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Feb 21, 2011 at 20:45 vote accept Michał Oszmaniec
Feb 19, 2011 at 22:55 answer added Omer timeline score: 12
Feb 19, 2011 at 22:36 comment added Anthony Quas It's probably helpful to define $X$ and $Y$ (especially they're not the same as $x$ and $y$): $Y=x+y$. I would call this the level of the point. The top level is level 0 and the bottom level in $N-1$. $X$ is then given by $X=x-(x+y)/2$. The $X$ coordinate in level $i$ goes from $-i/2$ to $i/2$.
Feb 19, 2011 at 20:44 comment added fedja It does follow if you can show that the boundary values are concave on each side. Unfortunately, I failed to show that so far. I'll think of it more in the evening.
Feb 19, 2011 at 20:16 comment added George Lowther That argument sounds rather tricky, but I think it is clear that letting $\mathbb{P}(X,Y)$ be the probability of hitting 0 from point (X,Y) then, for given Y, this is symmetric in X. That is is increasing for $X\le0$ should follow from the iterative definition of $\mathbb{P}(\cdot)$.
Feb 19, 2011 at 18:24 history edited Michał Oszmaniec CC BY-SA 2.5
deleted 2 characters in body; edited tags
Feb 19, 2011 at 17:09 history asked Michał Oszmaniec CC BY-SA 2.5