as the maximum distance between $v$ and any infected nodes in the graph. Define the Jordan infection centers of a graph to be the nodes with the minimum infection eccentricity given ${\bf Y}.$ In Figure \[figure:infectioneccentricity\], nodes $v_3,$ $v_{10},$ $v_{13}$ and $v_{14}$ are observed to be infected. The infection eccentricities of $v_1,v_2,v_3,v_4$ are $2,3,4,5,$ respectively, and the Jordan infection center is $v_1.$

We will show that the source associated with the optimal sample path is a node with the minimum infection eccentricity. We derive this result using three steps: first, assuming the information source is $v_r,$ we analyze $t^*_{v_r}$ such that $$t^*_{v_r}=\displaystyle\arg_{t}\max_{t,
{\bf X}[0,t]}\Pr({\bf X}[0,t]|v^*=v_r),$$ i.e., $t^*_{v_r}$ is the time duration of the optimal sample path in which $v_r$ is the information source. It turns out that $t^*_{v_r}$ equals to the infection eccentricity of node $v_r.$ Considering Figure \[figure:infectioneccentricity\] if the source is $v_1$, then the time duration of the optimal sample path starting from $v_1$ is $2.$

In the second step, we consider two neighboring nodes, say nodes $v_1$ and $v_2.$ We will prove that if $\tilde{e}(v_1)<\tilde{e}(v_2),$ then the optimal sample path rooted at $v_1$ occurs with a higher probability than the optimal sample path rooted at $v_2.$

Finally, 