I am attempting to derive the update equations for the conjugate to the Dirichlet distribution, as outlined here: https://mathoverflow.net/questions/20399/conjugate-prior-of-the-dirichlet-distribution
However, parameter update equation I calculate does not match the one suggested there.
My derivation is shown below: \begin{align} f({\theta}|{\alpha}) &= Dir({\theta}|{\alpha})\\ &=\frac{1}{B({\alpha})}\exp(\phi({\alpha})^{T}u({\theta})) \end{align} where, \begin{align} \phi({\alpha})^{T} &= [\alpha_1-1,\cdots,\alpha_K-1]\\ u({\theta}) &= [\ln(\theta_1),\cdots,\ln(\theta_K)]^{T}\\ B({\alpha}) &= \frac{\prod_{i=1}^{K}\Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^{K}\alpha_i\right)} \end{align}
Thus, \begin{align} f({\theta}|{\alpha}) &= \frac{1}{B({\alpha})}\exp\left(\sum_{i=1}^{K}\alpha_{i}\ln(\theta_i)-\ln(\theta_i)\right) \end{align}
The exponential family conjugate has form, \begin{align} p({\alpha}|{\nu},\eta) &\propto \frac{1}{B({\alpha})^{\eta}}\exp(\phi({\alpha})^{T}{\nu})\\ &= \frac{1}{B({\alpha})^{\eta}}\exp\left(\sum_{i=1}^{K}\alpha_{i}\nu_{i}-\sum_{i=1}^{K}\nu_i\right)\\ &\propto \frac{1}{B({\alpha})^{\eta}}\exp\left(\sum_{i=1}^{K}\alpha_{i}\nu_{i}\right) \end{align}
Now the posterior update on ${\alpha}$ given ${\theta}$, \begin{align} p({\alpha}|{\theta},{\nu},\eta) &\propto p({\alpha},{\theta}|{\nu},\eta)\\ &= f({\theta}|{\alpha})p({\alpha}|{\nu},\eta)\\ &\propto \left[\frac{1}{B({\alpha})}\exp\left(\sum_{i=1}^{K}\alpha_{i}\ln(\theta_i)-\ln(\theta_i)\right)\right]\times\nonumber\\ &\phantom{{}\propto} \left[\frac{1}{B({\alpha})^{\eta}}\exp\left(\sum_{i=1}^{K}\alpha_{i}\nu_{i}\right) \right]\\ &= \frac{1}{B({\alpha})^{\eta+1}}\exp\left(\sum_{i=1}^{K}\alpha_{i}\ln(\theta_i) + \alpha_{i}\nu_{i}-\ln(\theta_i)\right)\\ \end{align}
Therefore, I get the ${\eta^{t+1}} = {\eta^t} + 1$ update. However, the update on $\nu$ does not match. If we could drop the $-\ln(\theta_i)$, we would get update ${\nu_i^{t+1}} = {\nu_i^t} + \ln(\theta_i)$, which does not match the suggested ${\nu_i^{t+1}} = {\nu_i^t} - \ln(\theta_i)$.
And a follow-up: is there an intuitive meaning behind ${\eta}$ and $\nu$ in this conjugate? ${\eta}$ seems to indicate the level of confidence in the prior, and $\nu$ imposes asymmetry, but some more discussion on this would be appreciated.