General Statement
-----------------


Suppose we have a sequence of identically distributed but dependent random variables $(X\_n)\_{n\in \mathbb{N}}$ which take values on $\{0,\dots,m\}$ for some $m \in \mathbb{N}$ (suppose for all $n$, $X\_n \sim X$). Assume further that the $(X\_n)\_n$ are such that the correlation between $X\_{n}$ and $X\_{n+m}$ is independent on $n$ for all fixed $m$. We define the sequence $(Y\_n)\_n$ by:
$$
Y\_n := \sum\_{k=0}^{m-1} \delta\{X\_{n+k} > k\},
$$
where we use the notation 
$$\delta\{X\_n > n\} := \begin{cases} 1 & \mbox{ if } X\_n > n \\ 0 & \mbox{ otherwise} \end{cases}.$$
Note that the sequence $(Y\_n)\_n$ are also identically distributed and dependent (use the notation $Y$ with for all $n$: $Y\_n \sim Y$).
I have reason to believe that:
$$
\mbox{Var}(Y) \leq \mbox{Var}(X)
$$


My special case
---------------


In the case I'm interested in, $(X\_n)\_n$ is a Markov Process.


Motivation
----------


In case we only have $2$ possible values for $X$, say that it takes values in $\{0,m\}$, we have $\delta\{X\_n > n\} = X\_n/m$ and thus we find:
$$
Y = \sum\_{k=0}^{m-1} \frac{X\_k}{m}
$$
which incurrs:
\begin{align\*}
\mbox{Var}(Y)
&= \frac{1}{m^2}\sum\_{k,l=0}^{m-1} \mbox{Cov}(X\_k,X\_l)\\
&= \frac{1}{m^2}\sum\_{k,l=0}^{m-1} \mbox{Var}(X)\mbox{Corr}(X\_k,X\_l)\\
&\leq \mbox{Var}(X).
\end{align\*}
we see that the statement indeed holds, moreover we didn't use that $(X\_n)\_n$ is a Markov Chain. In the more dimensional case I have done simulations which all seem to be indicating that this inequality seems to hold in general but I haven't been able to prove it.



Show that $\mbox{Var}(\sum_{k=0}^{\infty} \delta\{L_{t-k} > k\}) \leq \mbox{Var}(L)$
$\newcommand{\E}{\mathbb{E}}$
You practically answered your own question.


Denote by $A\_k$ the event "$X\_0 > k$". Then $X\_0=\sum\_{k=0}^{m-1} \delta(A\_k)$.


Denote by $B\_k$ the event "$X\_k > k$". Then $Y\_0=\sum\_{k=0}^{m-1} \delta(B\_k)$.


$A\_k$ and $B\_k$ have the same probabilities, hence $\E[X\_0]=\E[Y\_0]$. We need to show that $\E[X^2\_0]\ge \E[Y^2\_0]$.


But $\E[\delta(A\_k)\delta(A\_l)]\ge \E[\delta(B\_k)\delta(B\_l)]$ simply because $A\_k \subset A\_l$ (for $k<l$). In other words the correlation is maximized when the sequence of sets is monotone.


The only requirement is that the $X\_n$'s are identically distributed, they sequence doesn't have to be shift invariant. In the case of Markov chain, you need to start from the stationary distribution.


