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Let $x$ and $z$ be real-valued random vectors, where $x = g(z)$, and $g$ is an invertible, continuous and differentiable transformation of $z$. Then $p_z(z) = p_x(g(z)) \lvert det(\frac{\partial g(z)}{\partial z})\rvert $. However, the equation 3 of Proposition 1 in Generative Adversarial Nets seems to suggest $p_z(z)dz = p_x(x)dx$, without the determinant. I wonder how to derive this equality.

EDIT: Thanks to Chill2Macht's comment, I read and watched the resources regarding change of variables. However, I am still not sure where 'det' went in $p_z(z)dz = p_x(x)dx$. For example, let $x = \mu + \sigma z$, where $\mu$ and $\sigma$ are constants. Then $\frac{\partial x}{\partial z} = \sigma$, and $p(x)dx = \sigma p(g(z)) dz$. Using their derivation, $\sigma$ should not be present.

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Answering my own question: $p_z(g^{-1}(x))$ is not a probability density function of $x$ because it doesn't integrate to 1. $p_z(g^{-1}(x))\lvert det(\frac{\partial(g^{-1}(x))}{\partial x})\rvert$ is the probability density function of $x$, and I think in the GAN paper, the $p_g(x)$ represents this expression.

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