This is a preliminary version.
Just a short proof that the stationary distributin is unique (under the right assumptions).
I’ll use the same symbols for distributions the their underlying *pdf*s and which of both
is meant can be inferred from context.

## Uniqueness Proof

Suppose we have two stationary distributions $\pi_0, \pi_1$ of some Markov chain $K$, i.e. for $i=0,1$ we have

\[\pi_i(y) = \int_X \pi_i(x) \ K(x,y) \ dx.\]Now consider their ratio $\frac{\pi_0}{\pi_1}$ and assume it’s minimized at $x’$ (the minimum exists if the underlying space is compact for instance), i.e.

\[x' := \mathop{\arg \min}_{x} \frac{\pi_0(x)}{\pi_1(x)}.\]With this in hand we clearly have $\pi_0(x’) = \frac{\pi_0(x’)}{\pi_1(x’)} \pi_1(x’)$ (basically multiplying by a “nutritious” one.). Computing the difference and applying the characteristic property given in the first equation above yields

\[0 \ = \ \pi_0(x') - \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x') \ = \ \int \underbrace{ K(x,x')}_{>0} \cdot \underbrace{ \Big[ \pi_0(x) - \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x) \Big]}_{\geq 0 \ \text{(since $x'$ is min)}} \ dx.\]By definition of $x’$ as the minimizer of the quotient we conclude that second integrand must be non-negative. Assuming the transition kernel isn’t vanishing it implies further that the second integrand must in fact be zero, and thus

\[\pi_0(x) = \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x).\]Therefore both candidates merely differ by a scaling factor. However since we deal with normalized probability densities here we can deduce that the scaling factor is $1$ which concludes the uniqueness proof.