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

According to the source I'm reading (https://ufal.mff.cuni.cz/~jurcicek/NPFL108-BI-2014LS/2013-05-28-ondrej-dusek-probit-ep.pdf), the answer is: $$Z_i = \Phi\left( \frac{y_i\cdot \sum_{j=1}^d m_j x_{ij}}{\sqrt{\sum_{j=1}^d v_j x_{ij}^2 +1}} \right)$$ but I'm unsure exactly how the author got to this - if anyone can help it would be much appreciated.


Hint #1 from me is to notice that you have $E (\Phi (y_iw^Tx) )$. Figure out the distribution of $y_iw^Tx$. Then you have an integral in one variable, instead of $d $.

update and hint #2 Let $t_i=y_iw^Tx_i $ this is normal with mean $\mu_i=y_i\sum_jm_jx_{ij} $ and variance $\sigma_i^2=\sum_jv_jx_{ij}^2$. The integral you are left with is $$I=\int_{-\infty}^{\infty}\int_{-\infty}^{t_i}\frac {1}{2\pi\sigma_i}\exp\left(-\frac {1}{2}y^2-\frac{(t_i-\mu_i)^2}{2\sigma_i^2}\right)dydt_i $$ Notice this is a density of a bivariate normal distribution, and the limits of integration mean you can write this as a probability $$I=Pr (Y<T_i) $$

Simplify the probability statement to have only $1$ normal random variable.

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