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2019 Community Moderator ElectionHow to predict the probability of an event?Behavioral Differences between Standard Autoencoder and Variational AutoencoderHow are the positions of the output nodes determined in the Kohonen - Self Organizing Maps algorithm?How do i use the Gaussian function with a Naive Bayes Classifier?Right Way to Input Text Data in Keras Auto Encoderwhat's the difference between autoencoder and autoassociative neural networks?Wasserstein distance between Gaussian and the empirical distributionHow to convert an array of numbers into probability values?Help solving Bigram Model with the following probabilitiesHow to use Adaptive Rejection Sampling to Update Alpha in iGMM
$begingroup$
I am trying to build a Variational Autoencoder. I was looking at various codes online and found most of them in some way or another copy Francois Chollet (Google researchers) code.
Now my main question with this code is this part:
As you can clearly see the $log(sigma)$ is the output from a Dense layer. Where did this assumption come from (the output is $log(sigma)$ and not $sigma$? How is it possible that we generate a random normal distribution with standard deviation $sigma$ like this? Is it due to the way the computer generates Normal Distributions?
probability autoencoder gaussian
asked 15 hours ago
DuttaADuttaA
506318
$endgroup$
add a comment |
$begingroup$
I am trying to build a Variational Autoencoder. I was looking at various codes online and found most of them in some way or another copy Francois Chollet (Google researchers) code.
Now my main question with this code is this part:
As you can clearly see the $log(sigma)$ is the output from a Dense layer. Where did this assumption come from (the output is $log(sigma)$ and not $sigma$? How is it possible that we generate a random normal distribution with standard deviation $sigma$ like this? Is it due to the way the computer generates Normal Distributions?
probability autoencoder gaussian
asked 15 hours ago
DuttaADuttaA
506318
$endgroup$
$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago
add a comment |
$begingroup$
I am trying to build a Variational Autoencoder. I was looking at various codes online and found most of them in some way or another copy Francois Chollet (Google researchers) code.
Now my main question with this code is this part:
As you can clearly see the $log(sigma)$ is the output from a Dense layer. Where did this assumption come from (the output is $log(sigma)$ and not $sigma$? How is it possible that we generate a random normal distribution with standard deviation $sigma$ like this? Is it due to the way the computer generates Normal Distributions?
probability autoencoder gaussian
asked 15 hours ago
DuttaADuttaA
506318
$endgroup$
I am trying to build a Variational Autoencoder. I was looking at various codes online and found most of them in some way or another copy Francois Chollet (Google researchers) code.
Now my main question with this code is this part:
As you can clearly see the $log(sigma)$ is the output from a Dense layer. Where did this assumption come from (the output is $log(sigma)$ and not $sigma$? How is it possible that we generate a random normal distribution with standard deviation $sigma$ like this? Is it due to the way the computer generates Normal Distributions?
probability autoencoder gaussian
probability autoencoder gaussian
asked 15 hours ago
DuttaADuttaA
506318
asked 15 hours ago
DuttaADuttaA
506318
edited 14 hours ago
DuttaA
asked 15 hours ago
DuttaADuttaA
506318
asked 15 hours ago
DuttaADuttaA
506318
asked 15 hours ago
DuttaADuttaA
506318
506318
$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago
add a comment |
$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago
add a comment |
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$begingroup$
@Esmailian you can see that K_log_sigma = dense(latent_dim) which is the same as z_mean taken in the above line
$endgroup$
– DuttaA
14 hours ago
$begingroup$
@Esmailian ok sorry..i'll edit that part out but the question will now be why is it $log(sigma)$ and not $sigma$
$endgroup$
– DuttaA
14 hours ago