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??
machine-learning deep-learning keras
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Alex DeftAlex Deft
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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??
machine-learning deep-learning keras
asked 1 min ago
Alex DeftAlex Deft
1
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.
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add a comment |
$begingroup$
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??
machine-learning deep-learning keras
asked 1 min ago
Alex DeftAlex Deft
1
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
machine-learning deep-learning keras
asked 1 min ago
Alex DeftAlex Deft
1
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.
asked 1 min ago
Alex DeftAlex Deft
1
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.
asked 1 min ago
Alex DeftAlex Deft
1
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.
asked 1 min ago
Alex DeftAlex Deft
1
asked 1 min ago
Alex DeftAlex Deft
1
1
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.
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.
Alex Deft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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