(b) What is the probability that a particle reaches the upper boundary n before the lower boundary m?


 Bryan Cody Page
 1 years ago
 Views:
Transcription
1 MATH 529 The Boudary Problem The drukard s walk (or boudary problem) is oe of the most famous problems i the theory of radom walks. Oe versio of the problem is described as follows: Suppose a particle begis at a positive iteger height o the y axis. O each uit step time iterval, the particle moves up 1 uit with positive probability p, dow 1 uit with positive probability q, or remais at the same height with probability r = 1 p q, where r < 1. We shall call this process a simple radom walk. The height after k steps ca be writte as h k = + X X k, where X i =1 Up 1 Dow are i.i.d. (a) For m < <, what is the probability that a particle ever reaches oe of the boudary heights of or m? (b) What is the probability that a particle reaches the upper boudary before the lower boudary m? (c) What is the average umber of steps eeded to reach either boudary of or m? After we solve these problems, other results ca be obtaied by takig limits as or as m, ad also by geeralizig to processes that ump more tha 1 uit at a time. 1. The Probability of Hittig a Boudary We iitially assume that the particle begis at a iteger height, where 0. The by vertical traslatio, we ca obtai results for m. We first aim to show that a radom particle will reach a boudary of 0 or with probability 1. But this result does ot mea that every sigle path reaches a boudary. Ideed, a path that starts at height 5 ad goes up, dow, up, dow, up, dow, ad ifiitum, will ever reach a boudary of 0 or 10, but istead will always oscillate betwee heights 5 ad 6. However, the probability of this sigle path occurrig is (p q) = 0. I fact, eve =1 though there are a ucoutable umber of paths that always remai iside the boudaries, the probability of this collectio of paths is equal to 0. But rather tha show this result directly, we shall show that the probability of hittig a boudary equals 1 by usig the techique of differece equatios. We ow derive the probability of hittig a boudary. We let B 0 deote the set of paths that cause a simple radom walk to reach a boudary of 0 or whe startig at height, ad let e = P( B 0 ) for 0. The 0 e 1. Moreover, if a walk begis at height 0 or at height, the it automatically reaches a boudary without eve takig a step, thus we have the boudary coditios e 0 = 1 = e (1.1)
2 For 0 < <, ay walk that hits a boudary must either go up, go dow, or remai costat o the first step; thus, we ca apply the Law of Total Probability to obtai: P( B 0 ) = P X 1 =1 ( ) P B 0 ( X 1 =1) + P X 1 = 1 ( ) +P( X 1 = 0) P B 0 ω 1 = 0 = p P +1 B 0 ( ) P B 0 ( ) + q P ( 1B 0 ) + r P ( B 0 ). ( X 1 = 1) That is, for 0 < <, e = p e +1 + q e 1 + r e. (1.2) Because e = (p + q + r)e, we ca substitute i (1.2) to obtai (p + q + r)e = p e +1 + q e 1 + r e, or p (e e +1 ) = q (e 1 e ) (1.3) for 0 < <. (Coveietly, the terms ivolvig r have dropped out.) Note that if p = 0 ad q > 0, the e 1 = e for 0 < <. Thus, 1 = e 0 = e 1 = e 2 =... = e. We see that P( B 0 ) = 1 for all 0. Likewise, P( B 0 ) = 1 for all if q = 0 ad p > 0. So hereby we ca assume that p 0 ad q 0. Next we defie the forward differece f by f = e e +1 for 0 <. The Equatio (1.3) becomes p f = q f 1, or f = q p f 1 (1.4) But also f 1 = (q / p) f 2. By recursio, we ca write f i terms of f 0 as f = q f p 0 (1.5) We ow ca evaluate the sum of the forward differeces to solve for f 0 :
3 0 = e 0 e = (e 0 e 1 ) + (e 1 e 2 ) +...+(e 1 e ) 1 = f i = i=0 q f p 0 1 i= 0 f 0 = 1 (q / p) f 0 1 q / p i if p q (1.6) Hece, f 0 must equal 0. The by Equatio (1.5), 0 = f = e e +1, ad thus e = e +1, for 0 <. Thus, 1 = e 0 = e 1 = e 2 =... = e 1 = e. We therefore state: Theorem 1. Let 0. There is probability 1 that a simple radom walk begiig at height will reach a boudary of 0 or. A simple radom walk that begis at height ad eds at boudaries m ad is equivalet to a simple radom walk that begis at height m ad eds at boudaries 0 or m. That is, B m = m B 0 m for m. Applyig Theorem 1 with iitial height m ad upper boudary m, we ca state: Corollary 1. Let m. There is probability 1 that a simple radom walk begiig at height will reach a boudary of m or. 2. The Boudary Problem We ow kow that simple radom walks reach a boudary with probability 1. We ext wish to determie the probability P m of reachig the top boudary before the lower boudary m. So we ow let U m be the set of paths that cause a simple radom walk to reach height before height m whe startig at height ad discrete time 0. We ow derive the probability of hittig the upper boudary before the lower boudary. To derive the result for m = 0, we eed oly modify the boudary coditios of our differece equatios argumet. Here we let U 0 be the set of paths that cause a simple radom walk to reach height before height 0 whe startig at height, ad let e = P 0 = P( U 0 ). The 0 e 1. Now if a walk begis at height, the it reaches first with probability 1; but if it begis at height 0, the it reaches first with probability 0. Thus, e 0 = 0 ad e = 1. (2.1) Sice a walk that hits first, for 0 < <, must either go up, go dow, or remai costat o the first step, we may apply the Law of Total Probability to obtai:
4 ( ) + P X 1 = 1 ( ) P( U 0 ) = P ( X 1 =1 ) P U 0 ω 1 =1 +P( X 1 = 0) P U 0 ω 1 = 0 ( ) + q P ( 1U 0 ) + r P ( U 0 ) = p P +1 U 0 = p e +1 + q e 1 + r e. The as with Equatio (1.3), we obtai, for 0 < <, ( ) ( ) P U 0 ω 1 = 1 p (e e +1 ) = q (e 1 e ), (2.2) Agai, if p = 0 ad q > 0, the e 1 = e for 0 < <. Thus, 0 = e 0 = e 1 = e 2 =... e 1 = 0. We see that P 0 = 0, for all 0 <, ad P 0 = e = 1. However if q = 0 ad p > 0, the e = e +1 for 0 < <. Thus, e 1 = e 2 =... = e 1 = e = 1. We see that P 0 = 1, for all 1, ad 0 P 0 = e0 = 0. I other words, if p = 0, the there is o chace of reachig the top boudary first uless we start at the top boudary. Ad if q = 0, the we are certai to reach the top boudary first uless we start at the bottom boudary. (Sice we still are assumig that r < 1, the p ad q caot both be 0.) Sice these trivial cases are resolved, we ow may assume that p 0 ad q 0. We agai let f = e e +1 be the forward differece for 0 <. Equatio (2.2) ad recursio, we have The by f = q f p 0. (2.3) Evaluatig the sum of these forward differeces we obtai: 1 = e 0 e = (e 0 e 1 )+ (e 1 e 2 )+...+(e 1 e ) 1 1 q = f i = i f 0 p i=0 i=0 f 0 = 1 (q / p) f 0 1 q / p if p q. (2.4)
5 Thus, f 0 = 1 1 q / p 1 (q / p) if p q. (2.5) Now by evaluatig the partial sum of the forward differeces, we obtai 1 e = (e 0 e ) = f i i=0 f 0 = 1 (q / p) f 0 1 q / p 1 q = i f 0 p i=0 / = 1 (q / p) if p q 1 (q / p) if p q. (2.6) Moreover, the simple radom walks that begis at height ad reach height before height m are equivalet to the simple radom walks that begis at height m ad reach height m before height 0. Thus, the probability of reachig the upper boudary first remais the same after subtractig m from each height. That is, P m = mp 0 m. Makig these substitutios ito Equatio (2.6), we ca state: Theorem 2. For simple radom walks begiig at height, with m, the probability of reachig height before height m is give by m m P m = 1 (q / p) m 1 (q / p) m if p q. The probability of reachig height m before height is give by Q m =1 P m = m (q / p) m (q / p) m 1 (q / p) m if p q. But if p = 0, the P m = 0 for m < ad P m = 1. A closedform formula is ot kow for the geeral case whe the walks do ot move up ad dow with the same heights. For that case, we will use the matrix methods to fid umerical values for P m.
6 3. Average Number of Steps Because we kow that simple radom walks hit a boudary with probability 1, we also ca obtai the average of the umber of steps S m eeded to hit a boudary. Because ( ) = P m = 1, the average umber of steps the ca be expressed by the usual P S m < defiitio of the mea of a discrete radom variable: E S m = ( ) k P S m = k, k =k 0 where k 0 = Mi{ m, }. But rather tha compute this sum directly to fid the average, we istead ca use the differece equatios argumet with a variatio of the boudary coditios. Let S m be the umber of steps eeded for a simple radom walk to reach a boudary of height or m whe startig at height. I particular, let S 0 be the umber of steps eeded for a simple radom walk to reach either height or height 0 whe startig at height, with 0, ad let e = E S 0 [ ]. The 0 e, ad if a walk begis at height or at height 0, the it reaches a boudary i 0 steps; thus, e 0 = 0 = e. Now suppose that 0 < <. Assume p = 0 so that paths ever go upward. Sice we assume that r < 1, the q > 0 ad the paths evetually will drift dowward ad reach the lower boudary. Sice it will take dowward steps to reach 0, the actual umber of steps required to do so forms a egative biomial radom variable. I this case, the average umber of steps e equals the egative biomial average of /q. Similarly, if q = 0, the the average umber of steps eeded to reach the (top) boudary is e = ( ) / p. Heceforth, we may assume that either p or q equals 0. So ow for 0 < <, we ca express E[ S 0 ] as E[ S 0 ] = 1 + p e +1 + q e 1 + r e. (Oe step must be take, the we proceed from there.) Applyig a similar differece equatios argumet, we obtai E S 0 = ( ) 2p 1 (q / p) p q 1 (q / p) p q if p q. (3.1)
7 Now if a walk begis at height ad eds at boudaries of m or, with m, the we ca subtract m from each height to obtai a equivalet walk that begis at height m ad eds at boudaries of 0 or m. The umber of steps eeded to reach a upper boudary remais the same alog each path, thus so does the average umber m of steps. That is, E[ S m ] = E[ m S 0 ]. Makig this substitutio ito (3.1), we have Theorem 3. For simple radom walks begiig at height, with m, the average umber of steps eeded to reach a boudary of height or height m is give by E S m = ( )( m) 2p m p q 1 q / p 1 q / p ( ) m ( ) m m p q if p q. But if p = 0, the E[ S m ] = ( m) / q for m < ad E[ S m ] = 0. As with the probability of hittig the top boudary first, a geeral formula is ot kow for the average umber of steps i the case whe the walks do ot move up ad dow with the same heights. For that case, we agai will use matrix methods. Disclaimer: The differece equatios argumet to derive (3.1) assumes a priori that the average is fiite i order to subtract averages ad create the forward differeces. If we make this assumptio (see Addedum), the we ca also apply Wald s Idetity to obtai the expected value of T = S 0. The fial height is either 0 or. Thus, P 0 = E[hT ] = + E[X X T ] = + E[X 1 ] E[T ] = + (p q)e[t ]. (3.2) I this case of p = q, Wald s Idetity does ot help for (3.2) simply becomes =. But for p q, (3.1) becomes 1 (q / p) 1 (q / p) = P 0 = + (p q)e[t ], which gives E[T ] = p q 1 (q / p) 1 (q / p) p q.
8 4. Geeral Average Edig Height Usig the formula for P m, we easily ca fid the average height of the simple radom walk upo stoppig at a boudary of or m. Here we let F m deote the fial height upo stoppig. Sice F m is either or m, the average edig height is give by Simplifyig this expressio, we obtai: E[ F m ] = P m + m 1 ( P m ). Theorem 4. For simple radom walks begiig at height, with m, the average height upo stoppig at boudaries of or m is give by E F m = ( m)(q / p) m m(q / p) m 1 (q / p) m if p q. But if p = 0, the E[ F m ] = m for m <, ad E[ F m ] =. Although our formulas have bee derived for simple radom walks that go up or dow 1 uit at a time, they apply wheever the walks go up ad dow the same amout a o each step. Ideed, we eed oly divide by this value of a ad the apply the formulas. I this case, the average fial height will be a E F m [ ], where,, ad m are the scaleddow iitial coditios. Example. You start with $5000 ad proceed to bet $50 o red at the roulette wheel util you reach $6000 or drop to $3000. (a) What is the probability of reachig $6000 before droppig to $3000? (b) What is the average umber of bets that will be made i this situatio? (c) What is the average amout of moey that remais i this situatio? Solutio. By dividig by $50, the problem is equivalet to startig with = $100 ad bettig $1 at a time util you reach = $120 or drop to m = $60. Usig p = 18/38 ad q = 20/38, we have (q / p) 40 (a) 100 P 60 = , ad 1 (q / p) 120 (b) E[ 100 S 60 ] = 60 p q 1 (q / p) (q / p) p q
9 So you have about a 12% chace of reachig the goal of $6000 before droppig to $3000. O the other had, you probably ca play for a log time sice the average umber of bets would be over 623. (c) The average edig fortue is 120 $ P 60 + $ Q 60 $ About 12% of the time we reach $6000 ad quit, ad about 88% of the time we drop to $3000 ad quit. O average, we are left with $ Addedum I order to use Wald s Idetity or differece equatios to derive the average time eeded to hit a boudary, we first eed to kow that the average time is fiite. So assume that q > p. Now igore costat steps ad cosider oly those paths that move up or dow with probabilities!p = p / (p + q) ad!q = q / (p + q). The!q >!p. I this case, we have previously see that the probability of hittig height 0 from height for the first time i 2i + (ocostat) steps is give by P( T! 2i + = 2i + ) =!p i!q i+. 2i + i Thus, for q > p, the average umber of (ocostat) steps eeded to drop to height 0 is E[!T ] = (2i + )P(T = 2i + ) =!q (2i + )! (!p!q) i. (1) i! (i + )! i=0 i=0 We ote that the maximum value of!p!q is 1/4 whe!p =!q =1/ 2, ad!p!q <1/ 4 whe q > p. Usig the Ratio Test, we ca show that the sum (1) is fiite as follows: (2i )! (!p!q) i+1 (i +1)! (i +1+ )! i! (i + )! (2i + + 2)(2i + +1) = i (2i + )! (!p!q) (i +1)(i +1+ ) (!p!q) 4!p!q <1. i Now let N i ~ geo(p + q) be the umber of steps required to obtai the ith ocostat step. The total umber of steps required to drop to 0 is the N N!T ; thus, E[N N!T ] = E[N 1 ] E[! T ] = 1 p + q E[! T ] <. Thus, for q > p, the average umber of steps eeded to reach a height of 0 or must be less tha or equal to the average umber of steps eeded to reach height 0, which is fiite. Next suppose that p > q. By symmetry, the average umber of steps eeded to go up to height must be fiite. Thus, the average umber of steps eeded to reach a height of 0 or must also be fiite.
10 Exercises 1. Cosider a oedimesioal radom walk that begis at height ad goes up or dow oe uit at a time. You wish to reach = + 3 before droppig to height 0 with probability (a) Derive the miimum height at which you ca begi. For this with p = q ad r = 0.2, what is the average umber of steps eeded to reach a boudary of 0 or = + 3, ad what is the average edig height? (b) Derive the miimum height at which you ca begi if p = 0.66 ad r = For this, what is the average umber of steps eeded to reach a boudary ad what is the average edig height? (c) If p = 0.25 ad r = 0, fid the maximum possible probability of reachig = + 3 before droppig to 0 (regardless of iitial height ). (d) Regardless of startig height, what is the maximum possible average umber of steps eeded to reach 0 or = + 3 whe (i) p = q = 0.4 ad whe (ii) p = 0.66, r = 0.12? (i.e., i these cases, what happes to the average as?) 2. Complete the differece equatios argumet to derive the formula for E[ S 0 ] i the case of p = q. 3. Let!p = p / (p + q),!q = q / (p + q) ad let! S0 be the umber of ocostat steps required to hit a boudary. Use (3.1) with!p ad!q to show that E! S0 = (p + q)e S Cosider a reflective lower boudary for which paths must move upward 1 uit o the subsequet step after hittig the lower boudary of 0. For 0 < <, derive the average umber of steps eeded to reach a upper boudary of whe startig at height. Bous Problem: I the case of p = q, argue idepedetly that the average umber of steps eeded to hit a boudary must be fiite.
Section 11.8: Power Series
Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i
More informationCS / MCS 401 Homework 3 grader solutions
CS / MCS 401 Homework 3 grader solutios assigmet due July 6, 016 writte by Jāis Lazovskis maximum poits: 33 Some questios from CLRS. Questios marked with a asterisk were ot graded. 1 Use the defiitio of
More informationRecurrence Relations
Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial()); } Let t be the umber of multiplicatios eeded to calculate factorial(). The
More informationHow to Maximize a Function without Really Trying
How to Maximize a Fuctio without Really Tryig MARK FLANAGAN School of Electrical, Electroic ad Commuicatios Egieerig Uiversity College Dubli We will prove a famous elemetary iequality called The Rearragemet
More informationHOMEWORK 2 SOLUTIONS
HOMEWORK SOLUTIONS CSE 55 RANDOMIZED AND APPROXIMATION ALGORITHMS 1. Questio 1. a) The larger the value of k is, the smaller the expected umber of days util we get all the coupos we eed. I fact if = k
More informationThe Random Walk For Dummies
The Radom Walk For Dummies Richard A Mote Abstract We look at the priciples goverig the oedimesioal discrete radom walk First we review five basic cocepts of probability theory The we cosider the Beroulli
More informationRADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify
Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationSEQUENCES AND SERIES
9 SEQUENCES AND SERIES INTRODUCTION Sequeces have may importat applicatios i several spheres of huma activities Whe a collectio of objects is arraged i a defiite order such that it has a idetified first
More informationRademacher Complexity
EECS 598: Statistical Learig Theory, Witer 204 Topic 0 Rademacher Complexity Lecturer: Clayto Scott Scribe: Ya Deg, Kevi Moo Disclaimer: These otes have ot bee subjected to the usual scrutiy reserved for
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationThe Poisson Distribution
MATH 382 The Poisso Distributio Dr. Neal, WKU Oe of the importat distributios i probabilistic modelig is the Poisso Process X t that couts the umber of occurreces over a period of t uits of time. This
More informationIntegrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number
MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios
More informationNUMERICAL METHODS FOR SOLVING EQUATIONS
Mathematics Revisio Guides Numerical Methods for Solvig Equatios Page 1 of 11 M.K. HOME TUITION Mathematics Revisio Guides Level: GCSE Higher Tier NUMERICAL METHODS FOR SOLVING EQUATIONS Versio:. Date:
More information( ) (( ) ) ANSWERS TO EXERCISES IN APPENDIX B. Section B.1 VECTORS AND SETS. Exercise B.11: Convex sets. are convex, , hence. and. (a) Let.
Joh Riley 8 Jue 03 ANSWERS TO EXERCISES IN APPENDIX B Sectio B VECTORS AND SETS Exercise B: Covex sets (a) Let 0 x, x X, X, hece 0 x, x X ad 0 x, x X Sice X ad X are covex, x X ad x X The x X X, which
More informationMath 4400/6400 Homework #7 solutions
MATH 4400 problems. Math 4400/6400 Homewor #7 solutios 1. Let p be a prime umber. Show that the order of 1 + p modulo p 2 is exactly p. Hit: Expad (1 + p) p by the biomial theorem, ad recall from MATH
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationsin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =
60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece
More informationCHAPTER 5. Theory and Solution Using Matrix Techniques
A SERIES OF CLASS NOTES FOR 20052006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL
More informationIP Reference guide for integer programming formulations.
IP Referece guide for iteger programmig formulatios. by James B. Orli for 15.053 ad 15.058 This documet is iteded as a compact (or relatively compact) guide to the formulatio of iteger programs. For more
More informationDiscrete Mathematics and Probability Theory Spring 2016 Rao and Walrand Note 19
CS 70 Discrete Mathematics ad Probability Theory Sprig 2016 Rao ad Walrad Note 19 Some Importat Distributios Recall our basic probabilistic experimet of tossig a biased coi times. This is a very simple
More informationMATH 10550, EXAM 3 SOLUTIONS
MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,
More informationSection 7 Fundamentals of Sequences and Series
ectio Fudametals of equeces ad eries. Defiitio ad examples of sequeces A sequece ca be thought of as a ifiite list of umbers. 0, , 0, , 0...,,,,,,. (iii),,,,... Defiitio: A sequece is a fuctio which
More informationProblems from 9th edition of Probability and Statistical Inference by Hogg, Tanis and Zimmerman:
Math 224 Fall 2017 Homework 4 Drew Armstrog Problems from 9th editio of Probability ad Statistical Iferece by Hogg, Tais ad Zimmerma: Sectio 2.3, Exercises 16(a,d),18. Sectio 2.4, Exercises 13, 14. Sectio
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationCALCULATION OF FIBONACCI VECTORS
CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationAnalysis of Algorithms. Introduction. Contents
Itroductio The focus of this module is mathematical aspects of algorithms. Our mai focus is aalysis of algorithms, which meas evaluatig efficiecy of algorithms by aalytical ad mathematical methods. We
More informationChapter 3. Strong convergence. 3.1 Definition of almost sure convergence
Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationTEACHER CERTIFICATION STUDY GUIDE
COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra
More informationChapter 6 Principles of Data Reduction
Chapter 6 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 0 Chapter 6 Priciples of Data Reductio Sectio 6. Itroductio Goal: To summarize or reduce the data X, X,, X to get iformatio about a
More informationSeries III. Chapter Alternating Series
Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with
More informationOnce we have a sequence of numbers, the next thing to do is to sum them up. Given a sequence (a n ) n=1
. Ifiite Series Oce we have a sequece of umbers, the ext thig to do is to sum them up. Give a sequece a be a sequece: ca we give a sesible meaig to the followig expressio? a = a a a a While summig ifiitely
More informationP.3 Polynomials and Special products
Precalc Fall 2016 Sectios P.3, 1.2, 1.3, P.4, 1.4, P.2 (radicals/ratioal expoets), 1.5, 1.6, 1.7, 1.8, 1.1, 2.1, 2.2 I Polyomial defiitio (p. 28) a x + a x +... + a x + a x 1 1 0 1 1 0 a x + a x +... +
More informationIn algebra one spends much time finding common denominators and thus simplifying rational expressions. For example:
74 The Method of Partial Fractios I algebra oe speds much time fidig commo deomiators ad thus simplifyig ratioal epressios For eample: + + + 6 5 + = + = = + + + + + ( )( ) 5 It may the seem odd to be watig
More informationMa 530 Infinite Series I
Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li
More informationEcon 325/327 Notes on Sample Mean, Sample Proportion, Central Limit Theorem, Chisquare Distribution, Student s t distribution 1.
Eco 325/327 Notes o Sample Mea, Sample Proportio, Cetral Limit Theorem, Chisquare Distributio, Studet s t distributio 1 Sample Mea By Hiro Kasahara We cosider a radom sample from a populatio. Defiitio
More informationThe standard deviation of the mean
Physics 6C Fall 20 The stadard deviatio of the mea These otes provide some clarificatio o the distictio betwee the stadard deviatio ad the stadard deviatio of the mea.. The sample mea ad variace Cosider
More information11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.
11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although
More informationFundamental Theorem of Algebra. Yvonne Lai March 2010
Fudametal Theorem of Algebra Yvoe Lai March 010 We prove the Fudametal Theorem of Algebra: Fudametal Theorem of Algebra. Let f be a ocostat polyomial with real coefficiets. The f has at least oe complex
More informationINFINITE SEQUENCES AND SERIES
11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS
More informationSubject: Differential Equations & Mathematical ModelingIII
Power Series Solutios of Differetial Equatios about Sigular poits Subject: Differetial Equatios & Mathematical ModeligIII Lesso: Power series solutios of differetial equatios about Sigular poits Lesso
More informationModel Theory 2016, Exercises, Second batch, covering Weeks 57, with Solutions
Model Theory 2016, Exercises, Secod batch, coverig Weeks 57, with Solutios 3 Exercises from the Notes Exercise 7.6. Show that if T is a theory i a coutable laguage L, haso fiite model, ad is ℵ 0 categorical,
More informationPRACTICE FINAL/STUDY GUIDE SOLUTIONS
Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)
More informationMathematics 170B Selected HW Solutions.
Mathematics 17B Selected HW Solutios. F 4. Suppose X is B(,p). (a)fidthemometgeeratigfuctiom (s)of(x p)/ p(1 p). Write q = 1 p. The MGF of X is (pe s + q), sice X ca be writte as the sum of idepedet Beroulli
More informationLecture 4 The Simple Random Walk
Lecture 4: The Simple Radom Walk 1 of 9 Course: M36K Itro to Stochastic Processes Term: Fall 014 Istructor: Gorda Zitkovic Lecture 4 The Simple Radom Walk We have defied ad costructed a radom walk {X }
More informationProbability, Expectation Value and Uncertainty
Chapter 1 Probability, Expectatio Value ad Ucertaity We have see that the physically observable properties of a quatum system are represeted by Hermitea operators (also referred to as observables ) such
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 19 11/17/2008 LAWS OF LARGE NUMBERS II THE STRONG LAW OF LARGE NUMBERS
MASSACHUSTTS INSTITUT OF TCHNOLOGY 6.436J/5.085J Fall 2008 Lecture 9 /7/2008 LAWS OF LARG NUMBRS II Cotets. The strog law of large umbers 2. The Cheroff boud TH STRONG LAW OF LARG NUMBRS While the weak
More informationSolutions to Final Exam Review Problems
. Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the
More informationARIMA Models. Dan Saunders. y t = φy t 1 + ɛ t
ARIMA Models Da Sauders I will discuss models with a depedet variable y t, a potetially edogeous error term ɛ t, ad a exogeous error term η t, each with a subscript t deotig time. With just these three
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationFIXED POINTS OF nvalued MULTIMAPS OF THE CIRCLE
FIXED POINTS OF VALUED MULTIMAPS OF THE CIRCLE Robert F. Brow Departmet of Mathematics Uiversity of Califoria Los Ageles, CA 900951555 email: rfb@math.ucla.edu November 15, 2005 Abstract A multifuctio
More informationLecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting
Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would
More informationTrue Nature of Potential Energy of a Hydrogen Atom
True Nature of Potetial Eergy of a Hydroge Atom Koshu Suto Key words: Bohr Radius, Potetial Eergy, Rest Mass Eergy, Classical Electro Radius PACS codes: 365Sq, 365w, 33+p Abstract I cosiderig the potetial
More informationLecture 9: Hierarchy Theorems
IAS/PCMI Summer Sessio 2000 Clay Mathematics Udergraduate Program Basic Course o Computatioal Complexity Lecture 9: Hierarchy Theorems David Mix Barrigto ad Alexis Maciel July 27, 2000 Most of this lecture
More informationMath 475, Problem Set #12: Answers
Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe
More informationSolutions to HW Assignment 1
Solutios to HW: 1 Course: Theory of Probability II Page: 1 of 6 Uiversity of Texas at Austi Solutios to HW Assigmet 1 Problem 1.1. Let Ω, F, {F } 0, P) be a filtered probability space ad T a stoppig time.
More informationMatsubaraGreen s Functions
MatsubaraGree s Fuctios Time Orderig : Cosider the followig operator If H = H the we ca trivially factorise this as, E(s = e s(h+ E(s = e sh e s I geeral this is ot true. However for practical applicatio
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationBinomial Distribution
0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 1 2 3 4 5 6 7 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Overview Example: coi tossed three times Defiitio Formula Recall that a r.v. is discrete if there are either a fiite umber of possible
More informationREVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.
REVIEW, MATH 00. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a)
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More informationw (1) ˆx w (1) x (1) /ρ and w (2) ˆx w (2) x (2) /ρ.
2 5. Weighted umber of late jobs 5.1. Release dates ad due dates: maximimizig the weight of otime jobs Oce we add release dates, miimizig the umber of late jobs becomes a sigificatly harder problem. For
More informationA Proof of Birkhoff s Ergodic Theorem
A Proof of Birkhoff s Ergodic Theorem Joseph Hora September 2, 205 Itroductio I Fall 203, I was learig the basics of ergodic theory, ad I came across this theorem. Oe of my supervisors, Athoy Quas, showed
More informationTest One (Answer Key)
CS395/Ma395 (Sprig 2005) Test Oe Name: Page 1 Test Oe (Aswer Key) CS395/Ma395: Aalysis of Algorithms This is a closed book, closed otes, 70 miute examiatio. It is worth 100 poits. There are twelve (12)
More informationA Simplified Binet Formula for kgeneralized Fibonacci Numbers
A Simplified Biet Formula for kgeeralized Fiboacci Numbers Gregory P. B. Dresde Departmet of Mathematics Washigto ad Lee Uiversity Lexigto, VA 440 dresdeg@wlu.edu Zhaohui Du Shaghai, Chia zhao.hui.du@gmail.com
More informationOutput Analysis and RunLength Control
IEOR E4703: Mote Carlo Simulatio Columbia Uiversity c 2017 by Marti Haugh Output Aalysis ad RuLegth Cotrol I these otes we describe how the Cetral Limit Theorem ca be used to costruct approximate (1 α%
More informationECE 901 Lecture 12: Complexity Regularization and the Squared Loss
ECE 90 Lecture : Complexity Regularizatio ad the Squared Loss R. Nowak 5/7/009 I the previous lectures we made use of the Cheroff/Hoeffdig bouds for our aalysis of classifier errors. Hoeffdig s iequality
More informationSequences of Definite Integrals, Factorials and Double Factorials
47 6 Joural of Iteger Sequeces, Vol. 8 (5), Article 5.4.6 Sequeces of Defiite Itegrals, Factorials ad Double Factorials Thierry DaaPicard Departmet of Applied Mathematics Jerusalem College of Techology
More informationProbability theory and mathematical statistics:
N.I. Lobachevsky State Uiversity of Nizhi Novgorod Probability theory ad mathematical statistics: Law of Total Probability. Associate Professor A.V. Zorie Law of Total Probability. 1 / 14 Theorem Let H
More informationDIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS
DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS VERNER E. HOGGATT, JR. Sa Jose State Uiversity, Sa Jose, Califoria 95192 ad CALVIN T. LONG Washigto State Uiversity, Pullma, Washigto 99163
More information6. Uniform distribution mod 1
6. Uiform distributio mod 1 6.1 Uiform distributio ad Weyl s criterio Let x be a seuece of real umbers. We may decompose x as the sum of its iteger part [x ] = sup{m Z m x } (i.e. the largest iteger which
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationMaximum and Minimum Values
Sec 4.1 Maimum ad Miimum Values A. Absolute Maimum or Miimum / Etreme Values A fuctio Similarly, f has a Absolute Maimum at c if c f f has a Absolute Miimum at c if c f f for every poit i the domai. f
More informationMath 140A Elementary Analysis Homework Questions 1
Math 14A Elemetary Aalysis Homewor Questios 1 1 Itroductio 1.1 The Set N of Natural Numbers 1 Prove that 1 2 2 2 2 1 ( 1(2 1 for all atural umbers. 2 Prove that 3 11 (8 5 4 2 for all N. 4 (a Guess a formula
More informationDefinition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.
4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad
More informationA PROOF OF THE TWIN PRIME CONJECTURE AND OTHER POSSIBLE APPLICATIONS
A PROOF OF THE TWI PRIME COJECTURE AD OTHER POSSIBLE APPLICATIOS by PAUL S. BRUCKMA 38 Frot Street, #3 aaimo, BC V9R B8 (Caada) email : pbruckma@hotmail.com ABSTRACT : A elemetary proof of the Twi Prime
More information11.6 Absolute Convergence and the Ratio and Root Tests
.6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does
More informationP1 Chapter 8 :: Binomial Expansion
P Chapter 8 :: Biomial Expasio jfrost@tiffi.kigsto.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 6 th August 7 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework
More informationREVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.
the Further Mathematics etwork wwwfmetworkorguk V 07 The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More informationNotes on iteration and Newton s method. Iteration
Notes o iteratio ad Newto s method Iteratio Iteratio meas doig somethig over ad over. I our cotet, a iteratio is a sequece of umbers, vectors, fuctios, etc. geerated by a iteratio rule of the type 1 f
More informationMarkov Decision Processes
Markov Decisio Processes Defiitios; Statioary policies; Value improvemet algorithm, Policy improvemet algorithm, ad liear programmig for discouted cost ad average cost criteria. Markov Decisio Processes
More informationMath 609/597: Cryptography 1
Math 609/597: Cryptography 1 The SolovayStrasse Primality Test 12 October, 1993 Burt Roseberg Revised: 6 October, 2000 1 Itroductio We describe the SolovayStrasse primality test. There is quite a bit
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 5: SINGULARITIES.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 5: SINGULARITIES. ANDREW SALCH 1. The Jacobia criterio for osigularity. You have probably oticed by ow that some poits o varieties are smooth i a sese somethig
More informationPROBLEM SET I (Suggested Solutions)
Eco3Fall3 PROBLE SET I (Suggested Solutios). a) Cosider the followig: x x = x The quadratic form = T x x is the required oe i matrix form. Similarly, for the followig parts: x 5 b) x = = x c) x x x x
More informationMA131  Analysis 1. Workbook 2 Sequences I
MA3  Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationMA131  Analysis 1. Workbook 9 Series III
MA3  Aalysis Workbook 9 Series III Autum 004 Cotets 4.4 Series with Positive ad Negative Terms.............. 4.5 Alteratig Series.......................... 4.6 Geeral Series.............................
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationLinear chord diagrams with long chords
Liear chord diagrams with log chords Everett Sulliva Departmet of Mathematics Dartmouth College Haover New Hampshire, U.S.A. everett..sulliva@dartmouth.edu Submitted: Feb 7, 2017; Accepted: Oct 7, 2017;
More information14.1 Understanding Rational Exponents and Radicals
Name Class Date 14.1 Uderstadig Ratioal Expoets ad Radicals Essetial Questio: How are radicals ad ratioal expoets related? Resource Locker Explore 1 Uderstadig Iteger Expoets Recall that powers like are
More informationMark Lundstrom Spring SOLUTIONS: ECE 305 Homework: Week 5. Mark Lundstrom Purdue University
Mark udstrom Sprig 2015 SOUTIONS: ECE 305 Homework: Week 5 Mark udstrom Purdue Uiversity The followig problems cocer the Miority Carrier Diffusio Equatio (MCDE) for electros: Δ t = D Δ + G For all the
More informationSection 1 of Unit 03 (Pure Mathematics 3) Algebra
Sectio 1 of Uit 0 (Pure Mathematics ) Algebra Recommeded Prior Kowledge Studets should have studied the algebraic techiques i Pure Mathematics 1. Cotet This Sectio should be studied early i the course
More informationQBINOMIALS AND THE GREATEST COMMON DIVISOR. Keith R. Slavin 8474 SW Chevy Place, Beaverton, Oregon 97008, USA.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 2008, #A05 QBINOMIALS AND THE GREATEST COMMON DIVISOR Keith R. Slavi 8474 SW Chevy Place, Beaverto, Orego 97008, USA slavi@dsloly.et Received:
More informationSolutions to Final Exam
Solutios to Fial Exam 1. Three married couples are seated together at the couter at Moty s Blue Plate Dier, occupyig six cosecutive seats. How may arragemets are there with o wife sittig ext to her ow
More informationUNIT #5. Lesson #2 Arithmetic and Geometric Sequences. Lesson #3 Summation Notation. Lesson #4 Arithmetic Series. Lesson #5 Geometric Series
UNIT #5 SEQUENCES AND SERIES Lesso # Sequeces Lesso # Arithmetic ad Geometric Sequeces Lesso #3 Summatio Notatio Lesso #4 Arithmetic Series Lesso #5 Geometric Series Lesso #6 Mortgage Paymets COMMON CORE
More information