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Assume a random vector $\mathbf{x}=(x_1,\ldots,x_n)^\top$ that has finite second moments, i.e., $$\int\mathbf{x}\mathbf{x}^\top\rho(\mathbf{x})\,\text{d}\mathbf{x} < \infty.$$ Does it follow that also the random vector $\mathbf{x}_{1:k}\,|\,\mathbf{x}_{k+1:n}$, $1<k<n$, has finite second moments?

I found an interesting answer of Ben in Deriving the conditional distributions of a multivariate normal distribution. However, I am still not sure if that holds for the case described here.

I would be grateful for references and/or short explanations.

Thank you!

MTP
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    Consider the contrapositive: if one of those conditional distributions has divergent second moment, consider what that implies about the second moment of the joint distribution. – whuber May 19 '19 at 13:22
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    Hhmm, you are right. Writing $$\rho(\mathbf{x})=\rho(\mathbf{x}_{1:k}\,|\,\mathbf{x}_{k+1:n})\,\rho(\mathbf{x}_{k+1:n})$$ implies that also the joint distribution does not have second moment which is contradictory. Thanks! – MTP May 19 '19 at 15:30
  • @whuber Writing down details made me unsure again. Let us write $$\infty >\int\mathbf{x}\mathbf{x}^\top\rho(\mathbf{x})\,\text{d}\mathbf{x} = \int\left(\int\mathbf{x}\mathbf{x}^\top\rho(\mathbf{x}_{1:k}|\mathbf{x}_{k+1:n})\,\text{d}\mathbf{x}_{1:k}\right) \rho(\mathbf{x}_{k+1:n})\,\text{d}\mathbf{x}_{k+1:n}.$$ As far as I see, the fact that the whole (left) integral is finite does _not_ imply that the inner integral on the right hand side, as a function of $\mathbf{x}_{k+1:n}$, is finite _for every_ $\mathbf{x}_{k+1:n}$, which is what I wanted to show. Or, maybe, I am overseeing something? – MTP May 25 '19 at 08:26
  • You do not need to show the inner integral is finite everywhere. This is where considering the contrapositive helps: when you assume some conditional distributions have infinite second moments, you realize this is OK only provided this occurs at a set of measure zero, for otherwise the full joint distribution must have some infinite second moments. – whuber May 25 '19 at 19:27

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