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Apparently, no. This is a generalization where the relationship between $k$ and $i$ is stochastic (i.e. defined by a distribution and not deterministic). To recover the original bits-back argument, we would need to pick a value $i_k$ for each possible value of $k$ and force $p(i=i_k \vert x, k)=1$ and 0 if $i \neq i_k$. Naturally, we would need to extend the ...

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