##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 $n$. 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.

##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.

##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 $n$, $X_n \sim X$). 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.

##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 $n$. 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.