#posterior-variance

Question 1: Single choice

mediumAI-checkedProbabilistic & Graphical Models

A noise-free GP has $k(x, x) = k(x_1, x_1) = k(x_2, x_2) = 1$ and $k(x, x_1) = 0.9$ . Observing $f(x_1) = 1.2$ gives $f(x) \mid f(x_1) \sim \mathcal{N}(1.08,\ 0.19)$ . You then also observe $f(x_2) = 1.4$ , where $k(x, x_2) = 0.8$ . How does the new posterior variance of $f(x)$ compare with $0.19$ ?

It is at most 0.19 whatever $k(x_1, x_2)$ is, since conditioning on more data never increases it1
It is larger than 0.19, because the second value 1.4 disagrees with 1.08, the current mean of $f(x)$ 2
Even its direction cannot be determined without knowing $k(x_1, x_2)$ 3
It depends on how close 1.4 is to 1.2, the first observed value4

Timed test