TBan-I

Nei seguente esercizio useremo le notazioni della statistica dei valori estremi usate nel corso.

Exercise 1: The Gumbel Distribution

In the spirit of the central limit theorem, you look for a scaling form:

$E_{\min }=a_M+b_{M}z$

The constants $a_M$ and $b_M$ absorb the dependence on $M$, while the random variable $z$ is distributed according to a probability distribution that does not depend on $M$.

In the Gaussian case, expand $A(E)$ around $a_M$:

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

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:

$P(z)=\exp (z)\exp (-e^z)$

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:

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