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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.

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  • $\begingroup$ why are $Y$'s identically distributed? $\endgroup$
    – Fedor Petrov
    Commented Jan 24, 2017 at 14:30
  • $\begingroup$ I added a condition on $(X_n)_n$, namely that the correlation between $X_n$ and $X_{n+m}$ doesn't depend on $n$. $\endgroup$
    – HolyMonk
    Commented Jan 24, 2017 at 14:33
  • $\begingroup$ It still seems to be not enough. Assume That $m=3$, $X,T,Z$ are independent i.i.d and your sequence is $ZTT\,TZX\,ZXT\,TZX\, ZXT\,...$. Then $X_n$ and $X_{n+3}$ are always independent, but $Y_1$ and $Y_3$ may have different distribution. $\endgroup$
    – Fedor Petrov
    Commented Jan 24, 2017 at 14:41
  • $\begingroup$ I don't want to have that $X_{n+m}$ and $X_n$ are independent, but I want to have that for every $m$, the map $n \mapsto \mbox{Corr}(X_{n+m}, X_n)$ is constant $\endgroup$
    – HolyMonk
    Commented Jan 24, 2017 at 15:05
  • $\begingroup$ The idea is to work with an ergodic process and view this far enough in time that we can assume that convergence to the stationary distribution has already occurred. $\endgroup$
    – HolyMonk
    Commented Jan 24, 2017 at 15:06

1 Answer 1

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$\newcommand{\E}{\mathbb{E}}$ 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.

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  • $\begingroup$ Nice, didn't think of that trick with $X_0$ $\endgroup$
    – HolyMonk
    Commented Jan 24, 2017 at 20:25

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