Why a Diagonal Covariance Matrix Does Not Prove Exchangeability
Summary
Exchangeability requires a random vector’s joint distribution to remain unchanged when its components are permuted. The discussion asks whether a diagonal covariance matrix is enough to establish that property and shows why it is not: zero covariances do not imply that the individual components have identical distributions. For example, independent variables can have different fixed values or distributions while their covariance remains zero.
For a jointly normal vector, the answer adds that exchangeability requires the covariance structure to be invariant under every permutation. This is stronger than ordinary matrix symmetry: the covariance entries must be interchangeable, so distinct lags cannot have distinct covariance values. The normality qualification matters because the covariance matrix alone does not describe a general non-normal joint distribution. The exchangeability criterion is a useful statistical concept for reasoning about time series and samples, but the brief answers do not provide a complete general characterization for non-normal variables.
Key ideas
- Exchangeability means the joint distribution is unchanged by every permutation of components.
- A diagonal covariance matrix alone does not imply that components have identical distributions.
- For jointly normal variables, permutation invariance requires the covariance structure to remain unchanged under permutations.
- Covariance-based conclusions may not extend to non-normal distributions.
Tags
Full text
# Exchangeability of random vector
# Exchangeability of random vector
I hope you can help me with this rather basic question that I asked myself. A random vector $(X_1,...,X_n)$ is said to be exchangeable if it has the same distribution as the permuted random vector $(X_{\pi(1)},...,X_{\pi(d)})$, for any permutation $(\pi{(1)},...\pi{(d)})$.
Can you say that this is the case if and only if the covariance matrix of the random vector $(X_1,...,X_n)$ is symmetric (edit: diagonal not symmetric)? Or is that not sufficient? I can't think of a counter example...
Thanks already, any hints are very much appreciated! :-)
## Answer by mark leeds (score 2, accepted)
https://quant.stackexchange.com/a/57229
@Wombat: I think it's best to think of it this way. Suppose you have a time series and the joint distribution of $n$ elements ( in a certain, specific order) of the series is normal with mean zero and some non-diagonal covariance matrix $\Sigma$.
Then, in this case, exchangeability means that you could take the $n$ elements in any order and the joint distribution of the $n$ elements stays the same. Note that the explanation below would only be true for the normal assumption case. In non-normal cases, maybe symmetric is needed. I don't know.
Anyway, getting back to the normal case, you would have a stronger condition than symmetry. Basically, you'll need all of the elements of the covariance matrix $\Sigma$ to be interchangeable without effecting the distribution. This implies a constant covariance value in each element of the matrix. That's stronger than symmetry because different lags have the same covariance.
## Answer by Valometrics.com (score 1)
https://quant.stackexchange.com/a/57226
let's take X1=1 and X2=2:
The covariance matrix is zero (X1 and X2 are independents) and X1, X2 have different distributions!!!Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.