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Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T13:11:08Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 15:11, 18 January 2026
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	Determine the limiting, <math>L</math>-independent distribution of <math>z</math>,		Determine the limiting, <math>L</math>-independent distribution of <math>z</math>,
	and identify the corresponding universality class.		and identify the corresponding universality class.


			== Exercise 3: Number of states close to the minimum ==

			In this exercise, we study a further consequence of extreme value statistics
			by characterizing not only the minimum energy, but also the number of states
			lying close to it.
			This quantity will play an important role in the discussion of the Random Energy Model.

			'''Definition of <math>n(x)</math>:'''
			Given a realization of the random energies
			<math>\{E_1, E_2, \ldots, E_M\}</math>,
			we define

			<center><math>
			n(x) = \#\{ i \mid E_{\min} < E_i < E_{\min} + x \},
			</math></center>

			that is, the number of states lying within an energy window <math>x</math>
			above the minimum.
			This is a random variable.
			We are interested in its disorder average,

			<center><math>
			\overline{n(x)} = \sum_{k=0}^{M-1} k \, \mathrm{Prob}[n(x)=k].
			</math></center>

			'''Final goal:'''
			Show that, in the large-<math>M</math> limit,
			provided the minimum belongs to the Gumbel universality class,

			<center><math>
			\overline{n(x)} = e^{x/b_M} - 1.
			</math></center>

			This result depends only on the scaling parameter <math>b_M</math>
			introduced in the extreme value analysis.

			'''Step 1: Exact manipulations'''
			The probability that exactly <math>k</math> states fall in the interval
			<math>(E_{\min}, E_{\min}+x)</math> can be written exactly as

			<center><math>
			\mathrm{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}.
			</math></center>

			Using the identity

			<center><math>
			\sum_{k=0}^{M-1} k \binom{M-1}{k} (A-B)^k B^{M-1-k}
			= (M-1)(A-B)A^{M-2},
			</math></center>

			one obtains

			<center><math>
			\overline{n(x)}
			= M(M-1)
			\int_{-\infty}^{\infty} dE \;
			p(E)\,[P(E+x)-P(E)](1-P(E))^{M-2}
			= -M \int_{-\infty}^{\infty} dE\,[P(E+x)-P(E)]
			\frac{dQ_{M-1}(E)}{dE},
			</math></center>

			where <math>Q_{M-1}(E) = [1-P(E)]^{M-1}</math>.

			'''Step 2: Gumbel limit'''
			From now on, we assume that the extreme value statistics of the energies
			is described by the Gumbel distribution.
			So far, no approximations have been made.

			For large <math>M</math>, we may use <math>Q_{M-1}(E) \simeq Q_M(E)</math>
			and its Gumbel scaling form,

			<center><math>
			-\frac{dQ_M(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}.
			</math></center>

			Introducing the scaling variable <math>z = (E-a_M)/b_M</math>,
			the main contribution to the integral comes from the region
			<math>E \approx a_M</math>, where
			<math>P(E) \approx e^{(E-a_M)/b_M}/M</math>.

			Compute the integral and show that

			<center><math>
			\overline{n(x)}
			= \bigl(e^{x/b_M}-1\bigr)
			\int_{-\infty}^{\infty} dz \, e^{2z-e^{z}}
			= e^{x/b_M}-1.
			</math></center>

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T13:03:00Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 15:03, 18 January 2026
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	while the scale of fluctuations is set by the typical minimum itself.		while the scale of fluctuations is set by the typical minimum itself.

	= Exercise 2: Weakest-link statistics and the Weibull law =		== Exercise 2: Weakest-link statistics and the Weibull law ==

	In this exercise, we consider a situation that is qualitatively different from the Gaussian case.		In this exercise, we consider a situation that is qualitatively different from the Gaussian case.

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T12:52:51Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 14:52, 18 January 2026
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	* Compute the typical breaking force <math>F_c^{\mathrm{typ}}</math> of the chain and determine its dependence on <math>L</math>. (Hint: use the condition <math>L\,P(x)=O(1)</math>.)		* Compute the typical breaking force <math>F_c^{\mathrm{typ}}</math> of the chain and determine its dependence on <math>L</math>. (Hint: use the condition <math>L\,P(x)=O(1)</math>.)

	* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables.		* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables. According to extreme value statistics, its distribution satisfies
	According to extreme value statistics, its distribution satisfies
	<center><math> Q_L(x) \equiv \mathrm{Prob}(F_c > x)		<center><math> Q_L(x) \equiv \mathrm{Prob}(F_c > x)
	\sim \exp\!\left[-L P(x)\right],		\sim \exp\!\left[-L P(x)\right],
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	P(x)=\int_0^x p(t)\,dt.		P(x)=\int_0^x p(t)\,dt.
	</math></center>		</math></center>
			Show that the appropriate scaling form is obtained by introducing
			<math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>,
			and defining the rescaled variable

	~~Show that the appropriate scaling form is obtained by introducing <math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>, and defining the rescaled variable~~ <center><math> z = \frac{x-a_L}{b_L}.</math></center>		<center><math>
			z = \frac{x-a_L}{b_L}.
			</math></center>

	Determine the limiting, <math>L</math>-independent distribution of <math>z</math>, and identify the corresponding ~~universal distribution~~.		Determine the limiting, <math>L</math>-independent distribution of <math>z</math>,
			and identify the corresponding universality class.

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T12:49:39Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 14:49, 18 January 2026
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	Show that the appropriate scaling form is obtained by introducing <math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>, and defining the rescaled variable <center><math> z = \frac{x-a_L}{b_L}.</math></center>		Show that the appropriate scaling form is obtained by introducing <math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>, and defining the rescaled variable <center><math> z = \frac{x-a_L}{b_L}.</math></center>

	Determine the limiting, <math>L</math>-independent distribution of <math>z</math>, and identify the corresponding ~~universality class~~.		Determine the limiting, <math>L</math>-independent distribution of <math>z</math>, and identify the corresponding universal distribution.

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T12:39:47Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 14:39, 18 January 2026
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	'''Questions:'''		'''Questions:'''

	* Compute the typical breaking force <math>F_c^{\mathrm{typ}}</math>		* Compute the typical breaking force <math>F_c^{\mathrm{typ}}</math> of the chain and determine its dependence on <math>L</math>. (Hint: use the condition <math>L\,P(x)=O(1)</math>.)
	of the chain and determine its dependence on <math>L</math>.
	(Hint: use the condition <math>L\,P(x)=O(1)</math>.)

	* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables.		* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables.

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T12:39:27Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 14:39, 18 January 2026
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	Throughout the exercise, we work in the limit of large <math>L</math>.		Throughout the exercise, we work in the limit of large <math>L</math>.

	The strength of each link is drawn from a Gamma distribution with shape parameter		The strength of each link is drawn from a Gamma distribution with shape parameter <math>\alpha > 0</math>:
	<math>\alpha > 0</math>:		<center><math> p(x) = \frac{x^{\alpha - 1}}{\Gamma(\alpha)}\, e^{-x},
	<center><math>
	p(x) = \frac{x^{\alpha - 1}}{\Gamma(\alpha)}\, e^{-x},
	\qquad x \ge 0.		\qquad x \ge 0.
	</math></center>		</math></center>

Rosso: /* Exercise 2: Weakest-link statistics and the Weibull law */

2026-01-18T12:38:56Z

Exercise 2: Weakest-link statistics and the Weibull law

← Older revision		Revision as of 14:38, 18 January 2026
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	* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables.		* The breaking force of the chain is equal to the minimum of the <math>L</math> random variables.
	According to extreme value statistics, its distribution satisfies		According to extreme value statistics, its distribution satisfies
	<center><math>		<center><math> Q_L(x) \equiv \mathrm{Prob}(F_c > x)
	Q_L(x) \equiv \mathrm{Prob}(F_c > x)
	\sim \exp\!\left[-L P(x)\right],		\sim \exp\!\left[-L P(x)\right],
	\qquad		\qquad
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	</math></center>		</math></center>

	Show that the appropriate scaling form is obtained by introducing		Show that the appropriate scaling form is obtained by introducing <math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>, and defining the rescaled variable <center><math> z = \frac{x-a_L}{b_L}.</math></center>
	<math>a_L = 0</math> and <math>b_L = F_c^{\mathrm{typ}}</math>,
	and defining the rescaled variable
	<center><math>
	z = \frac{x-a_L}{b_L}.
	</math></center>

	Determine the limiting, <math>L</math>-independent distribution of <math>z</math>,		Determine the limiting, <math>L</math>-independent distribution of <math>z</math>, and identify the corresponding universality class.
	and identify the corresponding universality class.

Rosso at 12:37, 18 January 2026

2026-01-18T12:37:11Z

@@ Line 1: / Line 1: @@
 We are interested in the asymptotic behavior of the cumulative distribution
 <math>P(E)</math> in the left tail <math>E\to -\infty</math>,
@@ Line 36: / Line 39: @@
 which is the expression used in the course to derive the scaling form of the minimum.

Rosso: Created page with "We are interested in the asymptotic behavior of the cumulative distribution $P(E)$ in the left tail $E\to -\infty$ , since the minimum is controlled by the regime where $M P(E)=O(1)$ . Starting from the Gaussian distribution with zero mean and variance $\sigma^2$ , we write the cumulative as $P(E) = \int_{-\infty}^E \frac{dx}{\sqrt{2\pi\sigma^2}} \, e^{-x^2/2\sigma^2}.$ Using integration by parts (o..."

2026-01-18T12:27:00Z

Created page with "We are interested in the asymptotic behavior of the cumulative distribution <math>P(E)</math> in the left tail <math>E\to -\infty</math>, since the minimum is controlled by the regime where <math>M P(E)=O(1)</math>. Starting from the Gaussian distribution with zero mean and variance <math>\sigma^2</math>, we write the cumulative as <center><math> P(E) = \int_{-\infty}^E \frac{dx}{\sqrt{2\pi\sigma^2}} \, e^{-x^2/2\sigma^2}. </math></center> Using integration by parts (o..."

New page

We are interested in the asymptotic behavior of the cumulative distribution
<math>P(E)</math> in the left tail <math>E\to -\infty</math>,
since the minimum is controlled by the regime where <math>M P(E)=O(1)</math>.

Starting from the Gaussian distribution with zero mean and variance <math>\sigma^2</math>,
we write the cumulative as
<center><math>
P(E) = \int_{-\infty}^E \frac{dx}{\sqrt{2\pi\sigma^2}}
\, e^{-x^2/2\sigma^2}.
</math></center>

Using integration by parts (or equivalently the change of variable
<math>t=x^2/2\sigma^2</math>), one finds
<center><math>
P(E) =
\frac{\sigma}{\sqrt{2\pi}\,|E|}\,e^{-E^2/2\sigma^2}
- \frac{1}{4\sqrt{\pi}}
\int_{E^2/2\sigma^2}^{\infty} \frac{dt}{t}\,e^{-t}.
</math></center>

For <math>E\to -\infty</math>, the second term is subleading, and the cumulative admits the asymptotic expansion
<center><math>
P(E) =
\frac{\sigma}{\sqrt{2\pi}\,|E|}\,e^{-E^2/2\sigma^2}
\left[1 + O\!\left(\frac{1}{E^2}\right)\right].
</math></center>

This result can be written in the form
<center><math>
P(E)=\exp\!\bigl(A(E)\bigr),
\qquad
A(E) = -\frac{E^2}{2\sigma^2}
- \log\!\left(\frac{\sqrt{2\pi}\,|E|}{\sigma}\right)
+ \ldots,
</math></center>

which is the expression used in the course to derive the scaling form of the minimum.