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If the Markov chain converged then $$\pi = Q* \pi$$where $ \pi$ is the posterior distribution and $Q$ is the transition distribution(it's a matrix in the discrete case).
I tried to apply that on the continuous case of Markov chain in this example where
the transition distribution is:
$$p(X_{t+1} | X_t=x_t) = \text{N}(\phi x_t, 1)$$ and the posterior(stationary) distribution is $$X_t \sim \text{N} \Bigg( 0, \frac{1}{1-\phi^2} \Bigg)$$
The product of them both doesn't equal the posterior.
That stationary distribution is correct. Using the law of total probability, you have:
$$\begin{equation} \begin{aligned} p(T_{t+1} = x) &= \int \limits_\mathbb{R} p(X_{t+1} = x | X_t = r) \cdot p(X_t = r) \ dr \\[6pt] &= \int \limits_{-\infty}^\infty \text{N}(x | \phi r, 1) \cdot \text{N} \bigg( r \bigg| 0, \frac{1}{1-\phi^2} \bigg) \ dr \\[6pt] &= \int \limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp \bigg( -\frac{1}{2} (x - \phi r)^2 \bigg) \cdot \sqrt{\frac{1-\phi^2}{2 \pi}} \exp \bigg( -\frac{1}{2} (1-\phi^2) r^2 \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{2 \pi} \int \limits_{-\infty}^\infty \exp \bigg( - \frac{1}{2} (x - \phi r)^2 - \frac{1}{2} (1-\phi^2) r^2 \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{2 \pi} \int \limits_{-\infty}^\infty \exp \bigg( - \frac{1}{2} \bigg[ (x - \phi r)^2 + (1-\phi^2) r^2 \bigg] \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{2 \pi} \int \limits_{-\infty}^\infty \exp \bigg( - \frac{1}{2} \bigg[ x^2 - 2 \phi x r + \phi^2 r^2 + r^2 - \phi^2 r^2 \bigg] \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{2 \pi} \int \limits_{-\infty}^\infty \exp \bigg( - \frac{1}{2} \bigg[ x^2 - 2 \phi x r + r^2 \bigg] \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{2 \pi} \int \limits_{-\infty}^\infty \exp \bigg( - \frac{1}{2} \bigg[ x^2 (1 - \phi^2) + (r-\phi x)^2 \bigg] \bigg) \ dr \\[6pt] &= \frac{\sqrt{1-\phi^2}}{\sqrt{2 \pi}} \exp \bigg( -\frac{1}{2} (1-\phi^2) x^2 \bigg) \int \limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp \bigg( - \frac{1}{2} (r-\phi x)^2 \bigg) \ dr \\[6pt] &= \text{N} \bigg( x \bigg| 0, \frac{1}{1-\phi^2} \bigg) \times \int \limits_{-\infty}^\infty \text{N} (r|\phi x,1) \ dr \\[6pt] &= \text{N} \bigg( x \bigg| 0, \frac{1}{1-\phi^2} \bigg). \\[6pt] \end{aligned} \end{equation}$$
