How is the standard deviation of VAE's obtained?
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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 yesterday
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 yesterday
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
yesterday
$begingroup$
@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
yesterday
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 yesterday
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 yesterday
DuttaADuttaA
506318
asked yesterday
DuttaADuttaA
506318
edited 19 hours ago
DuttaA
asked yesterday
DuttaADuttaA
506318
asked yesterday
DuttaADuttaA
506318
asked yesterday
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday