How can NN be thought of as p(x|z) in VAEs?












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So, you can feed random input (z) to a NN and make the output random, but that will be just f(p(z)), where f(.) is the deterministic NN. The other way is to think of the NN output as parameteres of a certain distribution and as such, the NN is P(x|z). BUT, that's not happening, the decoder of a VAE is producing an image (x), and it is not random! so, for a given z, it will produce the same x, all the time. Why do they think of it as P(x|z) in the derivations of VAEs??









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    So, you can feed random input (z) to a NN and make the output random, but that will be just f(p(z)), where f(.) is the deterministic NN. The other way is to think of the NN output as parameteres of a certain distribution and as such, the NN is P(x|z). BUT, that's not happening, the decoder of a VAE is producing an image (x), and it is not random! so, for a given z, it will produce the same x, all the time. Why do they think of it as P(x|z) in the derivations of VAEs??









    share







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    Alex Deft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      $begingroup$


      So, you can feed random input (z) to a NN and make the output random, but that will be just f(p(z)), where f(.) is the deterministic NN. The other way is to think of the NN output as parameteres of a certain distribution and as such, the NN is P(x|z). BUT, that's not happening, the decoder of a VAE is producing an image (x), and it is not random! so, for a given z, it will produce the same x, all the time. Why do they think of it as P(x|z) in the derivations of VAEs??









      share







      New contributor




      Alex Deft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      So, you can feed random input (z) to a NN and make the output random, but that will be just f(p(z)), where f(.) is the deterministic NN. The other way is to think of the NN output as parameteres of a certain distribution and as such, the NN is P(x|z). BUT, that's not happening, the decoder of a VAE is producing an image (x), and it is not random! so, for a given z, it will produce the same x, all the time. Why do they think of it as P(x|z) in the derivations of VAEs??







      machine-learning deep-learning keras





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      asked 1 min ago









      Alex DeftAlex Deft

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