TBan-I

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In the following exercises, we will use the notation from extreme value statistics as introduced in the course.

Exercise 1: The Gaussian case

Let us analyze in detail the case of a Gaussian distribution with zero mean and variance σ2. Using integration by parts, we can write :

P(E)=Edx2πσ2ex22σ2=12πE22σ2dttet=σ2π|E|eE22σ214πE22σ2dttet

The asymptotic expansion for E  is :

P(E)σ2π|E|eE22σ2+O(eE22σ2|E|2)


In general, the variable z=(EaM)/bM is distributed according to an M-independent distribution.

It is possible to generalize this result and classify the scaling forms into the Gumbel universality class:

  • Characteristics:
    • Applies when the tails of p(E) decay faster than any power law.
    • Examples: the Gaussian case discussed here or exponential distributions p(E)=exp(E)withE(,0).
  • Scaling Form: exp(z)exp(ez)


Exercise 2: The Weakest Link and the Weibull Distribution

Consider a chain of length L subjected to a tensile force F. Define Fc as the force required to break the chain. The goal of this exercise is to determine how Fc depends on L and to characterize its sample-to-sample fluctuations. Throughout the exercise, you work in the limit of large L.


Let x1,x2,,xL denote the strengths of the individual links. Assume that these are positive, identically distributed, and independent random variables. Consider the Gamma distribution with shape parameter α>0 and Γ(α) the Gamma function:

p(x)=xα1Γ(α)ex

Questions:

  • Compute the typical value Fctyp and discuss its dependence on L.


  • According to extreme value theory, the probability that the weakest link is smaller than x is
QL(x)exp[LP(x)]=exp[L0xp(t)dt]

Use the change of variable z=xaLbL with aL=0 and bL=Fctyp to find an L-independent distribution.

Exercise 3: number of states above the minimum

Definition of n(x):Given a realization of the random energies E1,E2,,EM, define

n(x)=#{iEmin<Ei<Emin+x}

that is, the number of random variables lying above the minimum

Emin

but less than

Emin+x

. This is itself a random variable. We are interested in its mean value:

n(x)=k=0M1kProb[n(x)=k]

The Final goal is to show that, for large M (when the extremes are described by the Gumbel distribution), you have:

n(x)=ex/bM1

Step 1: Exact manipulations: You start from the exact expression for the probability of k states in the interval:

Prob[n(x)=k]=M(M1k)dEp(E)[P(E+x)P(E)]k[1P(E+x)]Mk1

To compute n(x), you must sum over k. Use the identity

k=0M1k(M1k)(AB)kBM1k=(AB)ddAk=0M1(M1k)(AB)kBM1k=(M1)(AB)AM2

to arrive at the form:

n(x)=M(M1)dEp(E)[P(E+x)P(E)](1P(E))M2=MdE[P(E+x)P(E)]dQM1(E)dE

where QM1(E)=[1P(E)]M1.

Step 2: the Gumbel limit So far, no approximations have been made. To proceed, we use QM1(E)QM(E) and its asymptotics Gumbel form:

dQM1(E)dEdEexp(EaMbM)exp[exp(EaMbM)]dEbM=ezeezdz

where z=(EaM)/bM.

The main contribution to the integral comes from the region near EaM, where P(E)e(EaM)/bM/M.


Compute the integral and verify that you obtain:

n(x)=(ex/bM1)dze2zez=ex/bM1