Theoriae Causalitatis Principia Mathematica. Third Edition

Ilija Barukčić Theoriae causalitatis principia mathematica Ilija Barukčić Theoriae causalitatis principia mathematica –Causality – December 22, 2024 Printed by Books on Demand,Norderstedt, Germany History of publication: Die Kausalität FirstEdition published 1989 (FirstGerman Edition) ISBN 3-9802216-0-1 Second Edition published1997 (Second GermanEdition) ISBN3-9802216-4-4 Causality.New StatisticalMethods. Third Edition published 2005 (FirstEnglishEdition, BoD) ISBN 3-8334-3645-X Fourth Edition published2006 (Sec. EnglishEd., BoD) ISBN 3-8334-3645-X Fifth Edition published2008 (Third EnglishEdition, Lulu) Causality Volume I:ISBN 978-1-4092-2952 -0 Causality Volume II:ISBN 978 -1-4092 -2954 -4 Fifth Edition published2008, 13th Revision of the 5th Edition, May24th, 2009 Fifth Edition published2008, 14th Revision of the 5th Edition, June 14th, 2009 Fifth Edition published2008, 15th Revision published April 5th, 2010. Fifth Edition published2008, 16th Revision published May24th, 2010. Fifth Edition published2008, 17th Revision published August21st, 2010. Fifth Edition published2008, 18th Revision published Dec. 30th, 2010. Fifth Edition published 2008, 19th Revision published May, 1st, 2011. Lulu.com ©December 22, 2024 by Ilija Barukčić, Horandstrasse, Jever, Germany. All rights reserved. Alle Rechte vorbehalten. No partofthis publicationmay be reproduced, stored in aretrievalsystem, or transmitted in anyformorbyany means, electronic,mechanical, photocopying, recording, scanning, or otherwise,except as permitted under Section 107 or 108ofthe 1976 United States Copyright Act, without either the prior written permission of the author, or authorization through payment of the appropriate per-copyfee to the author. Herstellung und Verlag:Bod -Books on Demand. Norderstedt,Germany, 2021. Theoriae causalitatis principia mathematica. FirstEdition (April 23,2017; ISBN: 9783744815932) Second Edition (August 2, 2021; ISBN: 9783754331347 ) Third Edition (December 22,2024; ISBN: 9783769322385 ) Imagecredit: NASA,ESA,CSA,STScI, UnitedStates of America. This imagecaptured by Webb’sMIRI (Mid-InfraredInstrument) andreleased September 18, 2024 10:00AM(EDT), reveals the supermassiveblackholeatthe center of the large spiralgalaxy on the right.The blackholedraws in muchof the surrounding dust, creatingdistinctivelanes.Additionally, it exhibits Webb’s characteristic diffraction spikes,whichresult from the emitted light interacting with the telescope’s structure. Contents Part I Natureand causation 1Reality and causality ................................ 3 Causality ...........................................3 Anti-Causality....................................... 4 2Evolution and causation .............................7 Evolution and human brain............................7 Either adapt or become extinct .........................8 3Proof methods and causation .........................9 Induction and experiment..............................10 Deduction and logical fallacies.........................12 Proof methods andhuman knowledge................... 13 Proof by Thought Experiments.................... 16 Proof by Counterexample .........................16 Proof by modus ponens........................... 18 Proof by modus ponens contrapositivus ............. 21 Proof by modus sine .............................22 Proof by modus tollens ........................... 23 Proof by modus inversus..........................24 Proof by other methods ........................... 26 Part II Science and causation xiii xx Contents Boole and negation ..............................495 Marxand Engels andnegation..................... 496 Lexnegationis....................................... 497 Lexnegationisgeneralis...............................497 Part VI Causation 26 Mono-Causality ..................................... 507 Introduction ......................................... 508 Basics .............................................. 509 The difference between cause and not-cause......... 509 The difference between cause and effect .............510 The asymmetry of the causal relation ...............512 Theorems of mono-causality ........................... 517 The identityofcause and effect .................... 517 The coincidence of cause and effect .......... 517 Causa aequat effectum ..................... 519 David Hume’s problemofinduction ......... 525 One cause, one effect I..................... 526 The difference between cause and effect .............529 The contradiction between cause and effect .......... 531 One cause, one effect II .................... 532 Causal Relationship kand samplesize........ 535 The contradiction betweencause and effect ... 536 Causation and correlation .................. 542 27 Multi-Causality ..................................... 543 Introduction ......................................... 544 Theorems of multi-causality ........................... 545 One cause and achain of effects ................... 545 One cause and manyeffects....................... 548 Manycauses and one effect ....................... 551 Manycauses and manyeffects..................... 555 Manycauses and andachain of effects.............. 559 Achain of causes and one effect ................... 564 Achain of causes and manyeffects ................. 568 Achain of causes and achain of effects.............573 General relativity and causality ......................... 578 Part VII LawofNature Contents xxi 28 Law’sofnature ..................................... 585 Indeterminism....................................... 586 Determinism........................................587 The lawofnature relationship g........................589 29 Proof of God’sexistence .............................593 Science and ideology .................................594 Proof/Disproof of the existence of God .................. 595 General relativity and the existence of God...............598 Notice .................................................. 605 References ..........................................612 Author Index ...........................................629 Subject Index ...........................................633 Volume I The generaltheory of causality Part I Natureand causation Part II Science and causation Part III Basic Definitions 220 9Distributions The Variance of the Binomial Distribution KarlPearson presentedthe definition of the standard deviation of the binomial distribution as follows: “... the mean square error forany binomial distribution ... is identical with the value√npq”(see Pearson,1895,p.351) with the consequence that σ(X)2=N·p(X)·(1−p(X)). Numerus Legis Naturalis 51. In starkcontrasttoKarlPearson’s perspective,the variance of theBinomial distribution is given as: σ(X)2=E(X2)−(E(X))2=N2·p(X)·(1−p(X)) (464) where E(X)is theexpected value of X,and E(X2)is theexpected value of thesquareofX. Proof by Direct Proof. It holds that 1 =1and E(X)=E(X).For aBinomial random variable Xwith parameters N(number of trials) and p(X)(probabilityofsuccess), the expected value of Xis: E(X)=X·p(X)=N·p(X)(465) Dividing this equation by p(X), andunder the conditions of the Binomial distribution, we find that: X=N(466) Next, we calculate E(X2),the expected value of X2.For aBinomial distribution, thiscan be expressed as: E(X2)=X·X·p(X)=N·N·p(X)(467) Substituting these expressions into the variance formula: σ(X)2=E(X2)−(E(X))2(468) =(N·N·p(X))−(N·p(X))2(469) =N2·p(X)·(1−p(X)) (470) In general, the variance of the Binomial distribution is determined as: σ(X)2=N2·p(X)·(1−p(X)) (471) Quod eratdemonstrandum. 9Distributions 221 The CumulativeDistribution Function In General The CumulativeDistribution Function (CDF) of arandom variable X is afunction that givesthe probability that Xtakes avalue less thanor equal to agiven value x.Itisformallydefined as: FX(x)=p(X)=p(X≤x)(472) where FX(x)is the value of the CDF at x.The CDF is anon-decreasing functionbecause probabilities cannot decreaseasxincreases. Fora continuous random variable,the CDF is acontinuous function. Fora discreterandom variable,the CDF isastep function. Forcontinuous random variables,the CDFisrelatedtotheprobability densityfunction (PDF) fX(x)as: FX(x)=p(X)=p(X≤x)= x ∫ −∞ fX(t)dt (473) Foracontinuous random variable, the CDF is acontinuous function. Fordiscreterandom variables,the CDF isrelated to the probability mass function(PMF) pX(x)as: FX(x)=p(X)=p(X≤x)= t≤x pX(t)(474) Foradiscrete random variable, theCDF is astep function. The CDFsatisfies the following limit properties:limx→−∞ FX(x)= 0,limx→+∞FX(x)=1. These reflect the bounds of the CDF as it covers the entire rangeofX. Numerus Legis Naturalis 52. Under thegiven conditions, theprobability mass function satisfiesthe relationship: p(X<k)+p(X≥k)=p(X≤k)+p(X>k)=1(475) Proof by DirectProof. To begin, it holds that: p(X=k)=p(X=k)(476) 222 9Distributions This equalityisafundamental propertyofthe probability massfunction. Furthermore, we observe that: p(X=k)+p(X>k)=p(X=k)+p(X>k)(477) This reflects the additivenature of probabilities fordisjoint events. From this,itfollows that: p(X≥k)=p(X=k)+p(X>k)(478) This expression decomposes the cumulativeprobability forX≥k. Combining probabilitiesfor allpossible outcomes: p(X<k)+p(X≥k)=p(X<k)+p(X=k)+p(X>k)=1(479) Here, the total probabilityfor all disjoint eventssumsto1.Therefore, we conclude: p(X<k)+p(X≥k)=p(X≤k)+p(X>k)=1(480) This demonstratesthe completenessofthe probability distribution. Quod erat demonstrandum. From these relationships,wecan deduce the following equivalences: p(X≥k)=1−p(X<k)(481) This equality reflectsthatthe probability of Xbeing greater than or equal to kcomplementsthe probability of Xbeing strictlyless than k. Similarly,wehave: p(X≤k)=1−p(X>k)(482) This expresses the complementaryrelationshipbetween X≤kand X>k. NumerusLegis Naturalis 53. In general it is p(X≤(n−1)) =1−p(X>n−1)=1−p(X=n)=1−(pn)(483) Proof by Direct Proof. In general, it is p(X≤x)+p(X>x)=1(484) 9Distributions 223 Under conditions x=(n−1),Equation 484 becomes (see Equation 479, p. 222) p(X≤(n−1))+p(X>(n−1)) =1(485) In general, foradiscreterandom variable where nrepresents the sample of size, there are no outcomes greater than n.Therefore, the probability of exceeding n−1(i.e., being strictlygreater than n−1) is equivalent to the probabilityoftaking the value nor in other words p(X>n−1)=p(X=n)(486) Equation485 becomes p(X≤(n−1))+p(X=n)=1(487) As found before, itisp(X≥n)=p(X=n)=pn(see Equation 547,p. 235), Equation 487 becomes p(X≤(n−1))+pn=1(488) In general, it is p(X≤(n−1)) =1−(pn)(489) Quod eratdemonstrandum. 224 9Distributions Numerus Legis Naturalis 54. Under thespecific conditions of binomial distribution (see Equation 489,p.223),itis p(X≥n)=p(X=n)=1−p(X≤(n−1)) =pn(490) Proof by Direct Proof. Under the specific condition of thebinomial distribution, the number of trials nis suchthat nrepresents the maximum possiblenumber of successes(i.e., thetotal number of trials in the experiment), which implies (X≤n)=(X≤n)always. However, forthe binomial distribution with parameters n(number of trials) and p (probabilityofsuccessinasingletrial), the relationship p(X≥n)=p(X≥n)(491) translates formally to: p(X≥n)=p(X=n)+p(X>n)(492) In summary, abinomial random variable Xhas no possibility of exceeding n,the total number of trials. In brief,bythe definition of the binomial distribution, it is p(X>n)=0(493) Therefore, the equality p(X≥n)=p(X=n)=pn(494) holds under these specific conditions. Quod eratdemonstrandum. In general (see Equation 634,p.251), it is p(X≥n)=p(X=n)=(p)n=(1−q)n≈e−E(X)(495) Given Bernoullitrials with n=6138 (see Barukčić,2023c), the expectedsuccesses are k=6138,butthe observedsuccesses are k=6132. This indicates that in 6of 6138 trials,the event didnot occur as expected. Here, E(X=6)is the expected shortfall (events that did not occur). The probability can be approximated as: p(X≥n)=p(X=n)≈e−E(X)≈e−6=0.002478752177 (496) 9Distributions 225 Numerus Legis Naturalis 55. In general, itis p(X>0)=1−((1−p)n)(497) Proof by DirectProof. In general, it is p(X=0)+p(X>0)=1(498) As found before, itisp(X=0)=(1−p)n.Equation 498 becomes (1−p)n+p(X>0)=1(499) At the end, we obtain p(X>0)=1−((1−p)n)(500) Quod eratdemonstrandum. Numerus Legis Naturalis 56. In general, itis p(0<X<n)=1−(pn)−((1−p)n)(501) Proof by DirectProof. In general, it is p(X=0)+p(X=1)+p(X=2)+···+p(X=(n−1))+p(X=n)=1(502) We define p(0<X<n)=p(X=1)+p(X=2)+···+p(X=(n−1)) (503) Equation503 becomes p(X=0)+p(0<X<n)+p(X=n)=1(504) Rearranging equation 504,itis p(0<X<n)=1−p(X=0)−p(X=n)(505) As found before, it p(X=0)=(1−p)nis and p(X=n)=pn. p(0<X<n)=1−(pn)−((1−p)n)(506) Quod eratdemonstrandum. 226 9Distributions It is p(0<X<n)=p(1≤X≤(n−1)) =1−(pn)−((1−p)n)(507) The interval derived in Equation 507 is an exact interval, since the same is based directly on the binomial distribution rather than any approximation to the binomial distribution. JerzyNeyman (1894–1981) Historically, methods for calculating confidence intervals forbinomial proportions similar to p(1≤X≤(n−1)) =1−(pn)−((1−p)n)(508) began emerging in the1920s (see Wilson,1927), but it wasinthe early 1930s (see Clopper and Pearson,1934)that the foundational concepts ofconfidence intervals weresystematicallydeveloped.Thedevelopment of confidence intervaltheoryislargely attributed to JerzyNeyman, (see Neyman,1937)who, in 1937, provided the firstcomprehensiveand general account of the method. Using today’sformula of thenormal approximation, the successprobability pcan be estimated as p≈ˆp± zαˆp(1−ˆp) n,where ˆp≡ns nis the proportion of successes in aBernoulli trial processand servesasanestimator forpin the underlying Bernoulli distribution. Here: ns:Number of successes observed, n:Total number of trials, zα:Z-score corresponding to the desired confidence level(e.g., zα=1.96 for95% confidence). 9Distributions 227 The BinomialDistribution Under ExtremeConditions The behaviour of the binomialdistribution under extreme conditions, suchask=0ork=n−1ork=n,highlightsits sensitivity to boundary values. At k=n,the probabilitycorresponds to thesuccessofall trials, arare event unless p≈1. Fork=n−1, the distribution reflects scenarios whereall but one trialsucceed, often showing asignificant drop in probability. Theseedgecases are critical in understanding the tail behavior of the binomialdistributionand are often analyzed in reliability and risk assessments. The Binomial Distribution fork=0 Theoretically, it is possible that abinomial random variabledoes not occur at all in nBernoulli trials.Inthe binomialdistribution, the case k=0representsthe probabilitythat no successes occur in nindependent trials,eachwith success probability p.These circumstances are reflecting the scenariowhereall ntrials result in failure. This case is particularly relevant inapplications suchasrisk assessment or reliability testing, where the absence of aparticular outcomecarries significant importance. What consequences arisefrom this theoretical possibility? Numerus Legis Naturalis 57. Under conditions where k=0,itis n 0=1(509) Proof by DirectProof. n k=n! k!(n−k)!=n! 0!(n−0)!=n! 1·n!=n! n!=1(510) Quod eratdemonstrandum. Numerus Legis Naturalis 58. Under conditions where k=0,itis (1−p)n−k=(1−p)n(511) Proof by DirectProof. (1−p)n−k=(1−p)n−0=(1−p)n(512) Quod eratdemonstrandum. 228 9Distributions Numerus Legis Naturalis 59. Under conditions where k=0,itis pk=1(513) Proof by Direct Proof. pk=p0=1(514) Quod eratdemonstrandum. In the case k=0, the event of interesthas not occurred and we are confronted by conditions under which the event is deemed unlikelyor prevented by circumstances. As known, theabsence of an eventcan often signify as muchasits presence, depending upon thecontext. In probabilistic terms, (k=0)invites reflections what factors contributed to the nullresult, andhow doesthis absence reshape our understanding of the system under observation? The non-occurrence of something often remain silent yetprofoundlyconsequential. Numerus LegisNaturalis 60. Under conditions where k=0,the number of successes equals thetotal numberoftrials. It is p(X=0)=(1−p)n(515) Proofby Direct Proof. Under these circumstances, it is p(X=0)=n k=0p0(1−p)n−0(516) As proofed before, it is n 0=n! 0!(n−0)!=1,p0=1,und (1−p)n−0=(1−p)n(517) Equation 516 simplifies to: p(X=0)=n k=0p0(1−p)n−0=1·p0·(1−p)n−0=(1−p)n(518) Quod eratdemonstrandum. 9Distributions 229 The General Rule of Three The rule of threecan be generalized as follows. Numerus Legis Naturalis 61. The generalized form of theRuleof Three extends theestimation forthe upper limit of theprobability of rare events to arbitrary confidence levels α.The generalized form of the Rule of Three is given as: p≤−ln(α) n(519) where: •p:The upper bound on theevent probability. •α:The desiredconfidence level (e.g., for95% confidence, α= 0.05). •n:The number of independent trials. Proof by DirectProof. The Poisson distribution is used to model the probability of observing agiven number of eventsinafixed interval when the events occur independently. The probability massfunction (PMF) of the Poisson distribution is: p(X=k)=λke−λ k!,(520) where λis the expected number of events, kis the number of observed events, and eis the baseofthe natural logarithm. Under conditions, where k=0, it is p(X=(k=0)) =λke−λ k!=λ0e−λ 0! =e−λ(521) The probability of observing zero events when the expected number of events is λ,isgiven as p(X=0)=e−λ(522) The general rule of three providesasimpleapproximation forthe upper bound on λwhen no eventsare observed(X=0) with 100 ·(1−α)% confidence. To derive this,wesolve: 236 9Distributions The Binomial HypothesisTest The binomial testisanexact statistical procedure where the teststatistic follows the binomial distribution. This testisusedtoevaluatehypotheses about characteristics that have exactly twopossible outcomes (dichotomous traits,e.g., ‘Success’or‘Failure’, ‘Yes’or‘No’). The test determines whetherthe observed frequency distribution aligns with an expected or theoretical frequency distribution π0.Typically, the binomial testhelps decide whetherdeviations between observedand expected resultsare due to chance or indicate asignificant changeinthe underlying probabilities. In ahypothesis test, it is necessary to decide whether aclaim is true or not. The binomial testisbased on the binomial distribution and is usedtodetermine whether an observedvalue of successes significantlyexceeds an expected value of successesunder H0.A small p-value indicates sufficient evidence to reject H0in favour of HA. The right tailed testand the left tailed testare examples of one-tailed tests. The alternate hypothesis(HA)determines whether aright tailed test or aleft-tailed test is given. Aright tailed test (sometimes called an upper test)isatestwhere the alternative hypothesis HAstatement contains agreater than(>)symbol. The hypotheses forthe right-tailed binomial testare: H0:p≤p(X≤n−1)vs.HA:p>p(X≤n−1)(548) Decision Rule: If p-valueright tailed <α,then reject H0:p≤p(X≤n−1),where α is the significance level. Aleft-tailed test (sometimes called alower test) contains alessthan (<)symbol and statesthatthe true value of the parameter specified in the null hypothesisisless than the nullhypothesis claims. H0:p≥p(X≤n−1)vs.HA:p<p(X≤n−1)(549) Decision Rule: If p-valueleft tailed <α,then reject H0:p≥p(X≤n−1),where α is the significance level. 9Distributions 237 Right-Tailed Binomial Test inGeneral The right-tailed binomialtestevaluates whether pis greater than a specific value π0. Hypotheses foraRight-Tailed Test: •H0:p≤π0 The observed probability pis less than or equal to the theoretical probability π0. •HA:p>π 0 Theobservedprobability pis greaterthan the theoretical probability π0. whichisequivalent with: H0:π0≥pvs.HA:π0<p. Test Statistic: X∼Binomial(n,π 0).(550) p-Value:The probabilityofobserving Xor agreater result under H0 is: p-valueright tailed =p(X≥k)= i=n  i=kn iπi 0(1−π0)n−i(551) This calculation involvessumming over theprobabilities forall possible outcomes k=x,x+1,···,n. Decision Rule right-tailed test:Ifp-valueright <α,then reject H0:p≤π0,where αis the significance level. In aright-tailed test,wefocus on finding values that are greater than or equal to the observedresult.Ap-value in aright-tailed testhelps us understand howlikely it is to seeanoutcome at leastasextremeasthe one we observed, assuming the null hypothesis is true. If the p-value is very small, it means the result we observedisunlikelytohappen by chance, and we might think the null hypothesis is wrong. If the p-value is large,itsuggeststhe observedresult is not thatsurprising, so the null hypothesis could be correct. 238 9Distributions Alternative right-tailed p-Value Numerus Legis Naturalis 68. In general, theright-tailed p-value, p(X≥k),isequal to 1minus thecumulativeprobability up to k−1, whichisp(X≤(k−1)).Weobtain: p(X≥k)=1−p(X≤(k−1)) (552) Proof by Direct Proof. Forany discrete random variable X,the total probabilitymustsum to 1: p(X≤∞)=1(553) This means that the probabilitythat Xtakes anyvalue is 1, and it is split between the two regions: •p(X≤(k−1)):The probability that Xis less than or equaltok−1. •p(X≥k):The probabilitythat Xis greater than or equal to k. Since thesetwo events aremutually exclusive(no overlap), the total probability is thesum of the two probabilities: p(X≤(k−1))+p(X≥k)=1(554) Rearranging the above equation, we get(seeEquation 603,p.246): p(X≥k)=1−p(X≤(k−1)) (555) Quod eratdemonstrandum. Forthe case where k=n,the p-value becomes (see Equation 604, 246): p(X≥n)=p(X=n)=1−p(X≤n−1)=1−(1−πn 0)=πn 0≈e−E(X)(556) In general, thereisnopossibility of more than nsuccesses. Thisis because when k=n,the onlypossible outcome forXis exactly n, and p(X>n)=0. Thus, the probability of X≥nis the probability of observing exactly nsuccesses, whichcan be calculated using the complement of p(X≤n−1). 9Distributions 239 Right-tailed p-Value fork=0 Let Xbe abinomial random variable with parameters n(number of trials) and p(probabilityofsuccess). Numerus Legis Naturalis 69. TheHypothesesforaRight-Tailed Test are: H0:p≤π0vs.HA:p>π 0.For aright-tailed test wherethe observedvalue k=0,the p-value is theprobability of observing kor any value greater than kunder thenullhypothesis and given as p-valueright tailed =p(X≥0)=1(557) Proof by Direct Proof. The p-value foraright-tailed testisgiven by: p-valueright tailed =p(X≥k)= i=n  i=kn iπi 0(1−π0)n−i(558) Substituting k=0into the formula: p-valueright tailed =p(X≥0)= i=n  i=0n iπi 0(1−π0)n−i(559) Since the summation includesall possible outcomes of abinomial random variable, itevaluates to: p-valueright tailed =p(X≥0)=1(560) Quod eratdemonstrandum. The possibilitytoapproximate the binomial distribution with simpler formulas offerssignificant computational and conceptual advantages. Forlarge n,directlycalculating binomial probabilities involves computing factorials, which grow exponentiallyand can quicklybecome computationally expensiveorimpractical.Approximations,suchasthe normal or Poisson distributions,simplifycalculations while maintaining ahigh degree of accuracy under certain conditions. 240 9Distributions Right-tailed p-Value fork=1 Numerus LegisNaturalis 70. ForaBinomial distribution X∼ Binomial(n,p),the cumulative distribution function (CDF) forp(X≥1) can be derivedasfollows: p(X≥1)=1−p(X=0)(561) Proof by Direct Proof. The binomial probability mass function (PMF)isgiven by: p(X=k)=n kpk(1−p)n−k(562) Fork=0, it is: p(X=0)=n 0p0(1−p)n=1·(1−p)n=(1−p)n(563) In general, it is p(X=0)+p(X≥1)=1(564) The complement rule gives: p(X≥1)=1−p(X=0)(565) Substitute p(X=0)into the complement rule: p(X≥1)=1−(1−p)n(566) Quod eratdemonstrandum. ForaBernoullidistribution, it is n=1, and the equation becomes: p(X≥1)=1−(1−p)n=1−(1−p)1=p(567) Example. Forn=5and p=0.6, it is: p(X≥1)=1−(1−0.6)5=1−(0.4)5=1−0.01024 =0.98976. Thismeans theres a98.976% chanceofobservingat leastone success in 5trialswith asuccessprobability of 0.6 pertrial. 9Distributions 241 Right-tailed p-Value fork=n−1 The Hypotheses foraRight-TailedTest fork=n−1are: H0:p≤π0vs.HA:p>π 0(568) Numerus Legis Naturalis 71. Foraright-tailed test wherethe observedvalue is k=n−1,the p-value is theprobability of observing k=n−1or any value greater than k=n−1under thenull hypothesis and given as p(X≥n−1)=p(X=n−1)+p(X=n)=n·(1−π0) π0 +1·πn 0(569) Proof by Direct Proof. The probabilityfor exactly X=n−1 events is: p(X=n−1)=n n−1·πn−1 0·(1−π0)=n·πn−1 0·(1−π0)(570) The probability forexactly X=nevents is: p(X=n)=n n·πn 0·(1−π0)0=πn 0.(571) By definition, the right-tailed p-valueis: p-valueright tailed =p(X≥n−1)=p(X=n−1)+p(X=n)(572) Substituting the probabilities, the exact right-tailed p-value forX≥n−1 is given as: p-valueright tailed =p(X≥n−1)=p(X=n−1)+p(X=n)(573) =n·πn−1 0·(1−π0)+πn 0(574) =n·(1−π0) π0·πn 0+πn 0(575) =n·(1−π0) π0+1·πn 0(576) Quod eratdemonstrandum. 242 9Distributions In an investigation, 6events occurred out of n=6138 trials, where 0 events (see Barukčić, 2023c)were expected. The p-value (seeEquation 576)isgiven as: 6138 ·(1−1−6 6138 ) 1−6 6138  +1·1−6 6138 6138 =0.01731493293 (577) The null hypothesis(H0)isrejected. In the population, π0> 0.9990224829 (p-value (right-tailed) =0.0173). The exclusionrelationship is: pEXCL =1−6 6138=0.9990224829 (578) The 99.9999% confidence bound (see Equation 531), where α= 0.000001 and n=6138,isgiven by: 1−−ln(α) n≤p≤1(579) or as: 0.9977491837 ≤0.99902248 ≤1(580) Right-tailed p-Value fork=n The Hypotheses foraRight-Tailed Test fork=nare: H0:p≤π0vs.HA:p>π 0(581) Numerus Legis Naturalis 72. Foraright-tailed test wherethe observedvalue is k=n,the p-value is theprobability of observing k=n or any value greater than k=nunder thenull hypothesis and is given as p-valueright tailed =p(X≥n)=p(X=n)=πn 0(582) Proof by Direct Proof. The probability forexactly X=nevents is: p(X=n)=n n·πn 0·(1−π0)0=πn 0(583) 9Distributions 243 Since there are no values greater than nin abinomial distribution, the right-tailed p-value issolely: p-valueright tailed =p(X≥n)=p(X=n)(584) Substituting the probability, the exact right-tailed p-value forX≥nis: p-valueright tailed =p(X≥n)=πn 0(585) Quod eratdemonstrandum. In an investigation, 6events occurred out of n=6138 trials, where 0 events (see Barukčić, 2023c)were expected. Thep-value (seeEquation 585) is given as: p(X≥n)=p(X=n)=πn 0=1−6 6138 6138 =0.01731493293 (586) The null hypothesis(H0)isrejected. In the population, π0> 0.9990224829 (p-value (right-tailed) =0.0173). Left-Tailed Binomial Test inGeneral The left-tailed binomial testevaluates whether pis smaller than aspecific value π0. The hypotheses foraleft-tailed test are: •H0:p≥π0 The observed probability pis greater than or equal to the theoretical probability π0. •HA:p<π 0 The observedprobability pis smallerthan the theoretical probability π0. whichisequivalent with: H0:π0≤pvs.HA:π0>p. Test Statistic: X∼Binomial(n,π 0),(587) where Xis the number of successes in ntrials. p-Value:The p-valueleft tailed of alefttailed binomial testisthe probability of observing Xor asmaller resultunder H0and is calculated as: 244 9Distributions p-valueleft tailed =p(X≤k)= i=k  i=0n iπi 0(1−π0)n−i(588) Decision Rule left tailed:Ifp-valueleft tailed <α,then reject H0: p≥π0,where αis the significance level. Inverted Perspective Aglass can alwaysbeperceived as either halffullorhalfempty. Thus, the hypothesescan equivalently be expressedas: •H0:1−p≥1−π0, •HA:1−p<1−π0 The binomial distribution is discrete, the exact formulafor p(X≥k) is: p(X≥k)= i=n  i=kn iπi 0(1−π0)n−i(589) or equivalently: p(X≥k)=1−p(X<k)=1−p(X≤k−1)(590) Left-tailed p-Value fork=0 The Hypotheses foraLeft-TailedTest fork=0are: H0:p≥π0vs.HA:p<π 0(591) Numerus Legis Naturalis 73. Foraleft-tailed test wherethe observedvalue is k=0,the p-value is theprobability of observing k=0 or any value smaller than k=0under thenull hypothesis, and is given as: p-valuelefttailed =p(X≤0)=p(X=0)=(1−π0)n(592) Proof by Direct Proof. In general, the p-valueleft tailed is defined as: p-valueleft tailed =p(X≤k)= i=k  i=0n iπi 0(1−π0)n−i(593) 9Distributions 245 The summation foraleft-tailed p-value, p(X≤k),onlyincludes terms from i=0uptoi=k.This means itsums over all outcomes ifrom 0 (nosuccesses) up to i=k(exactlyksuccesses). Therefore, forthe case k=0, we obtain the p-value as: p-valueleft tailed =p(X≤k)= i=0  i=0n 0π0 0(1−π0)n−0(594) The probabilityfor exactly X=0events is: p(X=0)=n 0·π0 0·(1−π0)n=(1−π0)n(595) Since there are no values smaller than 0inabinomial distribution, the exact left-tailed p-value forX≤0issimply: p-valueleft tailed =p(X≤0)=p(X=0)=(1−π0)n=1−n·π0 nn(596) Quod eratdemonstrandum. Now, taking the limit as n→+∞,we have: p-valueleft tailed =p(X≤0)=limn→∞ 1−n·π0 nn=e−n·π0≈e−E(X)(597) Left-tailed p-Value fork=n−1 The Hypotheses foraLeft-TailedTest fork=n−1are: H0:p≥π0vs.HA:p<π 0(598) Numerus Legis Naturalis 74. Foraleft-tailed test wherethe observedvalue is k=n−1,the p-value is theprobability of observing k=n−1or any value smaller than k=n−1under thenull hypothesis, and is given as: p-valuelefttailed =p(X≤(k=n−1)) = i=n−1  i=0n iπi 0(1−π0)n−i=1−πn 0(599) 252 9Distributions Equation 585,p.243)itis: p(X≥n)=πn 0(640) Based on Equation 639 and Equation 640,itisequally(see Equation 495 on page224): p(X≥n)=p(X=n)=1−p(X≤n−1)=(p)n=(1−q)n≈e−E(X)(641) Approximation fork=n−1 Numerus LegisNaturalis 80. In general, it is p(X=n−1)=n·z2·e−E(X) Proof by Direct Proof. The binomial distribution is given by: p(X=k)=n kpk(1−p)n−k, where nis the number of trials, kis the number of successes, and pis the probabilityofsuccess. Substituting k=n−1, we have: p(X=n−1)=n n−1pn−1(1−p)n−(n−1). Simplifythe binomial coefficient,weobtain: n n−1=n! (n−1)!(n−(n−1))!=n! (n−1)!·1! =n. As next, simplify the powerof(1−p),itis: (1−p)n−(n−1)=(1−p)1=1−p. Furthermore, combine the results: p(X=n−1)=n·pn−1·(1−p). Rewriting the expression: 9Distributions 253 p(X=n−1)=n·pn p·(1−p), and: p(X=n−1)=n·1−p p·pn. Define zas: z2=1−p p where: •pis the probabilityofsuccess,with 0 <p<1, •1−prepresents the probabilityoffailure. Using the definition of z2,we obtain: p(X=n−1)=n·z2·pn. Expressing p=1−q,the equation simplifies as: p(X=n−1)=n·z2·(1−q)n. Further,rewrite: p(X=n−1)=n·z2·1−n·q nn . Define E(X)=n·q,and in the limit as n→∞,weapproximate: p(X=n−1)≈n·z2·e−E(X). Quod eratdemonstrandum. Example. Consider tossing acoin n=100 times with p=0.5. It is 100·e−50 = 1.92874985 ×10−20 and 0.5100 =7.88860905 ×10−31 Conclusion: -The cumulativeprobability up to k=n−1isvery close to 1, as p=0.5makes the binomialdistribution symmetric.-The probability of X=n−1(using the approximation) is extremelysmall due to the rapid decrease of e−E(X). 254 9Distributions P-Value Approximation fork=n In the following, the H0(null hypothesis) statesthatthe probability pis less than or equal to the cumulativeprobability p(X≤n−1), suggestingnosignificant differencefrom the expected outcome. HA (alternative hypothesis)assertsthatthe probability pis greater than the cumulativeprobability p(X≤n−1),indicating ameaningful deviation from the expected outcome. The hypotheses are formulated as follows: H0:p≤p(X≤n−1)vs.HA:p>p(X≤n−1)(642) The teststatistic is X,the observednumber of successes.Under the null hypothesis,the cumulative probability p(X≤n−1)is the thresholdfor p,and the testassesses whether pexceeds this threshold. The binomial distribution of X,denoted X∼Binomial(n,p),ischaracterized by n, the number of trials, and p,the probability of success in asingle trial. Thevalue of p(X≤n−1)is calculated as: p(X≤n−1)= n−1  k=0n kpk(1−p)n−k(643) Foraleft-tailed test (H0:p≤p(X≤n−1)), the p-valueleft tailed is: p-valueleft tailed =p(X≤xobs)=1−e−E(X)(644) where xobs is the observednumber of successes. Foraright-tailed test (HA:p>p(X≤n−1)), thep-valueright is (see Equation 495 on page 224): p-valueright tailed =p(X≥n)=p(X=n)=1−p(X≤xobs)=(p)n=(1−q)n≈e−E(X)(645) whichrepresents the probability of observing xobs or more successes. Example. Under thesecircumstances and more accurate, it is: p-valuert =(1−(1−pn)) =pn=1−6 6138 6138 =0.00247148903. In 6out of 6138 cases, an event occurredthat shouldnot have occurred (see Barukčić, 2023c). The right-tailed p-valueright is calculated approximatelyas: p-valueright =e−6=0.002478752177.The righttailed null hypothesis: H0:p≤p(X≤n−1)is rejected, leading to the acceptance of the alternativehypothesis: HA:p>p(X≤n−1)with p-valueright tailed=0.002478752177. 9Distributions 255 Stirling’sapproximation Poisson’s distribution Let us assume that the theoretical probabilityofobserving n·αfailures is given by the Poisson distribution formula: p(X=n·α)=(n·α)n·αe−(n·α) (n·α)!(646) where nis the number of trials,and αis the significance level. This is the expectednumber of failures under the nullhypothesis is very small and given as λ=n·α.The real or observednumber of failures is denoted by k,while the probabilityofobserving exactly kfailures is: p(X=k)=(n·α)ke−(n·α) k!(647) whichisthe probabilityofobserving exactly kfailures under the same assumptions. Thesetwo probabilities can be compared directly: p(X=n·α)<p(X=k),this suggeststhatthe observed number of failures kis more likelythan the expectednumber n·α. Decision Rule: If p(X=(n·α)) <p(X=k),this suggeststhatthe observednumber of failures kis more likelythan the expectednumber n·α.This could be evidence againstthe null hypothesis H0,suggesting that the observed data is significantlydifferent from what wasexpected. If p(X=n·α)>p(X=k),this suggeststhat the observednumber of failures kis less likelythan the expectednumber n·α,meaning the observeddata iscloser to the expected data under H0. In hypothesis testing, thiscomparison helps determine whether the observed data isconsistent with the null hypothesis H0,orifitsuggests asignificantdeviation thatmay warrant rejecting H0. Numerus Legis Naturalis 81. Stirling’sapproximation provides an estimation forthe Poisson distribution, expressedas: p(X=k)≈ 1 2π(n·(1−p))≈1 √2π·λ (648) Proof by Direct Proof. We are given the Poisson distribution probability mass function: 256 9Distributions p(X=n·(1−p))=(n·(1−p))n·(1−p)e−(n·(1−p)) (n·(1−p))!(649) while pisextremelyclose to1.For large n·(1−p),weuse Stirling’s approximation (see Stirling,1730)for factorials: n!≈√2πnn en(650) Substituting thisinto (n·(1−p))!, we get: (n·(1−p))!≈2π(n·(1−p))n·(1−p) en·(1−p)(651) SubstituteStirling’sapproximation into theoriginal Poisson formula: p(X=n·(1−p))≈ (n·(1−p))n·(1−p)e−(n·(1−p)) 2π(n·(1−p))n·(1−p) en·(1−p)(652) The terms (n·(1−p))n·(1−p)cancel out, leaving: p(X=n·(1−p))≈ 1 2π(n·(1−p))·e−(n·(1−p))+(n·(1−p))=1 2π(n·(1−p))(653) Forlarge λ=n·(1−p),the Poisson distribution probability mass function can be approximatedas: p(X=k)≈ 1 2π(n·(1−p))≈1 √2π·λ (654) Quod erat demonstrandum. Example. Set n=5500 and (1−p)=α=0.01.Wecalculate the approximation as: 1 2π(5500·0.01) =0.05379336612.Itisn·α= 5500 ·0.01 =55.Asnext, we use the Poisson distribution formula, known as: p(X=n·α)=(n·α)n·αe−(n·α) (n·α)!and substitute n·α=55 into the formula, yielding: (55)55e−55 (55)!=0.05371192363.This shows that the approximation is very close to the exact value computed using the Poisson distribution. 9Distributions 257 P-Value and significance level α Set the significance level αto: α=0.05. Based on the significance leveland thesamplesize n,wederivethe theoretical probabilityofasingleevent fork=nas follows. Numerus Legis Naturalis 82. Fork=n,itholds that: ptheoretical =eln(α) n Proof by DirectProof. In general, fork=n: p-valueright-tailed =p(X≤k)= x=n  x=nn npn(1−p)n−n=pn. Furthermore, the relationshipbetween thecalculated p-value and the significance levelis: p-valueright-tailed =pn theoretical =α. Taking the natural logarithm on bothsides gives: n·ln(ptheoretical)=ln(α). Solvingfor ln(ptheoretical),we obtain: ln(ptheoretical)=ln(α) n. Exponentiating both sides yields: ptheoretical =eln(α) n(655) Quod eratdemonstrandum. Example. Under conditions,where α=0.000001 and n=6138 ,it is: eln(0.000001) 6138 =0.9977517149 (656) whichclosely approximates(seeEquation 531, p. 231): 1−−ln (0.000001) 6138 =0.9977491837 (657) 258 9Distributions The Binomial HypothesisTest:Examples ExampleI.P-Values foraFair Coin Example A We areconsidering afair coin (p=0.5),with n=100 tosses and k=51 observedheads. The binomialdistribution’s probability mass function(PMF) is givenby: p(X=k)=n kpk(1−p)n−k(658) where: n=100 is the number of trials, k=51 is the number of successes (heads), p=0.5isthe (theoretical) probability of success(a head) in asingle trial. Left-Tailed Test The left-tailed hypothesis is: H0:p≥p(X≤51)vs.HA:p<p(X≤51)(659) Foraleft-tailed test,wecalculate the cumulativeprobability of observing k≤51,meaning the probability of obtaining 51 or fewerheads: pX ≤k)= x=k  x=0n xpx(1−p)n−x= k  x=0 p(X=x)(660) The p-valueleft-tailed is: p-valueleft-tailed =p(X≤51)=0.618. H0cannot be rejected. Right-Tailed Test The right-tailed hypothesis is: H0:p≤p(X≤51)vs.HA:p>p(X≤51)(661) Foraright-tailed test,wecalculatethe probability of observing k≥51, meaning 51 or moreheads: p(X≥k)= x=n  x=kn xpx(1−p)n−x= n  x=k p(X=x)(662) This can alsobeexpressedusingthe complementarycumulativeprobability: p(X≥k)=1−p(X<k)=1−p(X≤k−1)(663) 9Distributions 259 The p-valueright-tailed is: p-valueright-tailed =p(X≥51)=0.460. H0 cannot be rejected. Theseresults reflect theprobabilities of observing outcomes as extreme (or moreextreme) as k=51 under the null hypothesis of afair coin. Example B We are considering afair coin (p=0.5),with n=100 tosses and k=100 observedheads.The binomial distribution’s probability mass function (PMF) forn=kis given by: p(X=(k=n)) =n kpk(1−p)n−k=n npn(1−p)n−n=pn(664) where: •n=100 is the number of trials, •k=100 is the number of successes (heads), •p=0.5isthe theoretical probabilityofsuccess(ahead) in asingle trial. Left-Tailed Test The p-value forthe left-tailed testisthe probability of observing X≤k. Since k=100 (all heads), the left-tailed p-value is: p(X≤100)=p(X=100) The probability p(X=100)is given by the PMF: p(X=100)=100 100(0.5)100(1−0.5)0 p(X=100)=1·(0.5)100 Right-Tailed Test The p-value forthe right-tailed test is theprobability of observing X≥k.Since k=100,the right-tailed p-value is: p(X≥100)=p(X=100) 260 9Distributions From the previous calculation: p(X=100)=(0.5)100 Forn=100, k=100,and p=0.5, the probability is: p(X=100)=(0.5)100 ≈7.89 ×10−31. Therefore: •Left-tailed p-value: p(X≤100)=p(X=100)≈7.89 ×10−31, •Right-tailed p-value: p(X≥100)=p(X=100)≈7.89 ×10−31. General Casefor n=k=1 Forn=k=1, the PMF simplifies to: p(X=1)=1 1p1(1−p)0=p. Thus, fork=n=1, it is •Left-tailed p-value: p(X≤1)=p(X=1)=p, •Right-tailed p-value: p(X≥1)=p(X=1)=p. Hume’s problem of induction Howprobable isitthatevents observedn−1times will hold in general even forthe next instance (n)? Thisquestion lies at the heartof Hume’s problem of induction (see Hume,1739,Book 1, partIII, section 6, pp. 157-167), which challenges the logical justification of inductive reasoning. The p-value canbeauseful tool to assess the extent to which suchinductive inferences are supported by empirical evidence, though it cannot fullyguaranteethatthe event will occur at the trial naswell. In general, fork=n,itis p(X≥n)=p(X=n)=1−p(X<n)(665) =1−p(X≤(n−1)) (666) =1−(1−p(X=n))) (667) 9Distributions 261 Example II An investigationhas been performed. The observeddataincludes: •k=8successes, •2failures, •Total trials: n=10. The significance level is α=0.05. Right-Tailed p-Value Suppose we are testingthe following right-tailed hypotheses: •H0:π=0.4(the population success probabilityis0.4), •HA:π>0.4(the population successprobability is greater than 0.4). Probabilities forπ=0.4,n=10 The binomialdistribution forπ=0.4,n=10 is given as: p(X=i)=n iπi(1−π)n−i(668) The probabilities forp(X=k)from k=0tok=10 forthe given binomialdistribution (n=10,π=0.4): p(X=0)=10 00.40(1−0.4)10−0=0.0060466176 (669) p(X=1)=10 10.41(1−0.4)10−1=0.0403107840 (670) p(X=2)=10 20.42(1−0.4)10−2=0.1209323520 (671) p(X=3)=10 30.43(1−0.4)10−3=0.2149908480 (672) 268 9Distributions Z(X)is typically defined(see Kelley, 1924)as: Z(X)=X−µ σ,(702) where µ=E(X)and σ=σ(X)denote the expected value and standard deviationofthe randomvariable X,respectively. However, applying afixedglobal mean and standarddeviationacrossall trials,isnot alwaysand in everysingle instance fullyjustified. Underconditions were E(XR,t)and σ(XR,t)vary at eachindividualrun of the experiment t, eachindividual realization XR,tof the runofthe experiment thas aunique mean E(XR,t)and σ(XR,t)standard deviation. The standard normal variable Z(X) need toaccount forthesetrial-specific parameters under these circumstances. Foraspecific, single runortrial twith a unique realization XR,t,the formula forthe standard normal variablefor asingle runofanexperiment can be written as: Z(XR,t)=XR,t−E(XR,t) σ(XR,t)= EXR,t 2 EXR,t·EXR,t(703) where: •XR,tis the observedrealization of the random variable Xwithin the particular conditions of trial t, •E(XR,t)is the expected value or mean of Xas derived specifically within trial t, •σ(XR,t)represents the standard deviation of Xfortrial t,encapsulating its uniquevariability. The expression XR,t−E(XR,t) σ(XR,t)standardizes each XR,tto astandard normal variable (i.e., Z-score).IfXR,tfollows anormal distribution, then each terminthe summation willfollowastandard normal distribution too. Aggregating across ntrials, we obtain standard normal variable of the whole sample/population, Z(XR),as: Z(XR)= n  t=1 Z(XR,t)= n  t=1XR,t−E(XR,t) σ(XR,t)=X−µ σ(704) 9Distributions 269 The chi-squaredistribution is foundational in statistical hypothesis testing and variance estimation. Specifically,when Z1,Z2,...,Znare independent standardnormal variables, thesum of their squares, Z2 1+Z2 2+···+Z2 n,follows achi-squaredistribution with ndegrees of freedom. Thus,achi-squaredistribution with ndegrees of freedom, denoted χ2 n,isformally defined as the sum of the squares of nindependentstandard normal variables(see Sachs,1992,p.213). χ2 n= n  t=1 Z2 t= n  t=1XR,t−E(XR,t) σ(XR,t)2 = n  t=1 EXR,t EXR,t (705) In quantum theory, EXR,tis sometimes the expectation value of the local hiddenvariable. The tndistribution, commonlyknown as the Students t-distribution,isaprobabilitydistribution often used in statistics, especially forsmaller sample sizes or whenthe population variance is unknown and defined (see Sachs,1992,p.213) as: tn=Z χ2 n n (706) This distribution isparameterized by its degrees of freedom,denoted as n,whichaffects the shape of thedistribution. As n→∞, the t-distribution approaches the standard normal distribution, which is useful in hypothesistesting andconfidence intervalestimation. The t-distributionisparticularly useful forcases where the sample mean is used to estimate apopulation mean, and thepopulation standard deviation is unknown. The formula forthe t-distributionwith ndegrees of freedomis: tn= ¯ X−µ S √n (707) where ¯ Xis the samplemean, µis the population mean, Sis the sample standarddeviation, and nis thesample size.The sample standard deviation Sis given by: 270 9Distributions S= 1 n−1 n  i=1(Xi−¯ X)2(708) where: •nis the sample size, •Xirepresents eachindividual observation in the sample, •¯ X=1 n n  i=1 Xiis thesamplemean. This distribution wasoriginallydeveloped by William SealyGosset (see Gosset,1908b)under the pseudonym Student in 1908. The tdistribution playsacentral role in various statistical tests,suchasthe t-test forcomparing samplemeans. Somethingand its ownother The following theoremidentifies that E(UR,t)is akind of acounterparttoE(UR,t),highlighting an opposite relationshipbetween these twoentities i.e expectation values. Specifically,knowledgeofZand E(UR,t)the expected value of UR,tprovides insight into E(UR,t),which is sometimes treated as an indicator of an underlying, possiblylocal hidden parameter within acertainsystem. This parameter can be viewed as representing akind of Anti-UR,tor an alternativeexpression of UR,t, suggesting that the behaviour of one expectation inherentlydetermines the other.Inabroader scientific sense, this relationshiprevealsastructured dualitybetweenthesetwo entities,suggestingthat knowledgeof one aspect leads toanimplicitunderstanding of its complement. Numerus Legis Naturalis 83. In general, it holds that E(UR,t)=z(UR,t)2·E(UR,t)(709) Proof by Direct Proof. Starting from Axiom 1, whichstates that +1=+1(710) we extend thisidentity further in order to obtain UR,t=UR,t(711) and also, 9Distributions 271 UR,t−E(UR,t)=UR,t−E(UR,t)(712) Next, by normalizing the mean deviation, we arrive at UR,t−E(UR,t) σ(UR,t)=UR,t−E(UR,t) σ(UR,t)(713) leading us to define z(UR,t)as z(UR,t)≡UR,t−E(UR,t) σ(UR,t)(714) and hence, z(UR,t)2=U2 R,t·(1−p(UR,t))2 σ(UR,t)2(715) Equation715 simplifies as: z(UR,t)2=U2 R,t·(1−p(UR,t))2 U2 R,t·p(UR,t)·(1−p(UR,t))(716) Finally, z(UR,t)2is given as z(UR,t)2=(1−p(UR,t)) p(UR,t)(717) In general, it is (1−p(UR,t)) =z(UR,t)2·p(UR,t)(718) and 1 z(UR,t)2=p(UR,t) (1−p(UR,t))(719) The fraction 1−p(UR,t) p(UR,t)(see Equation 717)can be split as: z(UR,t)2=1−p(UR,t) p(UR,t)=1 p(UR,t)−p(UR,t) p(UR,t)=1 p(UR,t)−1(720) It is z(UR,t)2+1·p(UR,t)=1(721) The probabilityofasingle event is given as: 272 9Distributions p(UR,t)=1 z(UR,t)2+1(722) Equation 722 ensures that p(UR,t)>0for all real values of z(UR,t).The denominator z(UR,t)2+1isalwaysgreater than or equal to 1, as squares of real numbers are non-negative. Based on z-score, p(UR,t)cannot be zero, ensuring the probabilityremains strictly positive. Through further manipulation of Equation 715,weestablish equallythat z(UR,t)2=E(UR,t)2 σ(UR,t)2(723) and also that z(UR,t)2=E(UR,t)2 E(UR,t)·E(UR,t)(724) This resultisobtained consistentlyand we conclude that z(UR,t)2=E(UR,t) E(UR,t)(725) and therefore, it is generallytruethat E(UR,t)=z(UR,t)2·E(UR,t)(726) Quod eratdemonstrandum. The equation E(UR,t)=z(UR,t)2·E(UR,t)resembles Einsteins massenergy equivalence ER,t=c2·mR,t,asboth express aproportional relationshipbetweenenergyand anothersystemcharacteristic,mediated by asquared parameter. While Einsteinsequationrelates energy to mass with the constant c2,the former equation shows how E(UR,t)dependson E(UR,t),with z(UR,t)2acting as asystem-specific scaling factor.Both equations showaproportional scaling of parameters.Still, theydiffer in scope, withEinstein’sequation representing afundamental physical law, and the second equation representing aprobabilistic relationship specifictoasystem. 9Distributions 273 Variance and standardnormal variable The strategic significance of this proof lies in its ability to link the variance of arandom variable σ(UR,t)2with the z-value z(UR,t),anormalized measure often assumed or readilycalculable. By establishing a relationship betweenthe expectationvalues of UR,tand UR,t,the proof enables the calculation of variancethrough thez-score,which simplifies theprocess of determiningvariability in situations where direct variance calculation maybecomplexorinfeasible. This approachleverages thepower of expectation values and their statistical relationships, offering arobusttool foranalyzing thespread or uncertainty of random variables in probabilistic systems. In practical terms, the z-value serves as abridgetocompute variance, turning abstract statistical concepts into tangible,computable quantities. NumerusLegis Naturalis 84. In general, thefollowing holds: σ(UR,t)2=E(UR,t)·E(UR,t)=z(UR,t)2·E(UR,t)·E(UR,t)(727) Proof by DirectProof. Starting withAxiom 1, whichstates: +1=+1(728) it follows that this identity holds true in all instances.Consequently, we additionallyestablishthe equivalence: E(UR,t)=E(UR,t)(729) Referring to equation 726 from earlier,wesubstitute it into the equation above to yield: E(UR,t)=z(UR,t)2·E(UR,t)(730) Multiplyingbothsides of equation 730 by E(UR,t),wederivethe following expression: σ(UR,t)2=E(UR,t)·E(UR,t)=z(UR,t)2·E(UR,t)·E(UR,t)(731) whichisequivalent with the relationship: σ(UR,t)2=E(UR,t)·E(UR,t)=z(UR,t)2·E(UR,t)·E(UR,t)(732) Quod eratdemonstrandum. 274 9Distributions Normal andantinormal distribution Ilija Barukčić (1.10.1961) The probabilitydensity function of the normal distribution illustrated by Figure 11 is expressed as: p(X)=1 σ√2π e−1 2x−µ σ2 =1 σ√2π e−1 2Ex E(x),x∈R(733) where µis the mean (center)ofthe distribution, σis the standard deviation (spread) of the distribution, xis the random variable, xis the anti random variable. Fig. 11: Standard Normal Distribution. x p(X) +∞ −∞ The largerσbecomes,the flatter the curvesprogression (resulting in abroader curve and alower peak). The term 1 √2πwasintroduced by CarlFriedrich Gaussinthe early19th centuryaspartofhis work on probability theoryand statistics. Thistermoriginates from the process of normalizing the distribution, ensuring that the total area under the curve equals1,whichisafundamental requirement forany probability distribution. The presence of πconnects the probability theorytothe geometry(see Barukčić, 2023e). Assuming that the mathematical formula of the normal distribution is correct,whichhas not been verified atthis point,weobtain the probability densityfunction ofthe antinormal distribution expressed as: p(X)=1−p(X)=1−1 σ√2π e−1 2x−µ σ2 ,x∈R(734) 9Distributions 275 Fig. 12: Normal and Anti Normal Distribution. x p(X) +∞ −∞ Anti normal distribution Normal distribution The relationshipbetween the normal distribution and the anti-normal distribution is illustratedinthe previous figure purelyfor clarityand betterunderstanding whileatthe same timeserving as atemplatefor other distributions (Poisson and anti Poisson distribution et cetera (see Barukčić,2022c). Undercertainassumptions, thecausal relationship kcanbeexpressed using the standard normal variable Z. Numerus Legis Naturalis 85. The standardnormal variable Zand thesample size nare relatedas: n  i=1UR,t−EUR,t σUR,t·WR,t−EWR,t σWR,t n =z2 n(735) Proof by DirectProof. The proof begins by assuming the identity UR,t=WR,t(736) betweenthese two entities despite anyunderlying distinctions.Since UR,t=WR,t,their expected values are alsoequal: EUR,t=EWR,t(737) It follows that the expected squares of UR,tand WR,tare equal: EUR,t2=EWR,t2(738) 276 9Distributions Since UR,t=WR,t,their squared valuesare equal: U2 R,t=W2 R,t(739) The expected values of the squaredterms arealsoequal: EU2 R,t=EW2 R,t(740) Subtracting EUR,tfrom eachvariable yields: UR,t−EUR,t=WR,t−EUR,t(741) Given E(UR,t)=E(WR,t),the above expression implies: UR,t−EUR,t=WR,t−EWR,t(742) Dividing Equation 742 by σUR,tit is UR,t−EUR,t σUR,t=WR,t−EWR,t σUR,t(743) it is σUR,t=σWR,t.Equation 743 is rearranged as UR,t−EUR,t σUR,t=WR,t−EWR,t σWR,t(744) It is zUR,t=UR,t−EUR,t σUR,t(745) and zWR,t=WR,t−EWR,t σWR,t(746) Equation 744 becomes zUR,t=zWR,t(747) or zUR,t2=zWR,t2(748) Based on Equation 747 we rearrangeEquation 748 as: 9Distributions 277 zUR,t·zWR,t=zWR,t2(749) Under conditions of nruns of an experiment, we obtain n  i=1zUR,t·zWR,t= n  i=1zWR,t2=z2(750) Based on Equation 744,we obtain n  i=1UR,t−EUR,t σUR,t·WR,t−EWR,t σWR,t=z2(751) Dividing 751 by n, it is n  i=1UR,t−EUR,t σUR,t·WR,t−EWR,t σWR,t n =z2 n(752) There are circumstances,where thecausal relationshipkis given as: k=z2 n =    n  i=1UR,t−EUR,t σUR,t·WR,t−EWR,t σWR,t n(753) Quod eratdemonstrandum. Given the probabilitydensity function p(Z)of the standard normal distribution: p(Z)=1 √2π e−Z2 2,z∈R(754) we rearrangethis equation as: √2π·p(z)=e−z2 2(755) Take the natural logarithm, ln: ln(√2π·p(z)) =ln e−z2 2(756) Simplify equationas: 284 9Distributions Chi-SquareDistributionand Causal Relationship k The relationshipbetween the chi-squared distribution (seePearson, 1900b)and the causal relationship kreveals fundamental insights about dependency structures between random events.Thisconnection allows kto measurethe extent to whichobservedoccurrences can be treated as causallyrelated, thus playing apivotal role in hypothesis testing and causal inference. Numerus LegisNaturalis 88. In general, therelationship between thechi-squared distribution and thecausal relationship kis given by: kUR,t∩WR,t=2 χ2 EURt ·EWRt (782) Proof by DirectProof. We initiate the proof with Axiom 1, considered self-evident and requiring no additional justification: 1=1(783) MultiplyEquation 783 by the causal relationship kUR,t∩WR,t, yielding: kUR,t∩WR,t=kUR,t∩WR,t(784) We can express Equation 784 in detail as: kUR,t∩WR,t=UR,t·WR,t·pUR,t,WR,t−pUR,t·pWR,t U2 R,t·pUR,t·1−pUR,t·W2 R,t·pWR,t·1−pWR,t =σUR,t,WR,t σUR,t·σWR,t (785) Simplifying Equation 785 further,weobtain: kUR,t∩WR,t=EUR,t,WR,t−EUR,t·EWR,t E(URt)·EURt ·E(WRt)·EWRt (786) Squaring Equation 786,we have: kUR,t∩WR,t2=EUR,t,WR,t−EUR,t·EWR,t2 E(URt)·EURt ·E(WRt)·EWRt (787) Since the chi-squaredtermisdefined as 9Distributions 285 χ2=EUR,t·WR,t−EUR,t·EWR,t2 EUR,t·EWR,t(788) we can substitute this into Equation 787: kUR,t∩WR,t2=χ2 EURt ·EWRt (789) The relationshipbetween the causal relationshipand the chisquare distribution is given as: kUR,t∩WR,t=2 χ2 EURt ·EWRt (790) Quod eratdemonstrandum. KarlPearson (1857–1936) KarlPearson defined (see Pearson,1900b,1904)the phi coefficient φof a2×2contingency table as: φ=χ2 n(791) where χ2is the chi-squared statistic and nis the total sample size. Under the condition that EURt ·EWRt =n,itfollows that kUR,t∩WR,t=φ(792) ;however,this equivalence does not hold universally. 286 9Distributions Empirical Cumulative Distribution Function (ECDF) Traditional statistical methods, such as thet-testand other methods, require assumptions about the distributions of the data(e.g., normality). Serious problem arisefrequently in statistical analysis,especiallywhen the exact distribution of the data is unknown or whennormality is not givenfor sure. Distribution-free tests,alsoknown as non-parametric tests, are statistical methods that do not relyonassumptions about theunderlying distribution of the data. Examplesofdistribution-free testsinclude the Mann-WhitneyUtest(forcomparing twoindependent samples), the Wilcoxonsigned-rank test(forpaired samples), and the Kruskal-Wallistest(forcomparing more than twoindependent groups). However, methods basedonempirical cumulativedistribution functions without assumingany specificunderlying distribution forthe data could be of use too. The Empirical CumulativeDistribution Function (ECDF) as suchisanon-parametric estimator of the cumulativedistribution function(CDF) forasample of data. An ECDF might provide away (see Drion, 1952)toestimatethe cumulativeprobability foragivenvalue based on theobserveddata. In general, givenaset of nobservations, x1,x2,...,xn,the ECDF is defined as the proportion of the samplelessthan or equaltoaparticular value x.Mathematically,the ECDF at apoint xis: Fn(x)=1 n n  t=1 I(xt≤x), where: •Fn(x)is the empirical CDF at x, •xis afixedthreshold(not tied to individual trials), •xtare theobservedoutcomes of the random variables Xt •I(xt≤x)is an indicator function whichcheckswhether eachobservedoutcome xtis less than or equal to the fixed threshold x.If xt≤x,then I(xt≤x)=1, and I(xt≤x)=0otherwise. •nis the total number of observations, The ECDF is especially useful whenaspecificunderlying distribution is not assumed and when the cumulativeprobability is calculated from the data at hand. Rules forusing the Empirical CumulativeDistribution Function can be found in secondaryliterature. 9Distributions 287 Right tailed test In general, it isappropriate to distinguish among the expected value of therandom variable X,the singleexpected value of the random variable Xtat aspecific trial t,and the individual outcome xtof the random variable Xtat aspecific trial t.To reduce the complexity of mathematical notations and to simplify mathematical expressions, in this entirepublication p(Xt)is frequentlyused as ashorthandfor p(Xt=xt).Nonetheless,let Xdenote adiscrete random variable defined on afinite sample space, where the probability p(Xt=xt)represents the likelihood of Xttaking the value xtat the trial t=1,2,...,n.Under these deterministicconditions,each xtis fixed foreach twith its own p(Xt=xt)forall t. Numerus Legis Naturalis 89. The right-tailed pValue based on empirical cumulative distribution function is givenas: p-value (right-tailed) =n−kobs +1 n(793) Proof by Direct Proof. We beginour proof with the lawof identity,afundamentalprinciple in logic and mathematics that asserts: 1=1(794) This equality isuniversallyvalid and formsthe basis forestablishing further relationships and derivations in our proof. In the contextofa single trial t,weare dealing withone specific outcome xtof the single random variable Xt.Ingeneral, in asingle trial, the outcomeofthe random variable Xtis xt,whichisdeterminedbyits singleprobability p(Xt=xt)while theexpectedvalue forthis singletrial is defined as E(Xt).Building on the self-evident foundation as stated before, it logicallyfollows that it is equally E(Xt)=E(Xt)(795) However, the expected valuefor asingletrial E(Xt)whichrepresents the outcome xtweighted by the probability p(Xt=xt)is mathematically given as: E(Xt)=xt·p(Xt=xt)(796) 288 9Distributions under the assumption that p(Xt=xt)0. This approachemphasizes that the expected value is aweighted averageofpossible outcomes, and in thesingle-outcome deterministiccase, the weight is simply p(Xt= xt).Inbrief, it is p(Xt=xt)=E(Xt) xt (797) In the context of asingle trial, there is onlyone outcomeand we do not sum over possibleoutcomes.Nonetheless, it is necessary to consider multiple trials (overn), the total expected value E(X)is the sum of the single expected values foreachindividual trial. In otherwords, we sum the singleexpected values overall trials.The total expected value E(X)acrossall ntrials, where eachtrial has its ownoutcome xtand its associated probability p(Xt=xt)is given as: E(X)= n  t=1 xt·p(X=xt)(798) =E(X1)+E(X2)+···+E(Xn)(799) = n  t=1 E(Xt)(800) wherenis thesampleor populationsize.Underdeterministicconditions, where p(Xt=xt)=1, the expectation value itself reduces to the observedvalue: E(X)= n  t=1 xt·p(Xt=xt)= n  t=1 xt·1= n  t=1 xt(801) When the expectedvalue foreachtrial, E(Xt),isidentical across all trials t=1,...,n,the total expected value E(X)simplifies to: E(X)= n  t=1 xt·p(Xt=xt)= n  t=1 E(Xt)=n·E(Xt=1)(802) Under thesedeterministic conditions,where p(Xt=xt)is constant forall trials and eachtrial has afixeddeterministic outcomeXt=1, the expectation value simplifies to: 9Distributions 289 E(X)= n  t=1 E(Xt)(803) = n  t=1 xt·p(Xt=xt)(804) = n  t=1 1·p(Xt=xt)(805) =n·p(Xt=xt)(806) In the presence of uniformityacross trials,itfollows that: p(Xt=xt)=E(X) n(807) In general, under deterministic conditions where p(Xt=xt)=1for all trials and where Xt=1has afixeddeterministic outcomeateachtrial, we do expectthat kobs =n.However,due to errors in measurement or other sources of bias et cetera, the observedtotal kobs maydeviate from n. Still, howcan we,under the circumstances where kobs <nsuccesses are observedout of ntrials, assume with some probability that kobs =n? In other words, howcan we calculate the right-tailed p-value under these conditions? In general,the right-tailedp-value is defined as: p-value (right-tailed) =p(Xt≥kobs)(808) = n  xt=kobs p(Xt=xt)(809) =1−p(Xt≤(kobs −1)) (810) where: •kobs is theobservedvalue of the random variableX, •Xtis the random variable fortrial t, •p(Xt=xt)is the probabilitymass function (pmf) of Xt,which gives the probabilityofobserving aspecific outcome xtfortrial t, •The summation runs over all outcomes from the observedvalue kobs to the maximum possiblevalue n,representing the number of trials. 290 9Distributions E(X)= n  xt=kobs E(Xt)(811) = n  xt=kobs xt·p(Xt=xt)(812) = n  xt=kobs 1·1(813) = n  xt=kobs 1(814) The summation: n  xt=kobs 1(815) is essentiallysumming the value 1overeachvalue of xtfrom kobs to n. So, it countshow manyvalues of xtare in the rangefrom kobs to n. -The variable xtcan takeinteger values, and we sum 1for eachvalue of xtfrom kobs to n. -The resultofthe summation will simply be thetotal count of integers from kobs to n,inclusive. The integersfrom kobs to nform asequence: kobs,kobs +1,kobs +2,...,n. We need to count howmanyterms there are in this sequence. The sequence startsatkobs and ends at n.The difference between the first term kobs and the lastterm nis n−kobs,but since both endpoints are included inthe sequence, we need to add 1. Therefore, the total number of terms in the sequence is given by (seeTheorem 90,Equation 822): E(X)= n  xt=kobs 1=n−kobs +1(816) This is because: n−kobs gives the difference between the endpoints. Thus far, adding 1accounts forthe inclusion of both kobs and nin the sequence. The total probabilitymassmustsum to 1. Dividing by the totalnumber of trialsn,the normalized probability is: 9Distributions 291 p-value (right-tailed) =E(X) n =n−kobs +1 n(817) Quod eratdemonstrandum. Example. Mostofthe relationships developed in this publication are determined by the formula p∗=kobs n=1. In this case, we obtain the righttailed p-value as follows: p-value (right-tailed) =E(X) n=n−kobs+1 n=1−p∗+1 n(818) Forlargersample sizesn,the term 1 napproaches zero and becomes negligible,causing the right-tailed p-value to simplify to: p-value(right-tailed) =E(X) n=n−kobs+1 n=1−p∗+1 n≈1−p∗(819) Thedecision rulescan be summarized as follows: Test Type pV alue <α pV alue ≥α Left-Tailed Test HAaccepted H0accepted Right-Tailed Test HAaccepted H0accepted Left tailed test To formulate the left-tailed testanalogous to the right-tailed testdescribed before, the focus shiftstocalculating the probability that the observed value kobs or fewersuccesses occur.The left-tailed p-value is defined as: p-value (left-tailed) =p(X≤kobs)= kobs  xt=1(p(X)t=xt)(820) The totalprobabilityisnormalized by thetotal number of trials n, resulting in theleft-tailed p-value: p-value (left-tailed) =kobs n(821) 292 9Distributions The length of asequence Numerus Legis Naturalis 90. The lengthofthe sequence Nfrom t=kobs,kobs +1,...,(kobs +u)=nis given by: N=n−kobs +1(822) Proof by Direct Proof. We prove the length of the sequence t=kobs,kobs +1,...,nstep by step. The sequence consists of the indices: t=kobs,kobs +1,kobs +2,...,n(823) This sequence formsanarithmetic progression with: •Initial value a1=kobs, •Final value aN=n, •Step size d=1. The general formula forthe N-th termofanarithmetic progression is: aN=a1+(N−1)·d,(824) where: aNis the final termof the sequence, Nis the number of terms. Substituting the known values: aN=n,a1=kobs,d=1,we obtain: n=kobs +(N−1)·1.(825) Simplify: n=kobs +N−1(826) Rearranging toisolate N: N=n−kobs +1(827) Thus, the length of the sequence is: N=n−kobs +1(828) Quod eratdemonstrandum. Part IV Conditionalismand causation Part VI Causation Part VII LawofNature 612 Notice References Angrist,J.D., Imbens, G. W.,and Rubin, D. B. (1996). Identificationofcausal effects using instrumental variables. 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E.), 107, 110, 204 Aristotle, of Stagira (ca. 384 BCE-ca. 322 BCE), 32, 204, 379, 380 The doctrine of four causes, 32 Bar,CarlLudwig von(1836–1913), 97, 382, 383 Barukčić, Ilija (1961–), 58 Bayes, Reverend Thomas (1701–1761),74 Bell, John Stewart(1928–1990),87, 92, 93 Berkeley, Bishop George (1685–1753),48, 56 Bernoulli, Daniel (1700–1782), 138, 152 Bernoulli, Jacob (1654–1705), 180, 217, 295 Bernstein, Sergei(1880–1968),332 Bianchi, Luigi (1856–1928),195 Bohr,Niels Henrik David (1885–1962), 90–92 Bombelli, Raffaele (1526–1572),107, 110, 153 Boole, George (1815–1864), 107 Born, Max (1882–1970), 90, 153 Bravais, Auguste (1811–1863),61 Brouwer,Luitzen Egbertus Jan (1881–1966), 369 Bruno, Giordano (1548–1600),33, 358, 519 Carnap, Rudolf (1891–1970), 78 Cauchy,Augustin-Louis(1789–1857),122 Chebyshev, Pafnuty (1821–1894),162, 332 Christoffel, Elwin (1829–1900),195 Clopper,Charles J. (1899–1983),226 Cornfield, Jerome (1912–1979),300, 305 Cotes,Roger (1682–1716), 111 d’Holbach, Paul-HenriThiry(1723–1789), 34, 56 causal chain, 35 cause, 34 effect, 34 Darwin, Charles Robert (1809–1882), 7, 8, 115 DeGroot, Morris Herman (1931–1989), 129, 172, 173,179 Dirac, Paul Adrien Maurice(1902–1984), 107, 110, 206 Drude, Paul (1863–1906), 107, 110 Ducasse, CurtJohn (1881–1969), 410 Edgeworth, Francis Ysidro (1845–1926), 61 Einstein, Albert (1879–1955), 13,78, 85, 92, 150, 155, 358, 359, 516, 599, 601, 603 Does the moon existonlywhen youlook at the same?, 49 ImmanuelKant, 48 Objectivereality,49, 54 Reichenbach, 78 Euclid, of Alexandria (ca. 360-280 BCE), 124 Euler,Leonhard (1707–1783), 111, 118, 120, 248 Feuerbach, Ludwig Andreas (1804–1872), 57 Finsler,Paul (1896–1970), 142 Fisher,Sir Ronald Aylmer (1890–1962), 174, 180, 183,267, 300, 306, 371 Franz, John E., 396 Frege, FriedrichLudwig Gottlob (1848–1925), 107, 409, 422 Galton, Sir Francis (1822–1911), 61, 267 629 630 AUTHORINDEX Gauss, Carl FriedrichGauss (1777–1855), 174, 180, 267 Geyser,Gerhard Joseph Anton Maria (1869–1948), 544 Good, Irving John (1916–2009),79 Gosset, William Sealy(1876–1937),270 Hadamard, Jacques Salomon (1865–1963), 207 Haldane, John Burdon Sanderson (1892–1964), 61, 69, 70 Halley, Edmond (1656–1742),111 Hegel, GeorgWilhelmFriedrich (1770–1831), 52, 55, 56, 116, 117, 510, 520, 522, 523 -Kant’s scepticims, 52 Causality,57 Necessity,57 Heisenberg, Werner Karl(1901–1976), 90–92, 516, 586 Visit of EinsteininPrinceton, 516 Helmert, FriedrichRobert(1843–1917), 280, 391 Henle, FriedrichGustavJakob (1809–1885), 72 Hess, Karl(1945-), 93 Hessen, Johannes (1889–1971), 36, 526, 532, 544 Hesslow, Germund (1949-), 82 Heyting, Arend (1898–1980), 369 Hill, Sir Austin Bradford (1897–1991),72, 300 Hume, David (1711–1776),36, 40, 80, 411, 518, 524–526, 542 Problem of induction, 454, 525 Huygens, Christiaan (1629–1695),162, 167 Justice Matthews U.S. Supreme Court1884, 381, 394, 396 U.S. Supreme Court1884: Hayes v. Michigan Central R.Co., 111 U.S. 228, 381, 394, 396 Kant, Immanuel (1724–1804),43, 52, 103, 147, 598 Kelley,Truman Lee (1884–1961), 267 Khrennikov,Andrei (1958-), 93 Knuth, Donald E. (1938–),123 Koch,RobertHeinrichHermann (1843–1910), 72 Kohlrausch, R.(1809–1858),107, 110 Kolmogorov, Andrei Nikolaevich (1903–1987), 83, 124, 140, 143, 155, 162, 361, 372, 405, 412, 426 Korch, Helmut (1926–1998), 38, 40, 544 Kröber,Günter (1933–2012),87 Langii, Iohannis Christiani,426 Laplace,Pierre-Simon Marquis de (1749–1827), 89, 267, 587 Leibniz, GottfriedWilhelm (1646–1716), 56, 107, 115, 153, 512, 519 Levi-Civita,Tullio (1873–1941), 195 Lewis, David Kellogg (1941–2001), 41, 78 Lexis, Wilhelm (1837–1914), 267 Libri, Guillaume (1803–1869), 123 Lorentz, Hendrik Antoon (1853–1928), 134 Mackie, John Leslie (1917–1981), 410, 411 Markov, AndreyAndreyevich(1856–1922), 332 Marx, Karl(1818–1883), 52, 53,57, 116 Matthews, Justice, 381, 394, 396 Mercier,Charles (1851–1930), 82 Mill, John Stuart(1806-1873), 411 Mises,Richard Edler von(1883–1953), 161 Moivre, Abrahamde(1667–1754), 83,155, 157, 180, 267 Neumann, John von(1903–1957), 113, 148 Newton, Sir Isaac (1643–1727), 89, 118, 205 Neyman, Jerzy(1894–1981), 226 Nicod, Jean George Pierre (1893–1924), 368 Nieuwenhuizen, Theodorus Maria, 93 Noordhof, Paul JonathanPitt, 42 Pacioli, Luca (1447–1517), 108, 111 Pais, Abraham (1918–2000), 49 Peano, Giuseppe (1858–1932), 113 Pearl, Judea(1936-), 75,592 Counterfactuals andquantum theory, 76 Pearsonstill rules statistics, 69 Pearson, Egon S. (1895–1980), 226 Pearson, Karl (1857–1936), 60,70, 174, 180, 280, 284,285, 295, 391, 542 Peirce, CharlesSantiagoSanders (1839–1914), 267 Philo of Megara,409 Pisa, Leonardoof(1170–1240), 110 Planck, MaxKarlErnst Ludwig (1858–1947), 88, 107, 110, 205, 588 Poincaré,Jules Henri(1854–1912), 89 Polack, Fernando P, 369 Popper, SirKarlRaimund (1902–1994), 13, 15, 17, 593 Pythagoras, of Samos (ca. 570–ca. 495 BCE), 131 Pólya,György(1887–1985), 267, 337 Quetelet, LambertAdolphe Jacques (1796–1874), 267 Raedt, Hans de, 93 Recorde, Robert (1510–1558), 108, 111 AUTHOR INDEX 631 Reichenbach, Hans (1891–1953), 78, 89, 92, 515, 587 Ricci-Curbastro, Gregorio(1853–1925), 195 Riemann, Bernhard (1826–1866), 195 Robins, James M., 74 Rolle, Michel (1652–1719), 108, 111 Rota, Gian Carlo, 124 Russell, Bertrand (1872–1970), 409, 422 Russell, Bertrand Arthur William (1872–1970), 90 Salmon, WesleyCharles (1925–2001), 79–81 Schlick, FriedrichAlbertMoritz (1882–1936),588 Schrödinger,Erwin Rudolf Josef Alexander (1887–1961),115, 588 Schur,Issai (1875–1941), 207 Sheffer,HenryMaurice (1882–1964),367 Simpson, Thomas (1710–1761),174 Spinoza, Benedictus de (1632–1677),597 Spohn, Wolfgang Konrad, 42 Steno, Nicholas (1638–1686),138 Stiehler,Gottfried (1924–2007), 58 Stirling, James,1692–1770, 255 Suppes, Patrick(1922–2014),79, 80, 155 Tarski, Alfred (1901–1983), 422 Thales of Miletus (ca. 624/623–ca.548/545 BCE), 127 Thompson, Mary Elinore (1944-), vii Toohey, John Joseph, 11 Ulyanov, Vladimir Ilyich(1870–1924), 53 Uspensky,James Victor(1883–1947), 145, 295, 383 Uyomov, Avenir Ivanovich(1928–2012), 515 Venn, John (1834–1923), 426 Voigt, Woldemar (1850–1919), 195 Wallisii, Iohannis (1616–1703), 117, 118 Weber,W.E.(1804–1891), 107, 110 Weisio, Christiano, 426 Whitehead, AlfredNorth (1861–1947), 409 Whitworth, WilliamAllen (1840–1905), 162, 167 Widmann, Johannes(1460–1498), 108, 111 Williamson, Jon, 42 Wilson, Edwin Bidwell(1879–1964), 226 Wright, Sewall Green (1889–1988), 592 Yates, Frank (1902–1994), 373, 392, 415, 427 Yule, George Udny(1871–1951), 65, 70, 303, 305 Zesar,PatrickManuel, 507