Exchangeability and data smoothing If a non-i.i.d sequence of a continuous random variable that is exchangeable, is smoothed by taking rolling average, is the resulting sequence exchangeable? My intuition suggests that it is not. I'll appreciate solid explanations.
EDIT: My first question was about non-i.i.d data but I am wondering if exchangeability will hold on data smoothing even for i.i.d?
 A: Let $X=(X_1, \dotsc,X_n)$ be an exchangeable random vector. Then, specifically, its variance-covariance matrix (if it exists) must be (proportional to)
$$
  \Sigma = \begin{pmatrix} 1    & \rho & \dotsm & \rho \\
                           \rho & 1    & \dotsm & \rho \\
                             \ddots \\
                           \rho & \rho & \dotsm & 1 
            \end{pmatrix}
$$ with $\rho \ge -\frac1{n-1}$.
Let us transform $X$ linearly to get $m$ new variables $Y_j$:
$$
    Y = \begin{pmatrix} Y_1^T \\
                        Y_2^T \\
                        \vdots \\
                         Y_m^T
       \end{pmatrix} = A X
$$ where matrix $A$ has the vectors $a_i$ as rows. Then we can calculate
$$ \DeclareMathOperator{\cov}{\mathbb{C}ov}
   \cov AX = A \Sigma A^T =\Gamma
$$
And for $AX$ to be exchangeable it is necessary that all the diagonal elements $\Gamma_{ii}$ are equal and all the off-diagonal elements $\Gamma_{ij}, i\not= i$ are equal. Calculate
$$
   \Gamma_{ii} = a_i^T \Sigma a_j = \sum_{k,l} a_{1,k}\sigma_{k,l} a_{j,l} = \sum_k a_{i,k}^2 + \rho \sum_{k\not= l}a_{i,k}a_{i,l} \text{and} \\
\Gamma_{ij}=a_i^T \Sigma a_j = \sum_{k,l} a_{i,k} \sigma_{k,l} a_{j,l}=\sum_k a_{i,k} a_{j,l} +\rho\sum_{k\not=l} a_{i,k}a_{j,l}
$$
If this is to hold for all values of $\rho$, it must specifically hold for $\rho=0$ so that we get
$\Gamma_{ii}= \|a_i\|^2$ must be a constant and also $\Gamma_{ij} =a_i \cdot a_j$ also a constant. This would be necessary, but not sufficient, conditions for exchangeability of the $AX$. That would rule out the running mean you propose. 
