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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?
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.
