Timeline for Mean distance from origin after $N$ equal steps of Random-Walk in a $d$-dimensional space.
Current License: CC BY-SA 3.0
7 events
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| Sep 23, 2015 at 11:29 | comment | added | Titus | @leonbloy I read your exchange with Nate and wanted to learn more about Poissonization. To be clear - the variables $\{x_i, y_i\}$ obey a multinomial distribution, not Poisson, right? And if we approximate $n_i$ with $Pois(\lambda) = Pois(N/2d)$ aren't we implicitly assuming that the mean of $n_i$, $N/2d$, approaches a constant value as $N \to \infty$? This fails, and your solution is correct (and intuitive), so I'm wondering how to make this rigorous. | |
| Jul 8, 2015 at 13:52 | comment | added | DRG | Just to add, I'd be very grateful for a textbook or paper reference for this. Finding it very useful, but hard to credit! | |
| Sep 7, 2013 at 13:59 | comment | added | Picard Porath | @leonbloy:Can you give any reference to a publication where this formula was mentioned? Thanks | |
| Apr 11, 2012 at 14:00 | comment | added | Nate Eldredge | Oh right. I was confused and thinking of something else. Thanks for the clarification. | |
| Apr 11, 2012 at 13:54 | comment | added | leonbloy | @NateEldredge: the CLT is immediate only in 1D, in more dimensions $z_i$ is not the sum of $N$ variables but of $n_i$, which is itself a random variable (with $\sum n_i = N$), hence the CLT is not so clear here. Instead, it's clear that $x_1,y_1,x_2,y_2...$ is identical to a urns-and-balls ($2d$ urns, $N$ balls) model, which is equivalent ("Poissonization") to $2d$ iid Poisson variables conditioned to the value of their sum being $N$ (asymptotically, this conditioning turns irrelevant). | |
| Apr 11, 2012 at 2:27 | comment | added | Nate Eldredge | Perhaps I am missing something, but where did the Poisson variables come in? It seems to me that it is immediate from the CLT that $z_i$ is approximately normal, since it is the sum of $N$ iid random variables (with values $\pm 1$). Also, I think $x_i, y_i$ would be approximately normal, not Poisson (the distributions of the summands are not changing with $N$) and not independent of each other (since they must sum to $N$). Otherwise, I agree with the conclusion. | |
| Mar 11, 2012 at 15:55 | history | answered | leonbloy | CC BY-SA 3.0 |