TBan-I

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

Exercise 1: The Gumbel Distribution

Let's go back to the end of Lecture 1. In the Gaussian case, expand $A(E)$ around $a_{M}$:

$Q_{M}(E)\sim \exp \left(-MP(E)\right)\sim \exp \left(-Me^{A(a_{M})+A'(a_{M})\cdot (E-a_{M})}\right)$

Show that by setting

$a_{M}=E_{\min }^{\text{typ}}=-\sigma {\sqrt {2\log M}}+\ldots \quad {\text{and}}\quad b_{M}={\frac {1}{A'(a_{M})}}={\frac {\sigma }{\sqrt {2\log M}}}$

you find

$\exp(A(a_{M}))={\frac {1}{M}}\qquad {\text{and}}\qquad Q_{M}(E)\equiv {\text{Prob}}(E_{\min }>E)\sim \exp \!\!\left[-\exp \!\!\left({\frac {E-a_{M}}{b_{M}}}\right)\right]$

Therefore, the variable $z=(E-a_{M})/b_{M}$ 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 $p(E)=\exp(E)\quad {\text{with}}\quad E\in (-\infty ,0)$.

• Scaling Form: $\exp(z)\,\exp(-e^{z})$

Exercise 2: The Weakest Link and the Weibull Distribution

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

Let $x_{1},x_{2},\dots ,x_{L}$ 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 $\alpha >0$ and $\Gamma (\alpha )$ the Gamma function:

$p(x)={\frac {x^{\alpha -1}}{\Gamma (\alpha )}}e^{-x}$

According to extreme value theory, the probability that the weakest link is smaller than $x$ is

$Q_{L}(x)\sim \exp \!{\bigl [}-LP(x){\bigr ]}=\exp \!\!\left[-L\int _{0}^{x}p(t)\,dt\right]$

Use the change of variable $z={\frac {x-a_{L}}{b_{L}}}$ with $a_{L}=0$ and $b_{L}=F_{c}^{typ}$ 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 ${E_{1},E_{2},\ldots ,E_{M}}$, define

$n(x)=\#\{i\mid E_{\min }<E_{i}<E_{\min }+x\}$

that is, the number of random variables lying above the minimum $E_{\min }$ but less than $E_{\min }+x$. This is itself a random variable. We are interested in its mean value:

${\overline {n(x)}}=\sum _{k=0}^{M-1}k\,{\text{Prob}}[n(x)=k]$

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

${\overline {n(x)}}=e^{x/b_{M}}-1$

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

${\text{Prob}}[n(x)=k]=M{\binom {M-1}{k}}\int _{-\infty }^{\infty }dE\;p(E)\,[P(E+x)-P(E)]^{k}\,[1-P(E+x)]^{M-k-1}$

To compute ${\overline {n(x)}}$, you must sum over $k$. Use the identity

$\sum _{k=0}^{M-1}k{\binom {M-1}{k}}(A-B)^{k}B^{M-1-k}=(A-B){\frac {d}{dA}}\sum _{k=0}^{M-1}{\binom {M-1}{k}}(A-B)^{k}B^{M-1-k}=(M-1)(A-B)A^{M-2}$

to arrive at the form:

${\overline {n(x)}}=M(M-1)\int _{-\infty }^{\infty }dE\;p(E)\left[P(E+x)-P(E)\right](1-P(E))^{M-2}=M\int _{-\infty }^{\infty }dE\;\left[P(E+x)-P(E)\right]{\frac {dQ_{M-1}(E)}{dE}}$

where $Q_{M-1}(E)=[1-P(E)]^{M-1}$.

Step 2: the Gumbel limit So far, no approximations have been made. To proceed, we use $Q_{M-1}(E)\approx Q_{M}(E)$ and its asymptotics Gumbel form:

${\frac {dQ_{M-1}(E)}{dE}}\,dE\;\sim \;\exp \!\!\left({\frac {E-a_{M}}{b_{M}}}\right)\exp \!\!\left[-\exp \!\!\left({\frac {E-a_{M}}{b_{M}}}\right)\right]{\frac {dE}{b_{M}}}=e^{z}e^{-e^{z}}dz$

where $z=(E-a_{M})/b_{M}$.

The main contribution to the integral comes from the region near $E\approx a_{M}$, where $P(E)\approx e^{(E-a_{M})/b_{M}}/M$.

Compute the integral and verify that you obtain:

${\overline {n(x)}}={\bigl (}e^{x/b_{M}}-1{\bigr )}\int _{-\infty }^{\infty }dz\,e^{2z-e^{z}}=e^{x/b_{M}}-1$