Re-arranging of formulation for Generative Adversarial Networks (GAN) formulation for optimal $D$
I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
+ log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$.
I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.
So far I got:
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$
How do I proceed?
neural-networks
asked Dec 4 '18 at 10:40
havakokhavakok
510215
add a comment |
I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
+ log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$.
I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.
So far I got:
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$
How do I proceed?
neural-networks
asked Dec 4 '18 at 10:40
havakokhavakok
510215
add a comment |
I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
+ log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$.
I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.
So far I got:
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$
How do I proceed?
neural-networks
asked Dec 4 '18 at 10:40
havakokhavakok
510215
I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
+ log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$.
I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.
So far I got:
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$
How do I proceed?
neural-networks
neural-networks
asked Dec 4 '18 at 10:40
havakokhavakok
510215
asked Dec 4 '18 at 10:40
havakokhavakok
510215
asked Dec 4 '18 at 10:40
havakokhavakok
510215
asked Dec 4 '18 at 10:40
havakokhavakok
510215
asked Dec 4 '18 at 10:40
havakokhavakok
510215
510215
add a comment |
add a comment |
1 Answer
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The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
$$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
Now take logs to get the result you want.
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
$$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
Now take logs to get the result you want.
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
add a comment |
The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
$$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
Now take logs to get the result you want.
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
add a comment |
The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
$$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
Now take logs to get the result you want.
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
$$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
Now take logs to get the result you want.
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
answered Dec 4 '18 at 11:17
Matthew TowersMatthew Towers
7,46722344
7,46722344
add a comment |
add a comment |
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