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adityaramesh avatar adityaramesh commented on August 15, 2024 3

This is just the variational lower bound. It's straightforward to derive from Jensen's inequality, see e.g. this document.

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AaronHeee avatar AaronHeee commented on August 15, 2024 3

Hi @adityaramesh, thanks for your reply! Sorry that I am still confused how to use Jensen's inequality to derive Eq 1. My attempt is listed as below, but I am blocked at the last step -- do I miss something? Looking forward to your clarification, thanks!

image

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finbarrtimbers avatar finbarrtimbers commented on August 15, 2024 1

Hey folks- I was also confused by this, so I wrote up the derivation. I'd appreciate a second check over my math.

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kmaziarz avatar kmaziarz commented on August 15, 2024

Hi @AaronHeee, I'm also confused about the KL term. In particular, it seems odd that the KL term is under the expectation over z, as at that point all three variables (x, y and z) are already bound.

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j-cyoung avatar j-cyoung commented on August 15, 2024

Hey folks- I was also confused by this, so I wrote up the derivation. I'd appreciate a second check over my math.

Thanks for your sharing. But I'm confused about the exact form of distribution $q_\Phi$, which seems be either $q_\Phi(y,z|x)$ or $q_\Phi(z)$ in your derivation, while it is $q_\Phi(z|x)$ in the original formula.

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ahmdtaha avatar ahmdtaha commented on August 15, 2024

I have been looking into this equation for sometime. I wrote my thoughts in the attached main.pdf. I believe we need to convert a single-variable distribution $q_{\phi}(z|x)$ into a multi-variable distribution $q_{\phi}(y,z|x)$ to derive this equation. Yet, I am not sure if I made the right assumptions. I hope someone look into my derivation and share some feedback.

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blackcherry88 avatar blackcherry88 commented on August 15, 2024

there is a assumption that $y \text{, } z \text{ in } q_{\phi} \text{ are indepedent given x}$

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