What is the interpretation of the expectation notation in the GAN formulation?
$begingroup$
I'm confused about the expectation notation in the context of GAN loss functions.
The GAN loss for the discriminator is binary cross-entropy. ie: is this real or not.
real = $D(x)$ (ie: give the discriminator a real image)
fake = $D(G(z))$ (ie: generate a fake image and ask discriminator what it is)
Then the binary crossentropy is:
$$log(p) - log (1-p)$$
When used as a GAN loss we replace p with either "real class" or "fake class".
$$log(real) - log (1-fake)=\
log(D(x)) - log (1-D(G(z)))$$
So far, this is ok (i think haha).
But the actual formulation adds an expectation sign... which I don't understand why it's there.
$$E_{x~data}log(D(x)) - E_z log (1-D(G(z)))$$
machine-learning deep-learning statistics generative-models
asked 1 hour ago
xelaxela
14
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$endgroup$
add a comment |
$begingroup$
I'm confused about the expectation notation in the context of GAN loss functions.
The GAN loss for the discriminator is binary cross-entropy. ie: is this real or not.
real = $D(x)$ (ie: give the discriminator a real image)
fake = $D(G(z))$ (ie: generate a fake image and ask discriminator what it is)
Then the binary crossentropy is:
$$log(p) - log (1-p)$$
When used as a GAN loss we replace p with either "real class" or "fake class".
$$log(real) - log (1-fake)=\
log(D(x)) - log (1-D(G(z)))$$
So far, this is ok (i think haha).
But the actual formulation adds an expectation sign... which I don't understand why it's there.
$$E_{x~data}log(D(x)) - E_z log (1-D(G(z)))$$
machine-learning deep-learning statistics generative-models
asked 1 hour ago
xelaxela
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm confused about the expectation notation in the context of GAN loss functions.
The GAN loss for the discriminator is binary cross-entropy. ie: is this real or not.
real = $D(x)$ (ie: give the discriminator a real image)
fake = $D(G(z))$ (ie: generate a fake image and ask discriminator what it is)
Then the binary crossentropy is:
$$log(p) - log (1-p)$$
When used as a GAN loss we replace p with either "real class" or "fake class".
$$log(real) - log (1-fake)=\
log(D(x)) - log (1-D(G(z)))$$
So far, this is ok (i think haha).
But the actual formulation adds an expectation sign... which I don't understand why it's there.
$$E_{x~data}log(D(x)) - E_z log (1-D(G(z)))$$
machine-learning deep-learning statistics generative-models
asked 1 hour ago
xelaxela
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm confused about the expectation notation in the context of GAN loss functions.
The GAN loss for the discriminator is binary cross-entropy. ie: is this real or not.
real = $D(x)$ (ie: give the discriminator a real image)
fake = $D(G(z))$ (ie: generate a fake image and ask discriminator what it is)
Then the binary crossentropy is:
$$log(p) - log (1-p)$$
When used as a GAN loss we replace p with either "real class" or "fake class".
$$log(real) - log (1-fake)=\
log(D(x)) - log (1-D(G(z)))$$
So far, this is ok (i think haha).
But the actual formulation adds an expectation sign... which I don't understand why it's there.
$$E_{x~data}log(D(x)) - E_z log (1-D(G(z)))$$
machine-learning deep-learning statistics generative-models
machine-learning deep-learning statistics generative-models
asked 1 hour ago
xelaxela
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
xelaxela
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
xelaxela
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
xelaxela
14
asked 1 hour ago
xelaxela
14
14
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
xela is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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