Is $K_z( z, z^{\prime})=\sigma(w^Tz+b)\sigma(w^Tz^{\prime}+b)$ with nonlinear function $\sigma$, generally a valid covariance function for a GP? Is $\mathbf K_z(\mathbf z,\mathbf z^{\prime})=\sigma(\mathbf z^T\mathbf w+b)\sigma(\mathbf (z^{\prime})^T\mathbf w+b)$  with nonlinear function $\sigma$, generally a valid covariance function (kernel) for a Gaussian process?
In Yarin gal's paper "Dropout as a Bayesian Approximation: Appendix", section 3.1 (p.5), the author states that "It is trivial to show that [the below covariance function] defines a valid covariance function following [Tsuda et al., 2002]."
$$\mathbf K(\mathbf x_1,\mathbf x_2)=\int p(\mathbf w)p(\mathbf b)\sigma(\mathbf x_1^T\mathbf w+b)\sigma(\mathbf x_2^T\mathbf w+b)d\mathbf w db$$
However, due to my lack of understanding, I could not understand how the [Tsuda et al., 2002] relates to justifying the claim that the above covariance function is valid.
Specifically, [Tsuda et al., 2002]  deal with the following form of marginalized kernel and it seems it is the only notion that can be related to the covariance function.
$$\mathbf{K}(x,x^{\prime}) = \sum_{h\in\mathcal{H}}\sum_{h^{\prime}\in\mathcal{H}}p(h|x)p(h^{\prime}|x^{\prime})\mathbf{K}_z(z,z^{\prime})$$
Then, $\sigma(\mathbf w^T\mathbf x + b)\sigma(\mathbf w^T\mathbf y+b)$ would relate to $\mathbf{K}_z(z,z^{\prime})$ in [Tsuda et al., 2002]. [Tsuda et al., 2002] claims that $\mathbf{K}_z(z,z^{\prime})$ needs to be positive semidefinite. But it is hard for me to understand $\sigma(\mathbf w^T\mathbf x + b)\sigma(\mathbf w^T\mathbf y+b)$ (with any nonlinear function $\sigma$) is positive semidefinite.
Is the above approach correct? If not, what is the right approach to showing the above covariance function is valid?
 A: If we let $\phi(\mathbf x)=\sigma(\mathbf x^T\mathbf w+b)=\phi(\mathbf x)^T$, then $K(\mathbf x, \mathbf x')=\phi(\mathbf x)^T\phi(\mathbf x')$ defines a valid kernel.  The nonlinearity doesn't matter because it is just a transformation. If we look at the first integral, it's just an expected value of this wrt $\mathbf w$ and $b$. I don't think the other article is even needed.
Edit: This part is to explain why the expectation integral still defines a valid kernel. Assume $K(\mathbf x_1, \mathbf x_2)$ is a valid kernel as defined above. Let
$$K'(\mathbf x_1, \mathbf x_2)=\int p(\mathbf w)p(b) K(\mathbf x_1, \mathbf x_2)d\mathbf wdb$$
For validness, we need to have (for any $n$ and $\mathbf a$): $$\sum_{i=1}^n\sum_{j=1}^n a_i a_jK'(\mathbf x_i, \mathbf x_j)\geq 0$$
If we substitute the definition:
$$\begin{align}\sum_{i=1}^n\sum_{j=1}^n a_i a_jK'(\mathbf x_i, \mathbf x_j)&=\sum_{i=1}^n\sum_{j=1}^n a_i a_j\left(\int p(\mathbf w)p(b)K(\mathbf x_i, \mathbf x_j)d\mathbf wdb\right)\\&=\int p(\mathbf w)p(b)\left(\sum_{i=1}^n\sum_{j=1}^n a_i a_jK(\mathbf x_i, \mathbf x_j)\right)d\mathbf wdb\\&\geq0\end{align}$$
because the inside of the integral is always non-negative. We could easily change the order of integration and the summations, simply because the summation is bounded and finite (we could change the infinite summations in certain conditions but that's out of scope in this post).
One easier way to sense that this expectation represents a valid kernel is a well known property of the kernels:

*

*If $K_1$ and $K_2$ are valid kernels, their non-negatively weighted sum, i.e. $a_1K_1+a_2K_2$ is also a valid kernel.

