Sub-Gaussian concentration for reversible Markov chains with spectral gap
Setup.
Let $(X_i)_{i\ge1}$ be a stationary, $\pi$-reversible Markov chain on a measurable space with spectral gap $\gamma>0$.
Let $f:\mathcal X\to\mathbb R$ be square-integrable with respect to $\pi$ and assume the sign condition $f\le 0$ $\pi$-a.s.
Define the empirical average \begin{equation} f_n := \frac1n\sum_{i=1}^n f(X_i), \end{equation} and denote the first two (non-centered) moments \begin{equation} \mu_1 := \int f\,d\pi, \qquad \mu_2 := \int f^2\,d\pi. \end{equation} For convenience, let \begin{equation} v := 2\mu_2 - \mu_1^2. \end{equation}
Goal
I would like to obtain a sharp sub-Gaussian concentration bound for the right tail of $f_n$, i.e. \begin{equation} \mathbb P\!\left(f_n \ge \mu_1 + t\right) \le C\exp\!\left(-\frac{n t^2}{C'\sigma^2}\right), \end{equation} where $\sigma^2$ should depend explicitly on $\mu_2$ and the spectral gap $\gamma$, and interpolate between:
- the i.i.d. case ($\gamma=1$), and
- the poor mixing limit ($\gamma\to0$).
What I have so far
Let \begin{equation} \phi_n(\theta) := \mathbb E\!\left[\exp\!\left(\theta\sum_{i=1}^n f(X_i)\right)\right], \qquad \theta\ge0. \end{equation}
I derived the following upper bounds: \begin{equation} \phi_1(\theta) \le 1 + \mu_1\theta + \tfrac12\mu_2\theta^2, \end{equation} and, for general $n$, \begin{equation} \phi_n(\theta) \le \phi_1(\theta)^n + \phi_1(\theta)^{n-1}\,\mathrm{Var}\!\bigl(e^{\theta f(X_1)}\bigr) \left[\left(1+\frac{1-\gamma}{\phi_1(\theta)}\right)^{n-1} - 1\right]. \end{equation}
For small $\theta>0$, $ \mathrm{Var}\!\bigl(e^{\theta f(X_1)}\bigr)\sim \theta^2\mu_2$, and \begin{equation} \mathrm{Var}\!\bigl(e^{\theta f(X_1)}\bigr) \le \theta^2 v \quad\text{with}\quad v = 2\mu_2 - \mu_1^2. \end{equation}
From $\phi_1$ (or $\gamma=1$), one recovers the standard i.i.d. sub-Gaussian bound via the Chernoff method:
\begin{equation} \mathbb P(f(X_1)\ge\mu_1+t) \le \exp\!\left(-\frac{t^2}{2\mu_2}\right). \end{equation}
Case $n=2$
When $n=2$, the general bound simplifies to \begin{equation} \phi_2(\theta) \le \bigl(1+\mu_1\theta + \tfrac12\mu_2\theta^2\bigr)^2 + (1-\gamma)\theta^2 v. \end{equation}
Using a simple algebraic lemma:
Lemma.
If $a<0$, $b,c>0$, $X\ge0$, and $1+2aX\ge0$, then
$(1+aX+(b+c)X^2)^2 \ge (1+aX+bX^2)^2 + cX^2.$
Applying this with $a=\mu_1$, $b=\mu_2/2$, $c=(1-\gamma)v$, and $\theta\in[0,-1/(2\mu_1)]$ (recall $f\le0\Rightarrow \mu_1\le0$) gives \begin{equation} \phi_2(\theta) \le \bigl(1+\mu_1\theta + \theta^2(\tfrac{\mu_2}{2}+(1-\gamma)v)\bigr)^2 \le \exp\!\bigl(2\mu_1\theta + \theta^2\sigma_2^2\bigr), \end{equation} where \begin{equation} \sigma_2^2 = \mu_2 + 2(1-\gamma)v = \mu_2 + 2(1-\gamma)(2\mu_2 - \mu_1^2). \end{equation}
Hence, by Markov and Chernoff inequalities, \begin{equation} \mathbb P\bigl(f(X_1)+f(X_2)\ge 2\mu_1+2t\bigr) \le \inf_{\theta\in[0,-1/(2\mu_1)]} \exp(-2t\theta + \theta^2\sigma_2^2), \end{equation} yielding a sub-Gaussian tail with variance proxy $\sigma_2^2$.
Question
How can this analysis be extended to general $n$? While the first step using the algebraic lemma is valid for large $n\geq 2$, the variance will itself be a function of $\theta$ as it depends on $\phi_1(\theta)$, making the overall minimisation problem intractable.
Is the bound \begin{equation} \phi_n(\theta)\le \phi_1(\theta)^n + \phi_1(\theta)^{n-1}\,\mathrm{Var}\!\bigl(e^{\theta f(X_1)}\bigr) \left[\left(1+\frac{1-\gamma}{\phi_1(\theta)}\right)^{n-1}-1\right] \end{equation} sharp enough (for small $\theta$) to imply a sub-Gaussian bound of the form \begin{equation} \phi_n(\theta)\le \exp\!\bigl(n\mu_1\theta + n\theta^2 \tilde\sigma_n^2\bigr), \end{equation} leading to $\mathbb P(f_n\ge\mu_1+t)\le\exp(-n t^2/(C\tilde\sigma_n^2))$? If so, what would be an explicit expression for $\tilde\sigma_n^2$?
If not, what would be a better approach?
Any references or ideas showing how to bridge from this MGF bound to a clean sub-Gaussian concentration would be very helpful.