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We are given a simple path graph $P(V,E)$ with vertex set $V$ and edge set $E$, having $n=|V|$ nodes. Given an initial distribution $\mathbf{\mu}$ over $V$, let $d_t(\mathbf{\mu},\pi)$ be defined as $\|\mathbf{\mu}P^t-\mathbf{\pi}\|_2$, where $\pi$ is the stationary distribution and $P$ is the random walk transition matrix $D^{-1}A$, being $D$ the diagonal matrix with $D_{i,i}$ equal to the degree of node $i$ in $P$, and $A$ the adjacency matrix of $P$. Let $t_{\min}(\varepsilon)$ be the minimum integer value $t$ such that $d_t(\mathbf{\mu},\pi)<\varepsilon$ holds for any initial distribution $\mathbf{\mu}$.


Question: How can be prove that, for $n\to\infty$, both $t_{\min}(\epsilon)$ and the mixing time $t_{\mathrm{mix}}(\varepsilon)$ are equal to $\Theta\left(n^2\log\left(\frac{1}{\varepsilon}\right)\right)$ ?

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  • $\begingroup$ @AnthonyQuas Thank you for your comments! First of all, I mean $\Theta(n^2 \log(1/\varepsilon))$, because as far as I know, such upper bound is also tight up to a constant factor and I am actually mainly interested in a lower bound rather than an upper bound. Regarding your first comment, maybe I am missing something: it’s a just, unweighted, connected, undirected, single-edge path graph without self-loops, so it's unclear to me what you mean. $\endgroup$ Commented Nov 5 at 10:45
  • $\begingroup$ @AnthonyQuas, yes, but how can we prove that the Euclidean distance defined in the problem text has the same bound for $n\to\infty$? $\endgroup$ Commented Nov 5 at 23:38
  • $\begingroup$ @AnthonyQuas in my question I am just interested in the path graph, not in the complete graph. $\endgroup$ Commented Nov 6 at 9:31
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    $\begingroup$ I see now. Apologies for my confusion. $\endgroup$ Commented Nov 6 at 14:53

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