Why is the mixing time defined with $\frac{1}{4}$?

Question

Let $\{X_t\}_{t \in \mathbb{N}_0}$ be a discrete-time Markov chain with (countable) state space $\Omega$ and stationary distribution $\pi$.

Let $x,y \in \Omega$ and $P^t(x,y)$ be the probability of moving to state $y$ in $t$ time steps, given that we start at state $x$. The distance (using the total variation distance) between the distribution of the chain at time $t$ and the stationary distribution is given by $$d(t) = \max_{ x \in \Omega} ||P^t(x, \cdot)- \pi||_{TV},$$ where $$P^t(x,y):=\mathbb{P}(X_t=y \ | \ X_0=x).$$

Then, we define for $0< \varepsilon < 1$,

$$t_{mix}(\varepsilon) = \min\{t : d(t) \leq \varepsilon\}$$

and call it the mixing time of the Markov chain when $\varepsilon := \frac{1}{4}$.

My question is fairly simple, but I can't seem to find the answer anywhere:

Why is it used $\varepsilon:=\frac{1}{4}$? (and not $\varepsilon:=\frac{1}{5}$ for example)

Answer 1

We know that for a well-behaved Markov chain $\|P^t(x, \cdot) - \pi\|_{\text{TV}}$ decays exponentially: it is bounded by $C \alpha^t$ for constants $C>0$, $0<\alpha<1$. So $t_{\text{mix}}(\frac14)$ and $t_{\text{mix}}(\frac15)$ and $t_{\text{mix}}(\epsilon)$ for any constant $\epsilon>0$ are off from each other at worst by a constant factor, in that case. To some extent, picking $\frac14$ is arbitrary.

If we could pick $\frac12$, we'd definitely pick $\frac12$. There's problems with that choice: $\epsilon = \frac12$ is sufficiently little mixing that it's possible to achieve even if the Markov chain is not irreducible (and not planning on approaching $\pi$ in the first place). On the other hand, $t_{\text{mix}}(\epsilon)$ is infinite for all $\epsilon<\frac12$ if, say, you're taking a random walk on a disconnected graph.

(Relatedly, there's the bound on mixing time in terms of the conductance $\Phi^*$: $t_{\text{mix}}(\frac14) \ge \frac1{2\Phi^*}$. In general, $t_{\text{mix}}(\epsilon) \ge \frac{1 - 2\epsilon}{\Phi^*}$, which is only useful for $\epsilon < \frac12$.)

Since $\frac12$ is out, we decided to go with $\frac14$ at some point. I'm not aware of any reason why $\frac14$ is better than $\frac13$ or $\frac15$, except subjective preference and custom.