Average vs Worst-Case Hitting Time

Question

Consider a simple random walk on an undirected graph and let $H_{ij}$ be the hitting time from $i$ to $j$. How much bigger can $$ H_{\rm max} = \max_{i,j} H_{ij}, $$ be compared to $$ H_{\rm ave} = \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n H_{ij}.$$ For all the examples I can think of, these two quantities are of roughly the same order of magnitude.

To make this into a formal question, define $$\phi(n) = \max_{\mbox{undirected graphs with } n \mbox{ nodes }} \frac{H_{\rm max}}{H_{\rm ave}}.$$ How fast does $\phi(n)$ grow with $n$?

Answer 1

This is an addendum to the answer by Yuval Filmus. Indeed $\phi(n)=\Theta(n)$, and the upper bound is explained in that answer. I don't understand the argument for the lower bound given there, but a simpler graph will yield the required bound. Let $G$ consist of a clique of $n-1$ vertices, where one of them, say $x$, is connected to one additional vertex called $y$. Let $f(v):= E_v \tau_y=H(v,y)$ denote the expected hitting time from $v$ to $y$. Then, by conditioning on the first step and using symmetry, $$ \forall v\notin\{x,y\}, \quad f(v)=1+\frac1{n-2} f(x) +\frac{n-3}{n-2} f(v)$$ and $$f(x)=1+\frac{n-2}{n-1} f(v) \,.$$ Solving these linear equations gives $f(v)=(n-1)^2$ and $f(x)=n^2-3n+3$. On the other hand, only $n-1$ of the $n(n-1)$ mean hitting times are of order $n^2$, (namely when the target is $y$). The mean hitting time of every target vertex in the clique is less than $2n$, so $H_{\rm ave}=O(n)$.

Answer 2

The quantity $\phi(n)$ grows like $\Theta(n)$.

For the lower bound, consider the following graph: a clique on $m$ vertices connected to a path of length $k = C\log m$ for an appropriate $C$ (so $n = m + k \approx m$), in which each vertex of the path is also connected to the first vertex of the path (the one incident to the clique). The maximal hitting time is some large number $M = \exp(k)$. The hitting time of two vertices in the clique is around $m \ll M$. The hitting time of a vertex on the path and any other vertex is at most $M$. So the average hitting time is at most roughly $m + \frac{2kn}{n^2} M = O(kM/n)$. In fact, the hitting time of a vertex at distance $t$ from the clique is roughly $\exp(t)$ (for $t \gg \log m$), and if we take this into account then we get an average hitting time of roughly $m + \sum_{t=1}^k \frac{2n}{n^2} \exp(t) = O(M/n)$. This shows that $\phi(n) = \Omega(n)$.

For the upper bound, let $H_{ij}$ be the maximum hitting time. The triangle inequality shows that for each $k$, either $H_{ik} \geq H_{ij}/2$ or $H_{kj} \geq H_{ij}/2$. So at least $n/2$ of the hitting times are at least $H_{ij}$, showing that the average hitting time is at least $H_{ij}/n$. This shows that $\phi(n) = O(n)$.