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Given a list of identical and independently distributed Levy Stable 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.:

$$ M = \text{Max}(X_0, X_1, \dots, X_{n-1}) $$ $$ \text{Pr}( M > \sum_{j=0}^{n-1} X_j - M ) $$

Where, in Nolan'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.

From numerical simulations, it appears that for the region of $0 < \alpha < 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$. For $1 < \alpha < 2$ 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 $\gamma$.

I have tried making a heuristic argument to the in the form of:

$$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j > 0 )$$

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:

$$ \text{Pr}( X > x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha} $$ $$ c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi $$

but this leaves me nervous and a bit unsatisfied.

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.

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.

Note: There didn't seem to be much interest from this site so I've cross-posted to math.stackexchange.com here.

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2 Answers 2

When $\alpha<1$ your heuristic argument above is correct I think, at least for the totally asymmetric case that you considered ($\beta=1$). Totally asymmetric stable random variables with $\alpha<1$ are non-negative. Thus one of the random variables being larger than the sum of the rest implies that it must be the maximum. Thus, $$ P\left(M > \sum_{j=0}^{n-1} X_j - M \right) = \sum_{i=0}^{n-1}P\left( X_i > \sum_{j\neq i} X_j \right) = n P\left(X_0 > \sum_{j=1}^{n-1} X_j \right) $$ Now, if the tail asymptotics of $X$ are $P(X>x) \sim C_0 x^{-\alpha}$ then by conditioning on the sum $\sum_{j=1}^{n-1}$ we obtain that \begin{align*} \lim_{n\rightarrow\infty} n P\left(X_0 > \sum_{j=1}^{n-1} X_j \right) &= \lim_{n\rightarrow\infty} E\left[ n C_0 \left( \sum_{j=1}^{n-1} X_j \right)^{-\alpha} \right] \\ &= \lim_{n\rightarrow\infty} \frac{C_0 n}{n-1} E\left[ \left( \frac{1}{(n-1)^{1/\alpha}} \sum_{j=1}^{n-1} X_j \right)^{-\alpha} \right] \\ &= \lim_{n\rightarrow\infty}\frac{ C_0 n}{n-1} E\left[ X^{-\alpha} \right]\\ &= C_0 E\left[ X^{-\alpha} \right]. \end{align*} Note that the first equality can be justified by the dominated convergence theorem by using the fact that instead of using the tail asymptotics we use that there is an upper bound $P(X>x) \leq C' x^{-\alpha}$ for some constant $C'<\infty$.

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A similar argument can give a lower bound that matches your conjectured polynomial rate of decay when $\alpha >1$ as well. Indeed, \begin{align*}P\left( M > \sum_{j=0}^{n-1} X_j - M \right) &\geq \sum_{i=0}^{n-1} P(X_i > n, \, \max_{j\neq i} X_j \leq n, \, \sum_{j\neq i} X_j \leq n) \\& = n P(X_0>n)P\left( \max_{1\leq j\leq n-1} X_j \leq n,\, \sum_{j=1}^{n-1} X_j \leq n \right). \end{align*} The last probability tends to 1 as $n\rightarrow \infty$ and $n P(X_0>n) \sim C_0 n^{1-\alpha}$. I'm pretty sure that by being more careful you can actually get the correct asymptotics in a similar manner. –  Jon Peterson Jul 3 '13 at 14:11

When $\alpha>1$, the LLN implies that the sum is nearly $n\mathbb E X$, and $\mathbb E X$ depends on $\delta$. If $\mathbb E X<0$ the probability tends to 1. If $\mathbb E X>0$ the probability decays as you say.

When $\alpha<1$ the probability indeed depends to a constant. You can estimate $\mathbb P(X_0>X_1+\dots+X_{n-1})$, 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.

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.

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