Suppose that we have a random variable $\epsilon$ with density $q(\epsilon)$ and $w = t(\theta, \epsilon)$, where $t$ is a deterministic function of a constant $\theta$ and random variable $\epsilon$. Denote by $q(w|\theta)$ the marginal density over $w$. In a paper, I saw the following assumption:

$q(\epsilon) d\epsilon = q(w|\theta)dw$

Could you give me some intuition on where this assumption is satisfied?

I think this is not satisfied for the following Gaussian case. If we have

$w = \epsilon +\theta, \epsilon \sim \mathcal{N}(0,1)$

Then, we have $dw/d\epsilon = 1$. Therefore,

$q(\epsilon) d\epsilon \neq q(w|\theta)dw $

  • 1
    $\begingroup$ What paper was this? And what was the context? $\endgroup$
    – jbowman
    Jun 25, 2019 at 22:39
  • $\begingroup$ @jbowman This is a part of a reparameterization trick often used in variational inference. Here is the paper: arxiv.org/pdf/1505.05424.pdf $\endgroup$
    – KRL
    Jun 26, 2019 at 0:54


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