3
$\begingroup$

Look at Bernoulli percolation on $\mathbb{Z}^2$ with $p> p_c$ ($p$ can be arbitrarily close to 1).

I am interested in the probability that there exists an infinite cluster starting at $(0,0)$ and it stays above the graph $y=x^n$ where $n$ is a large even number.

Is there a relation between $p$ and $n$ so that this probability is bounded from below?

I would also hope that a similar result holds for $\mathbb Z^d$. And we can relax the condition on the "cone" to be its horizontal width grows to infinity.

$\endgroup$
2
  • 1
    $\begingroup$ By “inside” do you mean above or below? $\endgroup$ Commented Oct 15, 2019 at 4:47
  • $\begingroup$ "Inside" here would mean above the graph of $y=x^n$. $\endgroup$
    – Xiao
    Commented Oct 15, 2019 at 12:46

1 Answer 1

6
$\begingroup$

This is true for every positive $n$. I assume that for $n$ odd or fractional, the bounding curve is $y=|x|^n$. If $(0,0)$ is not connected to infinity, then there must be a blocking contour in the dual lattice. The exponential decay iof connectivity in subcritical percolation and the Borel Cantelli Lemma preclude that. See e.g. Grimmett’s book on Percolation. It is not enough for the width of tge cone to grow to infinity. The width must grow faster than logarithmically. A small constant times log will suffice if $p$ is close to 1.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .