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If: \begin{align} Y&=h(z,\theta)+\epsilon\\ \theta&\sim \text{Unknown Distribution}\\ \epsilon &\sim N(0,\sigma^2) \end{align} My book states: $$ f_{y|(\theta,\sigma)}=f_{\epsilon|\theta,\sigma}(y-h(z,\theta)|\theta,\sigma ) $$ I don't follow why.

(This is in context of something like Bayesian Regression).

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up vote 1 down vote accepted

$\epsilon \sim N(0,\sigma^2)$, so $y|\theta,\sigma = h(z,\theta) + \epsilon \sim N(h(z,\theta),\sigma^2)$.

The equation says that the density $f_{y|\theta,\sigma}$ at the point $y^*$ is precisely the same as the density $f_{\epsilon|\theta,\sigma}$ at the point $y^* - h(z,\theta)$. This is true because $f_{y|\theta,\sigma}$ is just a shifted version of $f_{\epsilon|\theta,\sigma}$. Does that make sense?

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