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Feb 19, 2013 at 23:44 comment added fedja There is one thing I fail to understand: what prohibits all vectors to be (almost) the same? Then the expectation of the length is $m$, not $\sqrt m$. Of course, then the variance is $0$, so that part is OK in this trivial example.
Feb 19, 2013 at 2:04 answer added Igor Rivin timeline score: 1
Feb 19, 2013 at 1:56 comment added Igor Rivin @Per: I am not sure I understand your question: the mean value of the length of the sum, is not the same as the length of the mean of the sum.
Feb 19, 2013 at 1:38 comment added Brendan McKay TOM, your reply to Per adds to my confusion. You refer to a finite set of vectors (all 0 except for a single 1) then you say "any vector of length 1" (of which there are infinitely many). You put your vectors in $R^n$, not $Z^n$, remember. What is the complete definition of $A$?
Feb 18, 2013 at 3:06 comment added TOM Per Alexandersson: there are n vectors with 1 as a coordinate and we have exponentially many, so I clearly mean any vector of legth 1. Joseph O'Rourke: it is - it is the about the average length of a random sum x_1+...+x_m, that is, it's euclidean length. Douglas Zare: I think I have really meant what I have written. Anyhow, how should I use the CLT in this case? I really need just a bound on the variance of the length of such a sum, nothing more.
Feb 17, 2013 at 19:58 comment added Douglas Zare It sounds like once you figure out what question you want to ask, the answer will be the Central Limit Theorem (for vector-valued summands, or for each coordinate if you don't mind losing a factor). If you mean something deeper than that, please clarify.
Feb 17, 2013 at 14:58 comment added Joseph O'Rourke Is not the question asking for statistics for the total travel of a random walk, each of whose steps is of length $1$?
Feb 17, 2013 at 14:23 comment added Per Alexandersson This seems strange; your unit vectors, are these (1,0,...) and (0,1,0,0,...) and so on, or just vectors of length 1? If the latter, the expected mean should be the zero vector then, no?
Feb 17, 2013 at 12:29 history asked TOM CC BY-SA 3.0