# Probability of the maximum (Levy Stable) arbitrary variable in a checklist being more than the amount of the remainder?

Initial blog post on Mathoverflow here.

Offered a checklist of the same and also individually dispersed Levy Stable arbitrary variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum goes beyond the amount of the remainder? 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 is symbols, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the essential backer, $\beta$ is the alter, $\gamma$ is the range parameter and also $\delta$ is the change. For simpleness, I have actually taken the alter parameter, $\beta$, to be 1 (maximally manipulated to the right) and also $\delta=0$ so every little thing has its setting focused in a period near 0.

From mathematical simulations, it shows up that for the area of $0 < \alpha < 1$, the probability merges to a constant, irregardless of $n$ or $\gamma$. Below is a story of this area for $n=500$, $0< \alpha < 1$, where each factor stands for the outcome of 10,000 arbitrary attracts. The chart looks specifically the very same for $n=100, 200, 300$ and also $400$. For $1 < \alpha < 2$ it shows up to go as $O(1/n^{\alpha - 1})$ (possibly?) irregardless of $n$ or $\gamma$. Below is a story of the probability for $\alpha \in (1.125, 1.3125)$ as a function of $n$. Keep in mind that it is a log - log story and also I have actually given the charts $1/x^{.125}$ and also $1/x^{.3125}$ for reference. It is tough to distinguish the chart unless you line them up, yet the suitable for each is a little bit off, and also it looks like if the (log - log) incline of the real information is steeper than my hunch for each and every. Each factor stands for 10,000 models. For $\alpha=1$ it is unclear (to me) what is taking place, yet it seems a lowering function depending on $n$ and also $\gamma$.

I have actually attempted making a heuristic argument to the in the kind 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 making use of formula is given by Nolan (pg. 27) for the parameters of the implied r.v. $U = X_0 - \sum_{j=1}^{n-1} X_j$ incorporated with the tail estimate :

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

yet this leaves me worried and also a little bit disappointed.

Simply for contrast, if $X_j$ were required consistent r.v.'s on the device period, this function would certainly decrease greatly promptly. I visualize comparable outcomes hold were the $X_j$ is Gaussian, though any kind of explanation on that particular factor would certainly be valued.

Obtaining shut kind remedies for this is possibly inconceivable, as there isn't also a shut kind remedy for the pdf of Levy - Stable arbitrary variables, yet obtaining bounds on what the probability is would certainly be handy. I would certainly value any kind of aid with concerns to just how to assess these sorts of inquiries as a whole such as basic approaches or referrals to various other operate in this location.

If this trouble is primary, I would substantially value any kind of reference to a book, tutorial or paper that would certainly aid me address troubles of this type.

UPDATE : George Lowther and also Shai Covo have actually addressed this inquiry listed below. I simply intended to offer a couple of even more images that contrast their response to several of the mathematical experiments that I did.

Below is the probability of the maximum component being bigger than the remainder for a checklist dimension of $n=100$ as a function of $\alpha$, $\alpha \in (0,1)$. Each factor stands for 10,000 simulations. Below are 2 charts for 2 values of $\alpha \in \{1.53125, 1.875\}$. Both have the function $(2/\pi) \sin(\pi \alpha / 2) \Gamma(\alpha) n (( \tan(\pi \alpha/2) (n^{1/\alpha} - n))^{-\alpha}$ with various prescalars before them to get them to align ($1/4$ and also $1/37$, specifically) laid over for reference.  As George Lowther appropriately mentioned, for the reasonably tiny $n$ being taken into consideration below, the result of the added $n^{1/\alpha}$ term (when $1 < \alpha < 2$) is non - minimal and also this is why my initial reference stories did not associate the outcomes of the simulations. As soon as the complete estimate is placed in, the fit is better.

When I navigate to it, I will certainly attempt and also upload some even more images for the instance when $\alpha=1$ as a function of $n$ and also $\gamma$.

0
2019-12-02 02:52:05
Source Share

So $M=\mbox{Max}(X_0,\ldots,X_{n-1})$ after that you recognize that :

$$\mathbb{P}(M\geq x)=1-F(x)^n$$

with $F(x)$ the Levy - secure collective circulation. Currently, you additionally have

$$\mathbb{P}(M\geq \sum X_i) = \int p(M\geq \sum X_i | \sum X_i = x) p(x) dx$$

Where $p(x)$ is the thickness function for Levy - secure circulations. (Since necessarily Levy - secure circulations are secure under summation of arbitrary Levy - secure variables.)

So it appears to me you have in concept all the active ingredients to attempt some asymptotics, also if you do not have a specific formula for all the circulations. Possibly you can attempt with some for which the formula exists, like Cauchy and also regular circulations.

Sorry for the sloppiness, simply attempting to illustration a strategy in the meantime.

0
2019-12-03 04:25:22
Source

Some approximation.

I take into consideration just the grandfather clause where the $X_i$ declare motor home is such that, by self - resemblance, $X_1 + \cdots + X_{n-1}$ is equivalent in circulation to $(n-1)^{1/\alpha}X_1$ where $\alpha$ is the affiliated index. After that, it appears valuable to start (in a similar way to what you did over) with $$P = n{\rm P}(X_0 > X_1 + \cdots + X_{n-1}),$$ where $P$ is the probability in the title. After that, remembering the self - resemblance argument, an uncomplicated application of the regulation of complete probability (conditioning on $X_0$) generates $$P = n\int F (s/(n - 1)^{1/\alpha } )f(s){\rm d}s,$$ where $F$ and also $f$ are the usual circulation and also thickness features of the $X_i$, specifically. See if you can proceed from below.

0
2019-12-03 02:26:04
Source

What you are requesting for is a grandfather clause of the joint circulation of the maximum and also amount of a series of arbitrary variables. The reality that the variables are secure streamlines this and also permits us to calculate the joint circulation in the restriction as n mosts likely to infinity. This restriction can be defined in regards to the joint circulation of the value of a secure Lévy process sometimes $t=1$ and also its maximum dive dimension over $t\le1$. Lévy procedures are appropriate continual with left restrictions (cadlag) and also have fixed independent increments. Secure Lévy procedures have increments with fixed circulations.

If $X_i$ each have the $\alpha$ - secure circulation represented by $S(\alpha,\beta,\gamma,\delta)=S(\alpha,\beta,\gamma,\delta;0)$ in the message you link, after that the amount $S=\sum_{i=0}^{n-1}X_i$ has the $S(\alpha,\beta,n^{1/\alpha}\gamma,\delta_n)$ circulation where $$\delta_n=\begin{cases} n\delta + \gamma\beta\tan\frac{\pi\alpha}{2}(n^{1/\alpha}-n),&\textrm{if }\alpha\not=1,\\ n\delta +\gamma\beta\frac{2}{\pi}n\log n,&\textrm{if }\alpha=1. \end{cases}\qquad\qquad{\rm(1)}$$ This is formula (1.9) on web page 20 of the connected notes. It will certainly be handy to rescale the amount to remove the dependancy on n. The arbitrary variable $Y=n^{-1/\alpha}(S-\delta_n)$ has the $S(\alpha,\beta,\gamma,0)$ circulation (by Proposition 1.16).

Currently, to connect this to Lévy procedures. Allow $Z_t$ be the Lévy procedure with $Z_1\sim Y$. We can take $X_i=n^{1/\alpha}(Z_{(i+1)/n}-Z_{i/n})+\delta_n/n$, which will certainly be independent with the called for $S(\alpha,\beta,\gamma,\delta)$ circulation. Actually, in this instance we have $Z_1=Y$ and also $S=n^{1/\alpha}Z_1+\delta_n$. Creating $M=\max_iX_i$ and also $\Delta Z^*_t$ for the maximum of $\Delta Z_s=Z(s)-Z(s-)$ over $s\le t$ offers $n^{-1/\alpha}(M-\delta_n/n)\to\Delta Z^*_1$ as n mosts likely to infinity. We can after that calculate the probability P that you are requesting for to leading order as n comes to be huge, \begin{align} P&\equiv\mathbb{P}\left(M > S-M\right)\\ &=\mathbb{P}\left(2n^{-1/\alpha}(M-\delta_n/n) > n^{-1/\alpha}(S-2\delta_n/n)\right)\\ &\sim\mathbb{P}\left(2\Delta Z^*_1 > Z_1+n^{-1/\alpha}\delta_n(1-2/n)\right) \end{align} So, $$P\sim\mathbb{P}\left(2\Delta Z^*_1 - Z_1 > n^{-1/\alpha}\delta_n\right)$$ and also the leading term in the asymptotics for P are established by the circulation of $2\Delta Z^*_1-Z_1$.

For $\alpha < 1$ it can be seen from (1) that $n^{-1/\alpha}\delta_n\to\gamma\beta\tan\frac{\pi\alpha}{2}$ as n mosts likely to infinity, so we get a limited and also favorable restriction $$P\to\mathbb{P}\left(2\Delta Z^*_1-Z_1 > \gamma\beta\tan\frac{\pi\alpha}{2}\right)$$ as $n\to\infty$. The tangent has actually just shown up in the expression below as a result of the certain parameterization that we are making use of. It is a little bit less complex if we revise this in regards to $\tilde Z_t=Z_t+\gamma\beta t\tan\frac{\pi\alpha}{2}$. After that, for the $\beta=1$ instance you are making use of, $\tilde Z_t$ has assistance $[0,\infty)$ (Lemma 1.10 in the connected message). This suggests that $\tilde Z$ is a secure subordinator and also the called for probability is $$P\to\mathbb{P}\left(2\Delta\tilde Z^*_1 > \tilde Z_1\right)=\frac{\sin(\pi\alpha)}{\pi\alpha}.$$ [I've been a little bit stealthy below, and also replaced in the expression that Shai Covo gave up his answer for $\mathbb{P}(2\Delta\tilde Z^*_1 > \tilde Z_1)$ and also streamlined a little bit making use of Euler's reflection formula. ]

On the various other hand, if $\alpha > 1$ after that (1) offers $n^{-1/\alpha}\delta_n\sim n^{1-1/\alpha}(\delta-\gamma\beta\tan\frac{\pi\alpha}{2})$ and also, if $\alpha=1$, after that $n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$. In this instance, P often tends to zero at a price depending on the tail of the circulation function of $2\Delta Z^*_1-Z_1$.

It is additionally feasible to make use of the Lévy-Khintchine representation to calculate the tail of the circulation function of $2\Delta Z^*_1-Z_1$. The dives of a Lévy procedure are defined by a Lévy action $\nu$ on the actual numbers, to make sure that the dives of dimension coming from a set $A$ take place according to a Poisson process of price $\nu(A)$. An $\alpha$ - secure procedure has Lévy gauge $d\nu(x)=c\vert x\vert^{-\alpha-1}\,dx$, where $c$ just relies on the indicator of $x$. We can back out the value of $c$ by computing the particular function of $Z_1$ making use of the Lévy - Khintchine formula (basically the Fourier change of $\nu$, see Wikipedia, which makes use of W for the Lévy action) and also comparing to expression (1.4) from your message. Experiencing this calculation offers $$d\nu(x)=\pi^{-1}\alpha\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta{\rm sign}(x))\vert x\vert^{-\alpha-1}\,dx.$$ The variety of dives of dimension more than $x$ is Poisson dispersed of price $\nu((x,\infty))$ offering, \begin{align}\mathbb{P}(\Delta Z^*_1> x) &= 1-e^{-\nu((x,\infty))}\\ &\sim\nu((x,\infty))\\ &=\pi^{-1}\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta)x^{-\alpha}. \end{align}\qquad{\rm(2)} Comparing with your connected message (Theorem 1.12), this is specifically the like the tail of $Z_1$! That is, $\mathbb{P}(\Delta Z^*_1 > x)\sim\mathbb{P}(Z_1 > x)$. This suggests that the tail of the circulation is totally as a result of the periodic large dive of the Lévy procedure. This does in fact represent experience. I have actually outlined Cauchy procedure example courses in the past, and also you do get some courses with a solitary large dive hushing all various other practices. We can additionally see why this need to hold true. The thickness of the dive circulation is sub - exponential, symmetrical to $\vert x\vert^{-\alpha-1}$. So one large dive of dimension $x$ has a much larger probability than n tiny dives of dimension $x/n$. It additionally recommends that, in your estimation of $X_i$, the examples adding to the probability come mostly from the instance where every one of them are moderately near the mean with the exception of one extreme (favorable or adverse) $X_i$.

So, from the argument over, the leading asymptotic of the probability P originates from the instance where $\Delta Z_1^*$ is large contrasted to $Z_1-\Delta Z_1^*$, or when the biggest adverse dive $\Delta(-Z)^*_1$ is huge contrasted to $Z_1+\Delta(-Z)^*_1$. This offers \begin{align} \mathbb{P}(2\Delta Z^*_1-Z_1 > x) &\sim\mathbb{P}(\Delta Z^*_1 > x) +\mathbb{P}(\Delta (-Z)^*_1 > x)\\ &\sim 2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)x^{-\alpha}, \end{align} where I replaced in (2 ). Connecting in the asymptotic kind for $x=n^{-1/\alpha}\delta_n$ computed over offers $$P\sim2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)\left(\delta-\gamma\beta\tan\frac{\pi\alpha}{2}\right)^{-\alpha}n^{1-\alpha}$$ for $\alpha > 1$. This validates your $O(1/n^{\alpha-1})$ hunch from the mathematical simulation! Additionally, making use of $x=n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$ for $\alpha=1$ offers $$P\sim\frac{1}{\beta\log n}.$$ Hopefully this is all proper, yet I can not dismiss making a math slip over. Modify : The restriction over for $\alpha > 1$ does not look great. It seems a little over the story in the initial blog post. Nonetheless, also for n = 1000, it needs to not be really near the asymptotic restriction defined. Changing the asymptotic restriction for $\delta_n$ by its specific kind reliable changes the $n^{1-\alpha}$ term for P by $(n^{1-1/\alpha}-1)^{-\alpha}$. For n = 1000 this about increases the computed number so, the extra exact value for $\delta_n$ actions you better far from the story. Something has actually failed someplace ...

0
2019-12-03 02:07:11
Source

My previous solution was pointless. The new solution listed below finishes George is solution for the $0 < \alpha < 1$ instance, highlighted in the first chart over.

For this instance, George gave the asymptotic estimate $P_n \sim {\rm P}(2 \Delta Z^*_1 - Z_1 > 0)$ as $n \to \infty$, where $P_n$ is the probability the OP requested for, $Z = \lbrace Z_t : t \geq 0 \rbrace$ is a purely secure L\' evy procedure of index $\alpha$, and also $\Delta Z^*_1$ is the biggest dive of $Z$ while interval $[0,1]$. It is hard to get ${\rm P}(2 \Delta Z^*_1 - Z_1 > 0)$ in shut kind, yet without a doubt feasible, and also actually the adhering to even more basic outcome is recognized : $${\rm P}\bigg(\frac{{\Delta Z_1^* }}{{Z_1 }} > \frac{1}{{1 + y}}\bigg) = \frac{{y^\alpha }}{{\Gamma (1 - \alpha )\Gamma (1 + \alpha )}}, \;\; y \in [0,1],$$ where $\Gamma$ is the gamma function. Allowing $y = 1$ hence offers $P_n \sim 1/[\Gamma(1 - \alpha) \Gamma(1 + \alpha)]$. This asymptotic expression concurs quite possibly with the chart over, as I validated by examining numerous values of $\alpha \in (0,1)$. (For instance, $\alpha = 1/2$ offers $P_n \sim 2/\pi \approx 0.6366$.)

A generalization. As opposed to simply taking into consideration the probability ${\rm P}(M > \sum\nolimits_{i = 0}^{n - 1} {X_i } - M)$, we can take into consideration the probability ${\rm P}(M > y^{-1}[\sum\nolimits_{i = 0}^{n - 1} {X_i } - M])$, $y \in (0,1]$. Because George is solution, this need to represent the formula offered over for ${\rm P}(\Delta Z_1^* / Z_1 > 1/(1+y))$. This can be more generalised, in numerous instructions, based upon recognized arise from the literary works (3 valuable referrals in this context are shown someplace in the remarks above/below).

MODIFY :

As George observed, the term $\Gamma(1 - \alpha) \Gamma(1 + \alpha)$ can be streamlined to $(\pi\alpha) / \sin(\pi\alpha)$. The previous expression represents $[\Gamma(1 - \alpha)]^k \Gamma(1 + k \alpha)$, which is included in the specific formula for ${\rm P}(\Delta Z_1^* / Z_1 > 1/(1+y))$, $y > 0$, of the kind $\sum\nolimits_{k = 1}^{\left\lceil y \right\rceil } {a_k \varphi _k (y)}$. The function $\varphi _k (y)$ is some $(k-1)$ - dimensional indispensable, therefore that formula is computationally really pricey currently for modest values of $y$.

0
2019-12-03 01:35:03
Source