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 $a_M$:

$Q_M(E)\sim \exp (-MP(E))\sim \exp (-M{e}^{A(a_M)+A^{\prime }(a_M)\cdot (E-a_M)})$

Show that by setting

$a_M=E_{\min }^{\text{typ}}=-\sigma \sqrt{2\log M}+\dots \text{and}{b}_M=\frac{1}{A^{\prime }(a_M)}=\frac{\sigma }{\sqrt{2\log M}}$

you find

$\exp (A(a_M))=\frac{1}{M}\text{and}{Q}_M(E)\equiv \text{Prob}(E_{\min }>E)\sim \exp [-\exp \left(\frac{E-a_M}{b_M}\right)]$

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 $p(E)=\exp (E)\text{with}E\in (-\mathrm{\infty },0)$.

• 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 $\mathrm{\Gamma }(\alpha )$ the Gamma function:

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

Questions:

• Compute the typical value $F_c^{typ}$ and discuss its dependence on $L$.

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

$Q_L(x)\sim \exp [-LP(x)]=\exp [-L\underset{0}{\overset{x}{\int }}p(t)dt]$

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 $E_1,E_2,\dots ,E_M$, define

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

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:

$\text{Prob}[n(x)=k]=M(\genfrac{}{}{0ex}{}{M-1}{k})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dEp(E)[P(E+x)-P(E){]}^k[1-P(E+x){]}^{M-k-1}$

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

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

to arrive at the form:

$\overline{n(x)}=M(M-1)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dEp(E)[P(E+x)-P(E)](1-P(E){)}^{M-2}=-M\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dE[P(E+x)-P(E)]\frac{d{Q}_{M-1}(E)}{dE}$

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:

$-\frac{d{Q}_{M-1}(E)}{dE}dE\sim \exp \left(\frac{E-a_M}{b_M}\right)\exp [-\exp \left(\frac{E-a_M}{b_M}\right)]\frac{dE}{b_M}=e^z{e}^{-e^z}dz$

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:

$\overline{n(x)}=(e^{x/b_M}-1)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dz{e}^{2z-e^z}=e^{x/b_M}-1$