# 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?

• In the question as written, $T$ can be any set equipped with a kernel. So $T$ can be replaced by $T \times K$ and the kernel value at $(t,\,t)$ be replaced by the value of the tensor product kernel at $([t, \,k], \, [t, \,k])$.
– Yves
Dec 11, 2019 at 9:05

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*}
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.