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parse/train/HklY120cYm/HklY120cYm.md CHANGED
@@ -68,7 +68,7 @@ $$
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  q ( \pmb { x } ) = q ( \pmb { z } ) \left| \operatorname* { d e t } \left( \frac { \partial f ( \pmb { z } ) } { \partial \pmb { z } } \right) \right| ^ { - 1 } ,
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  $$
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- where de t ∂f(z)∂z  is the determinant of the Jacobian and is computationally expensive to obtain in general. Inverse autoregressive flow (IAF) (Kingma et al., 2016) is a special normalizing flow with a simple Jacobian determinant. In IAF, $_ { z }$ has the same dimension as $_ { \textbf { \em x } }$ , and the transformation is based on an autoregressive network taking $_ z$ as the input: $x _ { t } = f ( z _ { \leq t } ; \vartheta )$ , where $\vartheta$ are parameters of the model. Note that the $t$ -th variable $x _ { t }$ only depends on previous and current latent variables $z _ { \leq t }$ , thus the Jacobian is a triangular matrix and the determinant is the product of the diagonal entries,
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  $$
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  \operatorname* { d e t } \left( \frac { \partial f ( \boldsymbol { z } ) } { \partial \boldsymbol { z } } \right) = \prod _ { t } \frac { \partial f ( \boldsymbol { z } _ { \le t } ) } { \partial \boldsymbol { z } _ { t } } ,
 
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  q ( \pmb { x } ) = q ( \pmb { z } ) \left| \operatorname* { d e t } \left( \frac { \partial f ( \pmb { z } ) } { \partial \pmb { z } } \right) \right| ^ { - 1 } ,
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  $$
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+ where de t ∂f(z)∂z  is the determinant of the Jacobian and is computationally expensive to obtain in general. Inverse autoregressive flow (IAF) (Kingma et al., 2016) is a special normalizing flow with a simple Jacobian determinant. In IAF, $_ { z }$ has the same dimension as $_ { \textbf { \em x } }$ , and the transformation is based on an autoregressive network taking $_ z$ as the input: $x _ { t } = f ( z _ { \leq t } ; \vartheta )$ , where $\vartheta$ are parameters of the model. Note that the $t$ -th variable $x _ { t }$ only depends on previous and current latent variables $z _ { \leq t }$ , thus the Jacobian is a triangular matrix and the determinant is the product of the diagonal entries,
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  $$
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  \operatorname* { d e t } \left( \frac { \partial f ( \boldsymbol { z } ) } { \partial \boldsymbol { z } } \right) = \prod _ { t } \frac { \partial f ( \boldsymbol { z } _ { \le t } ) } { \partial \boldsymbol { z } _ { t } } ,