Q$6(10$ points$)$ Suppose we use the Independent Cascade $($IC$)$ model to model the intluence

diffusion in a network. In the IC model, we have all the seed node$($s$)$ activated at round $0.$ For

each node $u$ firstly activated at round $t,$ at round $t+1,u$ tries to activate each of its inactive

out$-$neighbor $v($which means there is a directed edge $(u,v))$ with a success probability $p{p}_{uv}.$

Note that if $u$ fails to activate $v$ at round $t+1,u$ will not have another chance to activate $v$ in

the future rounds. The diffusion ends at a round when we do not have any newly activated

nodes. Given a seed set $S,$ to calculate the probability $p_v$ that node $v$ is activated in the

diffusion started by $S,$ one comes up with the following non$-$linear system.

$p_v=\{\begin{array}{ccc}1,& if& \text{v}\text{i}\text{n}\text{S}\\ 1-pro{d}_{(u,v)inE}(1-p_u**p{p}_{uv}),& if& u!\text{i}\text{n}\text{S}\end{array}$

The intuition is that if $v$ is not a seed, $p_v$ depends on each of its in$-$neighbor $u^{\text{'}}s($which means

there is an edge $(u,v)$ probability of being activated. Suppose we can solve this non$-$linear

system exactly, which means we can find all the $p_v$ to make all the equations hold. Can we use

the solution to this non$-$linear system to calculate each $p_u$ exactly? Please provide your

justification. $($Hint: you may want to use the following influence graph as an example.$)$