of an AR(1) whose regression coefficient is less than one in absolute value. Which is to say I am asked to show that 
$$Y_t = ρY_{t-1} + ε_t$$ is ergodic to the mean when $$|ρ| < 1$$
I know that the process is stationary when ρ is less than one in absolute value but I do not know how to show that this implies that 
$$\frac{1}{T}\sum Y_t$$ converges to the expected value of $$Y_t$$ as T tends to infinity.
Is it that I should invert the process to an infinite order MA process in order to show ergodicity?
 A: The stationary distribution is $\pi(y_t) = \text{Normal}(0, \sigma^2/(1-\rho^2))$, where $\sigma^2 = \text{Var}(\epsilon_t)$. I assume you can show this, or that you assume it is known. To show Ergodicity, we need to show that this is the distribution that the $n$th step transition distribution approaches, as $n \to \infty$.
Let $k(y_{t+1}|y_{t}) = \text{Normal}_{Y_{t+1}}(\rho y_{t},\sigma^2)$ be the one step transition density. Define higher order powers of $k$ as $k^n(y_{t+n}|y_{t}) = \int k^1(y_{t+n}|y_{t+n-1})k^{n-1}(y_{t+n-1}|y_{t})dy_{t-1}$. We can see that 
$$
k^n(y_{t+n}|y_{t}) = \text{Normal}(\rho^n y_{t}, \sigma^2[1 + \rho^2 + \rho^4 + \cdots + \rho^{2(n-1)}])
$$
because
$$
y_t = \rho^ny_{t-n} + \epsilon_t + \rho \epsilon_{t-1} + \cdots \rho^{n-1}\epsilon_{t-n+1}.
$$
To see that, just keep substituting using your first formula.
As $n\to \infty$, for any $y_{t}$, this goes to a normal distribution with mean $0$ and variance $\sigma^2/(1-\rho^2)$. This is the same distribution as $\pi(\cdot)$, the one we mentioned before.
Also, you don't need Ergodicity to hold to get a law of large numbers. It just needs to be irreducible, and have a stationary distribution (which it does).
