Integration against multivariable gaussian reduce to single variable gaussian

Question

Given the logistic function $f(x) = \frac{1}{1+e^{-x}}, w, \phi\in\mathbb{R}^n$ and $q\sim \mathcal{N(\mu, \Sigma)}$(i.e. normally distributed, with mean $\mu\in\mathbb{R}^n$ and covariance matrix $\Sigma\in\mathbb{R}^{n\times n}$), how can I show that $$\int f(w^T\phi)q(w)dw =\int f(a)p(a)da$$ with $p\sim \mathcal{N}(\hat{\mu}, \hat{\sigma}^2)$ and find $\hat{\mu}\in\mathbb{R}$ and $\hat{\sigma}\in\mathbb{R}$? I am stuck and don't know where to start.

Answer 1

Write $X=\mu+\sqrt{\Sigma}Z$ where $Z\sim N(0,I_n)$ denote $A=e^{-\langle\phi,\mu\rangle }$ and $B=\sqrt{\Sigma}\phi.$ For computing $$I=E\left(\frac{1}{1+Ae^{-\langle B,Z\rangle }}\right)$$ you have to compute the distribution of $Y=\langle B,Z\rangle $ that you can obtain from the integral $$E\left(e^{s\langle B,Z\rangle }\right)=e^{ \frac{s^2\|B\|^2}{2}}$$ saying that $Y\sim N(0,\|B\|^2)$ or $Y\sim \|B\|Z_1$ where $Z_1\sim N(0,1).$ Finally $$I=\int_{R}\frac{1}{1+Ae^{-\|B\|z}}e^{-z^2/2}\frac{dz}{\sqrt{2\pi}}$$ which is not computable by the elementary means. Note that $\|B\|^2=\phi^T\Sigma \phi:$ you do not need to compute $\sqrt{\Sigma}.$

Answer 2

Define the random vector $W\sim q\sim\mathcal{N}(\mu,\Sigma)$. Then by the law of the unconscious statistician we have $$\int f(w^T\phi)q(w)dw = \mathbb{E}[f(W^T\phi)].$$ However, $A:=W^T \phi$ is a linear transformation of a gaussian random vector and so it is gaussian as well and $A\sim\mathcal{N}(\phi^T\mu, \phi^T\Sigma\phi)$ which density we call $p$. Then $$\int f(w^T\phi)q(w)dw=\mathbb{E}[f(W^T\phi)]=\mathbb{E}[f(A)]=\int f(a)p(a)da.$$ where $p\sim\mathcal{N}(\underbrace{\phi^T\mu}_{\hat{\mu}}, \underbrace{\phi^T\Sigma\phi}_{\hat{\sigma}})$.