bio  website  perso.enslyon.fr/… 

location  Lyon  
age  37  
visits  member for  4 years, 3 months 
seen  1 hour ago  
stats  profile views  698 
Dec 7 
answered  Probability that a selfavoiding walk on $\mathbb{Z}^3$ closes to a polygon 
Nov 2 
comment 
When exactly and why matrix multiplication became a part of undergraduate curriculum?
Well line 2 is the product. 
Sep 22 
awarded  Yearling 
Sep 13 
comment 
Can we give any upper bound on $E[\max_{n \leq N} X_n]$ in terms of $\max_{n \leq N} E[X_n]$
Another counterexample: take $(X_n)$ iid of mean $0$, then $\max E[X_n] = 0$. But $E[\max X_n]$ can take pretty much any value; if the support of the distribution of $X$ is not bounded above, then as $N \to \infty$ it also goes to $+\infty$. 
Sep 8 
comment 
First passage percolation on a random geometric graph in the large connectivity limit
It should not be difficult to show that $h_\infty$ does not depend on the distribution as soon as $P$ has positive density and no atom at $0$: indeed, as soon as bonds with small weight percolate "enough", they should be the only ones present on the shortest path, at least asymptotically as $\rho\to\infty$. 
Sep 8 
comment 
First passage percolation on a random geometric graph in the large connectivity limit
Do you expect the underlying point process model to be relevant? Or would you think that something similar could happen e.g. for FPP on the square lattice? 
Mar 19 
comment 
Pairs of Permutations up to Simultaneous Conjugation
(2 years later) Is there an efficient algorithmic way to check if two pairs of permutations are simultaneously conjugated like this? 
Nov 7 
comment 
Embedding points in 2D based on distance estimates?
I believe fdp and sfdp implement something like that, where by default $l_{ij}=1$ but you can specify a length for an edge to set another value.

Sep 22 
awarded  Yearling 
Sep 18 
answered  Embedding points in 2D based on distance estimates? 
Aug 16 
answered  Estimate size of graph by taking random walks 
Jun 11 
answered  Conway's game of life for random initial position 
Jun 11 
comment 
Conway's game of life for random initial position
@helper Sure, I did say "there may be", and it is quite possible that indeed density would always go to zero. I was simply pointing out that what you said could not be enough, because it did not use the specifics of GoL. And in general it is quite hard to tell what happens for a given model ... 
Jun 11 
comment 
Conway's game of life for random initial position
@helper To give a more "physical" intuition: the time needed for a box to die out will typically be exponential in the volume of the box (that's what it takes for each cell to die at the same time), while the time for a neighbouring box to make you alive again is linear in the diameter (propagation fronts move linearly). So even if one of the regions were to die out (which it will), there may be plenty of "life reservoirs" in the vicinity to resuscitate it before they themselves die. 
Jun 11 
comment 
Conway's game of life for random initial position
@helper The argument is definitely wrong. There are processes (e.g. the contact process with high enough propagation rate, or "slices" of directed percolation) which die in any finite volume for the reason you mention, but nevertheless admit invariant distributions with positive asymptotic density. 
May 24 
answered  Independence using reflecting brownian motion 
May 24 
comment 
Independence using reflecting brownian motion
Wait, if you take 2d Brownian motion conditioned on the event that at time 1 its two coordinates have the same sign, or equivalently if you take two independent Brownian motions $B_1$ and $B_2$ and then set $X=B_1$ and $Y=\pm B_2$ where the sign is chosen so that $X$ and $Y$ have the same sign at time $1$, then $X$ and $Y$ have independent absolute values but are not independent... 
May 15 
comment 
Another colored balls puzzle (part II)
Good to know, at least my asymptotic analysis gives the right order of magnitude and even the right growth rate. Far from an exact result, but I believe that the approach is more robust than bijective methods like the one suggested by Ori. 
May 15 
revised 
Another colored balls puzzle (part II)
Result of the computation for case 1 
May 15 
comment 
Another colored balls puzzle (part II)
Very nice indeed! 