Comparing mixing time of lazy and non-lazy Markov chains Suppose we have a probability distribution $\pi : X \rightarrow [0,1]$ where $X$ is finite and let $Q : X \times X \rightarrow [0,1]$ be a Markov kernel that is reversible with respect to $\pi$. That is,
$$\pi(x) Q(x,y) = \pi(y) Q (y,x) $$
for all $x,y \in X$. Suppose we know the mixing time $t_x(Q, \varepsilon)$ of $Q$ to $\pi$ when started at $x$, defined as
$$ t_x(Q, \varepsilon) = \min \{ t \in \mathbb{N} : || Q^t(x, \cdot) - \pi ||_1 \leq \varepsilon \}. $$
Question: what can be said about the mixing time of the non-lazy version of this kernel?
That is, we can define $\tilde{Q}(x,y) = 0$ if $x=y$, and $\tilde{Q}(x,y)\propto Q(x,y)$ otherwise. Clearly $\tilde{Q}$ is still reversible with respect to $\pi$ and so has the same stationary distribution. So we can consider $t_x(\tilde{Q}, \varepsilon)$ and ask whether it is smaller (and by how much) than $t_x(Q, \varepsilon)$.
If anyone knew how to compare the two chains spectral gaps or log-Sobolev constants I would be particularly interested in that.
Motivation: I have a distribution $\pi$ and a Markov kernel $Q$ that mixes to $\pi$ quite slowly. However $Q $ is very often lazy, i.e. $Q(x,x)$ is close to $1$ for most $x$'s. I was hoping that there might be a way to show that the non-lazy version of my kernel mixes faster. 
 A: 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.  $\quad\square$
