Marginal probability in Gaussian Process Let $\mathbf{a} \sim \mathcal{GP}(\mathbf{m},\mathbf{C})$ where $\mathbf{a} \in \mathbb{R}^T$ is modeled as Gaussian process with mean $\mathbf{m} \in \mathbb{R}^T$ and prior covariance $\mathbf{C} \in \mathbb{R}^{T \times T}$. Then the marginal probability of $a_t$ being non-zero can be written as
\begin{align*}
p(a_t > 0) = \phi(m_t/\sqrt{C_{t,t}})
\end{align*}
where $\phi$ is normal cummulative distribution function. Now $\mathbf{a}$ is not a one dimension object, but a 2 dimensional object $\mathbf{A} \sim \mathcal{GP}(\mathbf{M},\mathbf{C}_1 \otimes \mathbf{C}_2) $ where $\mathbf{M} \in \mathbb{R}^{T \times K}$ and prior covariances $\mathbf{C}_1 \in \mathbb{R}^{T \times T}$ and $\mathbf{C}_2 \in \mathbb{R}^{K \times K}$. In this case, how do I calculate the marginal probability of $a_{t,k} > 0$. For one dimension, I can marginalize by just dropping the irrelvant variables, but is it true for more than one variable?
 A: Yes, by definition a random variable is Gaussian if and only if all linear functionals are normally distributed. With $e_t$ the $t$th standard Euclidean basis vector, we have $a_{t,k}=e_t'Ae_k=tr(Ae_ke_t')=\langle A, e_ke_t'\rangle$, i.e., $a_{t,k}$ is a linear functional of $A$ and is hence normally distributed. Its mean and variance are $\langle M, e_k e_t'\rangle=e_t'Me_k=m_{t,k}$, and variance computed analogously but more tediously as $(C_1\otimes C_2)_{(t,k),(t,k)}=(C_1)_{t,t}(C_2)_{k,k}$. 
We therefore have
\begin{align*}
    p(a_{t,k} > 0) = \phi\Big(m_{t,k}\Big/\sqrt{(C_1)_{t,t}(C_2)_{k,k}}\Big)
\end{align*}
As Yves remarks, the kernels may be defined over much more general spaces than Euclidean ones.  
A: Maybe this is not the answer you were looking for but if we assume that $\mathbf{a}$ is i.i.d and that $p(\mathbf{a}(t) > 0\vert t_{0}) = \phi(m_{t}\,C_{t,t}^{-\frac{1}{2}})$ then for each $\mathbf{a}\in\mathbf{A}$ that should be $\prod_{\mathbf{a}\in\mathbf{A}} \phi_{\mathbf{a}}(m_{t}\,C_{t,t}^{-\frac{1}{2}})$ if I'm not mistaken.
