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Apr 17, 2016 at 6:50 comment added Steve Thanks. I am interested in the case where $s = o(\sqrt{d}$, that is $s^2 / d$ goes to zero.
Apr 11, 2016 at 8:43 comment added Ben Barber What ranges of $s$ and $d$ are of interest? If say $s$ is some constant fraction of $d$ then the inner products are tightly concentrated in an interval of length $O(\sqrt d)$ around $0$. More generally, the inner products look like taking some number of steps of the symmetric random walk on $\mathbb Z$, with varying degrees of laziness.
Apr 8, 2016 at 8:10 comment added kodlu @FedorPetrov, thanks, that $d-s$ helps.
Apr 8, 2016 at 6:57 comment added Fedor Petrov Definitely $M_k$ does not depend on $\bf{v}$.
Apr 8, 2016 at 6:56 comment added Fedor Petrov Ah, $n-s$ should be of course $d-s$.
Apr 8, 2016 at 6:54 comment added Fedor Petrov @kodlu each other coordinate is 0, +1 or -1, this is what $1+2x$ is about: 1 is for 0, $x$ for +1, $x$ for -1.
S Apr 8, 2016 at 5:59 history suggested Minkov CC BY-SA 3.0
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Apr 8, 2016 at 4:54 review Suggested edits
S Apr 8, 2016 at 5:59
Apr 7, 2016 at 22:15 comment added kodlu It would be wonderful to have this explained in a full answer. I can see the $t$ and $t^{-1}$ account for +1 and -1, in up to $s$ positions (thus raise to power $s$) but how does the $(1+2x)$ term work?
Apr 7, 2016 at 21:23 comment added Fedor Petrov $M_k/2$ is a coefficient of $x^st^k$ in $(1+tx+x/t)^s (1+2x)^{n-s}$.
Apr 7, 2016 at 20:14 history asked Steve CC BY-SA 3.0