**Edit [following the comment by R W]**: As *R W* pointed out, the stationary distribution may change if you remove the laziness as you suggested. Still your motivation makes sense: can making the chain less lazy reduce the mixing significantly faster? What I wrote below answers that.

The "non-lazy version" of an aperiodic Markov kernel is not always aperiodic, so the convergence may fail.

Still for your motivation, you could ask whether "less-lazy versions" of $Q$ mix significantly faster. ~~I don't have a general answer, but~~ I argue that ~~at least the simplest form of~~ "non-lazification" does not help beyond the obvious linear speed-up. Below, I use the better definition
\begin{align*}
t_x(Q,\varepsilon) &:=
\min\{t_0\in\mathbb{N}:
\text{$\|Q^t(x,\cdot)-\pi\|<\varepsilon$ for all
$t\geq t_0$}
\} \;.
\end{align*}

Let $\hat{Q}$ be another irreducible and aperiodic Markov kernel with the property that
\begin{align*}
Q(x,y) &= p\hat{Q}(x,y)+(1-p)1_x(y)
\end{align*}
for some $0<p<1$. This is a *simple* "less-lazy version" of $Q$. The evolutions of $Q$ and $\hat{Q}$ can be coupled in a natural way as follows: let $\hat{Z}_0,\hat{Z}_1,\ldots$ be a Markov chain with kernel $\hat{Q}$ and let $B_1,B_2,\ldots$ be a sequence of independent Bernoulli random variables with $\mathbb{P}(B_k=1)=p$ and independent of $\hat{Z}_0,\hat{Z}_1,\ldots$. Let $N_t:=B_1+B_2+\cdots+B_t$. Define $Z_t:= \hat{Z}_{N_t}$. It is clear that $Z_0,Z_1,\ldots$ is a Markov chain with kernel $Q$.

**Lemma**.
Let $0<\varepsilon,\delta<1$ be arbitrary.
Then, for $t\geq\frac{1}{2\delta^2}\log\frac{1}{\varepsilon}$, we have $\mathbb{P}\big(N_t<(p-\delta)t\big)<\varepsilon$.

*Proof*. This is just rewriting Hoeffding's inequality. $\quad\square$.

**Proposition**. Let $0<\delta<p$ and $0<\gamma<1$ be arbitrary. Then, for every $0<\varepsilon<1$, we have
\begin{align*}
t_x(Q,(1+\gamma)\varepsilon) &\leq
\max\left(\frac{1}{p-\delta}t_x(\hat{Q},\varepsilon), \frac{1}{2\delta^2}\log\frac{1}{\gamma\varepsilon}\right) \;.
\end{align*}
So, choosing $\delta$ and $\gamma$ to be small, and ignoring the second term on the right-hand side (which is independent of the size of the state space), we find that the mixing time of $\hat{Q}$ is no less than about $p$ times the mixing time of $Q$. This is the linear speed-up one would expect.

*Proof*.
Let $n_0$ be such that
\begin{align*}
\|\hat{Q}^n(x,\cdot)-\pi\| &< \varepsilon
\end{align*}
for every $n\geq n_0$. Let $t_0:=\max(\frac{n_0}{p-\delta}, \frac{1}{2\delta^2}\log\frac{1}{\gamma\varepsilon})$. According to the above lemma, $\mathbb{P}(N_t<n_0)\leq\gamma\varepsilon$ for all $t\geq t_0$. Note that
\begin{align*}
Q^t(x,y) &= \sum_{n=0}^\infty\mathbb{P}(N_t=n)\hat{Q}^n(x,y) \;.
\end{align*}
Therefore, for every $t\geq t_0$,
\begin{align*}
\|Q^t(x,\cdot)-\pi\| &<
\|\sum_n\mathbb{P}(N_t=n)\hat{Q}^n(x,\cdot) - \pi\| \\
&\leq
\sum_{n<n_0}\mathbb{P}(N_t=n)\underbrace{\|\hat{Q}^n(x,\cdot) - \pi\|}_{\leq 1} + \sum_{n\geq n_0}\mathbb{P}(N_t=n)\underbrace{\|\hat{Q}^n(x,\cdot) - \pi\|}_{\leq\varepsilon} \\
&\leq
\mathbb{P}(N_t<n_0) + \varepsilon \\
&\leq
(1+\gamma)\varepsilon \;.
\end{align*}
This proves the claim. $\quad\square$

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