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Why is the ELBO of $p(x)=\int p(x|z)p(z)\mathrm{d}z$ easier to compute/estimate than the expression itself? Can we compute this quantity itself through sampling in the same way? I understanding that aggregating over data means taking log before summing over, but does it create any complication?

Hanhan Li
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To calculate $p(x) = \int p(x|z) p(z) dz$, you have to calculate it with all configurations of $p(x|z)$ and $p(z)$, which scales exponentially with time. Thus, it is easier to estimate $p(x)$ than to calculate it correctly.

Minh-Long Luu
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