3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Use of a running mean indicates that the sequencing of the variables have meaning, maybe time? To maintain exchangeability I would guess that at least you must treat the variables symmetrically, and that excludes a running mean. So probably the answer is no. $\endgroup$ – kjetil b halvorsen Mar 29 '18 at 22:13
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.