# What does this assumption mean regarding equal marginal densities?

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,

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

• What paper was this? And what was the context? – jbowman Jun 25 '19 at 22:39
• @jbowman This is a part of a reparameterization trick often used in variational inference. Here is the paper: arxiv.org/pdf/1505.05424.pdf – KRL Jun 26 '19 at 0:54