TBan-I - Revision history

Rosso: /* Exercise 1: The Gumbel Distribution */

2025-09-13T12:13:08Z

Exercise 1: The Gumbel Distribution

← Older revision		Revision as of 14:13, 13 September 2025
Line 1:		Line 1:
	In the following exercises, we will use the notation from extreme value statistics as introduced in the course.		In the following exercises, we will use the notation from extreme value statistics as introduced in the course.

	= Exercise 1: The ~~Gumbel Distribution~~ =		= Exercise 1: The Gaussian case =
	Let~~'s go back to~~ the ~~end~~ of ~~Lecture 1.~~		Let us analyze in detail the case of a Gaussian distribution with zero mean and variance <math>\sigma^2</math>. Using integration by parts, we can write :
	~~In the~~ Gaussian ~~case, expand~~ <math>~~A(E)</math> around <math>a_M~~</math>:		<center> <math>P(E) = \int_{-\infty}^E \frac{dx}{\sqrt{2 \pi \sigma^2}} \, e^{-\frac{x^2}{2 \sigma^2}} = \frac{1}{2\sqrt{\pi}} \int_{\frac{E^2}{2 \sigma^2}}^{\infty} \frac{dt}{\sqrt{t}} e^{-t}= \frac{\sigma}{\sqrt{2 \pi} \|E\|} e^{-\frac{E^2}{2 \sigma^2}} - \frac{1}{4\sqrt{\pi}} \int_{\frac{E^2}{2 \sigma^2}}^{\infty} \frac{dt}{t} e^{-t} </math> </center>
	<center> <math>~~Q_M~~(E) \~~sim~~ \~~exp~~\~~left(~~-~~M P(~~E)\~~right)~~ \~~sim~~ \~~exp~~\~~left(~~-M e^{~~A(a_M) +A'(a_M)~~\~~cdot (~~E-~~a_M)~~}\~~right)~~ </math> </center>		The asymptotic expansion for <math>E \to -\infty</math> is :
	~~Show that by setting~~		<center> <math>P(E) \approx \frac{\sigma}{\sqrt{2 \pi} \|E\|} e^{-\frac{E^2}{2 \sigma^2}} + O(\frac{e^{-\frac{E^2}{2 \sigma^2}}}{ \|E\|^2} ) </math> </center>
	<center> <math>~~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}~~}</math> </center>

	~~you find~~

	~~<center><math>~~
	~~\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]~~
	~~</math></center>~~

			In general, the variable <math>z = (E - a_M)/b_M</math> is distributed according to an ''M''-independent distribution.
	~~Therefore~~, the variable <math>z = (E - a_M)/b_M</math> 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''':		It is possible to generalize this result and classify the scaling forms into the '''Gumbel universality class''':

Rosso: /* Exercise 3: number of states above the minimum */

2025-08-31T13:17:32Z

Exercise 3: number of states above the minimum

← Older revision		Revision as of 15:17, 31 August 2025
Line 74:		Line 74:
	<center><math> \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} </math></center>		<center><math> \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} </math></center>
	to arrive at the form:		to arrive at the form:
	<center><math> \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{d Q_{M-1}(E)}{dE} </math></center>		<center><math> \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{d Q_{M-1}(E)}{dE} </math></center>

	where <math>Q_{M-1}(E) = [1-P(E)]^{M-1}</math>.		where <math>Q_{M-1}(E) = [1-P(E)]^{M-1}</math>.
Line 80:		Line 80:
	'''Step 2: the Gumbel limit ''' So far, no approximations have been made. To proceed, we use <math> Q_{M-1}(E)\approx Q_M(E)</math> and its asymptotics Gumbel form:		'''Step 2: the Gumbel limit ''' So far, no approximations have been made. To proceed, we use <math> Q_{M-1}(E)\approx Q_M(E)</math> and its asymptotics Gumbel form:
	<center><math>		<center><math>
	\frac{d Q_{M-1}(E)}{dE} \, dE		- \frac{d Q_{M-1}(E)}{dE} \, dE
	\;\sim\;		\;\sim\;
	\exp\!\!\left(\frac{E-a_M}{b_M}\right)		\exp\!\!\left(\frac{E-a_M}{b_M}\right)

Rosso: /* Exercise 2: The Weakest Link and the Weibull Distribution */

2025-08-31T13:15:33Z

Exercise 2: The Weakest Link and the Weibull Distribution

← Older revision		Revision as of 15:15, 31 August 2025
Line 45:		Line 45:
	</math></center>		</math></center>

			'''Questions:'''

	According to extreme value theory, the probability that the weakest link is smaller than <math>x</math> is		* Compute the typical value <math>F_c^{ typ}</math> and discuss its dependence on <math>L</math>.


			* According to extreme value theory, the probability that the weakest link is smaller than <math>x</math> is
	<center><math>		<center><math>
	Q_L(x) \sim \exp\!\bigl[-L P(x)\bigr]		Q_L(x) \sim \exp\!\bigl[-L P(x)\bigr]

Rosso: /* Exercise 2: The Weakest Link and the Weibull Distribution */

2025-08-31T13:13:30Z

Exercise 2: The Weakest Link and the Weibull Distribution

← Older revision		Revision as of 15:13, 31 August 2025
Line 51:		Line 51:
	= \exp\!\!\left[-L \int_0^x p(t) \, dt \right]		= \exp\!\!\left[-L \int_0^x p(t) \, dt \right]
	</math></center>		</math></center>
			Use the change of variable <math>z = \frac{x - a_L}{b_L}</math> with <math>a_L = 0</math> and <math>b_L = F_c^{typ}</math> to find an <math>L</math>-independent distribution.

	=Exercise 3: number of states above the minimum=		=Exercise 3: number of states above the minimum=

Rosso: /* Exercise 2: The Weakest Link and the Weibull Distribution */

2025-08-31T13:12:21Z

Exercise 2: The Weakest Link and the Weibull Distribution

← Older revision		Revision as of 15:12, 31 August 2025
Line 47:		Line 47:

	According to extreme value theory, the probability that the weakest link is smaller than <math>x</math> is		According to extreme value theory, the probability that the weakest link is smaller than <math>x</math> is
			<center><math>
			Q_L(x) \sim \exp\!\bigl[-L P(x)\bigr]
			= \exp\!\!\left[-L \int_0^x p(t) \, dt \right]
			</math></center>

	=Exercise 3: number of states above the minimum=		=Exercise 3: number of states above the minimum=

Rosso: /* Exercise 2: The Weakest Link and the Weibull Distribution */

2025-08-31T13:11:45Z

Exercise 2: The Weakest Link and the Weibull Distribution

← Older revision		Revision as of 15:11, 31 August 2025
Line 44:		Line 44:
	p(x) = \frac{x^{\alpha - 1}}{\Gamma(\alpha)} e^{-x}		p(x) = \frac{x^{\alpha - 1}}{\Gamma(\alpha)} e^{-x}
	</math></center>		</math></center>


			According to extreme value theory, the probability that the weakest link is smaller than <math>x</math> is

	=Exercise 3: number of states above the minimum=		=Exercise 3: number of states above the minimum=

Rosso: /* Exercise 3: The Weakest Link and the Weibull Distribution */

2025-08-31T13:10:36Z

Exercise 3: The Weakest Link and the Weibull Distribution

← Older revision		Revision as of 15:10, 31 August 2025
Line 29:		Line 29:


	== Exercise 3: The Weakest Link and the Weibull Distribution ==		== Exercise 2: The Weakest Link and the Weibull Distribution ==

	Consider a chain of length <math>L</math> subjected to a tensile force <math>F</math>.		Consider a chain of length <math>L</math> subjected to a tensile force <math>F</math>.
Line 35:		Line 35:
	The goal of this exercise is to determine how <math>F_c</math> depends on <math>L</math> and to characterize its sample-to-sample fluctuations.		The goal of this exercise is to determine how <math>F_c</math> depends on <math>L</math> and to characterize its sample-to-sample fluctuations.
	Throughout the exercise, you work in the limit of large <math>L</math>.		Throughout the exercise, you work in the limit of large <math>L</math>.


			Let <math>x_1, x_2, \dots, x_L</math> 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 <math>\alpha > 0</math> and <math>\Gamma(\alpha)</math> the Gamma function:

			<center><math>
			p(x) = \frac{x^{\alpha - 1}}{\Gamma(\alpha)} e^{-x}
			</math></center>

	=Exercise 3: number of states above the minimum=		=Exercise 3: number of states above the minimum=

Rosso: /* esercizio 2: The weakest link */

2025-08-31T13:09:53Z

esercizio 2: The weakest link

← Older revision		Revision as of 15:09, 31 August 2025
Line 28:		Line 28:
	* '''Scaling Form:''' <math> \exp(z)\,\exp(-e^{z}) </math>		* '''Scaling Form:''' <math> \exp(z)\,\exp(-e^{z}) </math>

	~~=esercizio 2: The weakest link=~~

			== Exercise 3: The Weakest Link and the Weibull Distribution ==

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

	=Exercise 3: number of states above the minimum=		=Exercise 3: number of states above the minimum=

Rosso: /* Exercise 1: The Gumbel Distribution */

2025-08-31T12:53:41Z

Exercise 1: The Gumbel Distribution

← Older revision		Revision as of 14:53, 31 August 2025
Line 26:		Line 26:
	** Examples: the Gaussian case discussed here or exponential distributions <math>p(E) = \exp(E) \quad \text{with} \quad E \in (-\infty, 0)</math>.		** Examples: the Gaussian case discussed here or exponential distributions <math>p(E) = \exp(E) \quad \text{with} \quad E \in (-\infty, 0)</math>.

	* '''Scaling Form:'''		* '''Scaling Form:''' <math> \exp(z)\,\exp(-e^{z}) </math>
	~~<center>~~<math> ~~P(z) =~~ \exp(z)\,\exp(-e^{z}) </math~~></center~~>

	=esercizio 2: The weakest link=		=esercizio 2: The weakest link=

Rosso: /* Exercise 1: The Gumbel Distribution */

2025-08-31T12:50:40Z

Exercise 1: The Gumbel Distribution

← Older revision		Revision as of 14:50, 31 August 2025
Line 2:		Line 2:

	= Exercise 1: The Gumbel Distribution =		= Exercise 1: The Gumbel Distribution =
			Let's go back to the end of Lecture 1.
	In the ~~spirit of the central limit theorem~~, ~~you look for a scaling form:~~		In the Gaussian case, expand <math>A(E)</math> around <math>a_M</math>:
			<center> <math>Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) </math> </center>
	~~<center>~~<math> ~~E_{\min} = a_M + b_M z~~ </math>~~</center>~~		Show that by setting
			<center> <math>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}}</math> </center>

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

	~~In the Gaussian case, expand <math>~~A(E)</math> ~~around <math>a_M~~</~~math~~>:

	By setting

	<center><math>
	a_M = E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} + \ldots
	\~~qquad~~\text{and}\~~qquad~~
	b_M = \frac{1}{A'(a_M)} = \frac{\sigma}{\sqrt{2 \log M}}
	</math></center>

	you find		you find


	<center><math>		<center><math>