Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:50:59Z http://mathoverflow.net/feeds/question/47487 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47487/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest? dorkusmonkey 2010-11-27T06:50:52Z 2010-11-29T20:40:45Z <p>Given a list of identical and independently distributed <a href="http://en.wikipedia.org/wiki/Stable_distribution" rel="nofollow">Levy Stable</a> random variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum exceeds the sum of the rest? i.e.:</p> <p>$$M = \text{Max}(X_0, X_1, \dots, X_{n-1})$$ $$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M )$$</p> <p>Where, in <a href="http://academic2.american.edu/~jpnolan/stable/chap1.pdf" rel="nofollow">Nolan</a>'s notation, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the critical exponent, $\beta$ is the skew, $\gamma$ is the scale parameter and $\delta$ is the shift. For simplicity, I have taken the skew parameter, $\beta$, to be 1 (maximally skewed to the right) and $\delta=0$ so everything has its mode centered in an interval near 0.</p> <p>From numerical simulations, it appears that for the region of $0 &lt; \alpha &lt; 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$. For $1 &lt; \alpha &lt; 2$ it appears to go as $O(1/n^{\alpha - 1})$ (maybe?) irregardless of $n$ or $\gamma$. For $\alpha=1$ it's not clear (to me) but appears to be a decreasing function dependent on $n$ and $\gamma$.</p> <p>I have tried making a heuristic argument to the in the form of:</p> <p>$$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j > 0 )$$</p> <p>Then using formula's provided by Nolan (pg. 27) for the parameters of the implicit r.v. $U = X_0 - \sum_{j=1}^{n-1} X_j$ combined with the tail approximation:</p> <p>$$\text{Pr}( X > x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha}$$ $$c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi$$</p> <p>but this leaves me nervous and a bit unsatisfied.</p> <p>Just for comparison, if $X_j$ were taken to be uniform r.v.'s on the unit interval, this function would decrease exponentially quickly. I imagine similar results hold were the $X_j$'s Gaussian, though any clarification on that point would be appreciated.</p> <p>Getting closed form solutions for this is probably out of the question, as there isn't even a closed form solution for the pdf of Levy-Stable random variables, but getting bounds on what the probability is would be helpful. I would appreciate any help with regards to how to analyze these types of questions in general such as general methods or references to other work in this area.</p> <p>Note: There didn't seem to be much interest from this site so I've cross-posted to math.stackexchange.com <a href="http://math.stackexchange.com/questions/12097/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/47487/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater/47718#47718 Answer by Omer for Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest? Omer 2010-11-29T20:40:45Z 2010-11-29T20:40:45Z <p>When $\alpha>1$, the LLN implies that the sum is nearly $n\E X$, and $\E X$ depends on $\delta$. If $\E X&lt;0$ the probability tends to 1. If $\E X>0$ the probability decays as you say. </p> <p>When $\alpha&lt;1$ the probability indeed depends to a constant. You can estimate $\P(X_0>X_1+\dots+X_{n-1})$, since the sum is just an independent stable variable. These events are almost disjoint. In the maximally skewed case, the variables are positive, and then the events are completely disjoint.</p> <p>The plimit probability can also be expressed as the probability that the maximal point in a certain non-homogenuous Poisson process is greater than the sum of the rest.</p>