What does E mean in the GAN equation?
0
$begingroup$
I don't understand what "E" means in the GAN equation.
It would be great, if someone could explain what it means. Is it entropy or expected value or something else? I couldn't manage to find the answer in the original article.
machine-learning gan generative-models
edited 2 mins ago
Ethan
606324
asked 7 hours ago
PetroPetro
1
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add a comment |
0
$begingroup$
I don't understand what "E" means in the GAN equation.
It would be great, if someone could explain what it means. Is it entropy or expected value or something else? I couldn't manage to find the answer in the original article.
machine-learning gan generative-models
edited 2 mins ago
Ethan
606324
asked 7 hours ago
PetroPetro
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago
add a comment |
0
0
0
$begingroup$
I don't understand what "E" means in the GAN equation.
It would be great, if someone could explain what it means. Is it entropy or expected value or something else? I couldn't manage to find the answer in the original article.
machine-learning gan generative-models
edited 2 mins ago
Ethan
606324
asked 7 hours ago
PetroPetro
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I don't understand what "E" means in the GAN equation.
It would be great, if someone could explain what it means. Is it entropy or expected value or something else? I couldn't manage to find the answer in the original article.
machine-learning gan generative-models
machine-learning gan generative-models
edited 2 mins ago
Ethan
606324
asked 7 hours ago
PetroPetro
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 mins ago
Ethan
606324
asked 7 hours ago
PetroPetro
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 mins ago
Ethan
606324
edited 2 mins ago
Ethan
606324
edited 2 mins ago
Ethan
606324
606324
asked 7 hours ago
PetroPetro
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
PetroPetro
1
asked 7 hours ago
PetroPetro
1
1
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Petro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago
add a comment |
1
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago
1
1
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago
add a comment |
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1
$begingroup$
Looks like the statistical expectation when x (y) is specified by the prior distribution p(x) (p(y))?
$endgroup$
– bradS
7 hours ago