I'm stuck on understanding a chain of reasoning in the paper Tweedie's Formula and Selection Bias by Efron (link here). The following equations come from the beginning of section 2, equations 2.1-2.4. We have the following setup $$ \eta \sim g(\cdot), \quad \text{and} \quad z|\eta \sim f_{\eta}(z) = e^{\eta z - \psi(\eta)}f_0(z) $$ Where $g$ is some unknown distribution, $z|\eta$ comes from an exponential family, $\psi(\eta)$ is the cumulant generating function so that the density integrates to 1, and $f_0(z)$ is the density of $z | \eta=0.$ Baye's rule gives the posterior density of $\eta | z$ as $$ g(\eta | z) = f_{\eta}(z)g(\eta) / f(z) $$ where $f(z)$ is the marginal density of $z,$ $$ f(z) = \int_{\mathcal{Z}}f_\eta(z)g(\eta)d\eta $$ Here $\mathcal{Z}$ is the domain of the exponential family.
Now here's the part I don't understand how to recreate: given the equations above, the author claims
$$ g(\eta | z) = e^{z\eta - \lambda(z)}[g(\eta)e^{-\psi(\eta)}] $$ Where $\lambda(z) = \log\left(\frac{f(z)}{f_0(z)}\right).$
If anyone can explain how the density of $\eta | z$ is being derived here I would greatly appreciate the help. Thanks.
Re-organize the right hand side,
$\begin{align} & e^{z\eta - \Psi(\eta)} e^{-\lambda(z)} g(\eta) \\ =& \mathbf{e^{z\eta - \Psi(\eta)} f_0(z)}/f(z) g(\eta)\\ =& \frac{\mathbf{f_{\eta}(z)}g(\eta)}{f(z)} , \end{align}$
which is equivalent to $g(\eta|z)$.
The reason for choosing exponential family distribution is that it enables a conjugate prior, e.g., $g(\eta)$, and a closed-form posterior, $g(\eta|z)$.