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I'm currently trying to calculate the moments of $$r(w) = \frac{1}{Z}\Phi(y_i\cdot w^T x_i) \prod_{j=1}^d \mathcal{N}(w_j|m_j^{\setminus i},v_j^{\setminus i}), $$ where the partition (or normaliser) function \begin{align}Z_i &= \int \Phi(y_i\cdot w^T x_i) \prod_{j=1}^d \mathcal{N}(w_j|m_j^{\setminus i},v_j^{\setminus i})dw\\ &= \Phi\left( \frac{y_i\cdot \sum_{j=1}^d m_j x_{ij}}{\sqrt{\sum_{j=1}^d v_j x_{ij}^2 +1}} \right).\end{align} The first and second order moments can be calculated as:

$$\mathbb{E}_r[w] = m_j^{\setminus i} + v_j^{\setminus i} \frac{\partial \log Z_i}{\partial m_j^{\setminus i}}, $$

$$VAR_r[w] = v_j^{\setminus i} - \left(v_j^{\setminus i}\right)^2 \left( \left(\frac{\partial \log Z_i}{\partial m_j^{\setminus i}}\right)^2 - 2 \frac{\partial \log Z_i}{\partial v_j^{\setminus i}}\right). $$

To calculate these moments, I need the log-partition differentiated with respect to both $m_j^{\setminus i}$ and $v_j^{\setminus i}$. How exactly would I calculate $\frac{\partial \log Z_i}{\partial m_j^{\setminus I}}$ and $\frac{\partial \log Z_i}{\partial v_j^{\setminus I}}$?

Note:$\Phi(.)$ is the probit function and $\mathcal{N}(.)$ a Gaussian distribution, so: $\Phi(y_i\cdot w^T x) = \int^{y_i\cdot w^T x}_{-\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}t^2} dt$ and $ \mathcal{N}(w_j|m_j^{\setminus i},v_j^{\setminus i}) = \prod_{j=1}^d \frac{1}{\sqrt{2\pi v_j^{\setminus i}}} e^{-\frac{1}{2}\left(\frac{w_j-m_j^{\setminus i}}{v_j^{\setminus i}}\right)^2} dw.$

I have been reading(https://ufal.mff.cuni.cz/~jurcicek/NPFL108-BI-2014LS/2013-05-28-ondrej-dusek-probit-ep.pdf).

George
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  • tell me if this helps: https://stats.stackexchange.com/questions/61080/how-can-i-calculate-int-infty-infty-phi-left-fracw-ab-right-phiw – Taylor Nov 21 '17 at 01:04
  • Not particularly no. If I was just interested in taking derivatives then maybe yes but I need to take log-derivatives. – George Nov 21 '17 at 11:07
  • oh I thought you were interested in the other result. The chain rule doesn't work? – Taylor Nov 21 '17 at 14:00

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