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### Homework5

Course: MATH 548, Spring 2012
School: UNC
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UNC - MATH - 548
W8(7.37)(a) Find a recurrence for n 2. Hint: Without the special restriction,for strings of length 2 or more there would (normally) be nine possibilities forthe last 2 entries: 00, 01, , 22. Here the endings 00 and 11 are beingdisallowed. This leaves
UNC - MATH - 548
W9(7.48) For each of Parts (a) and (g), do only Steps (i) and (ii) below: Findthe generating function H(x) for the sequence determined by the recurrence andthe initial conditions, as on p. 235. (We will not discuss the partial fractionstechnique, whic
UNC - MATH - 548
W 10aBijective Proof Requirements: For n 0, let A n and B n be sets ofcombinatorial objects. Set an := |An| and bn := |Bn|. If you are asked toprovide a bijective proof that an = bn for n 0, you are to do the followingfor general n 0: Define a functio
UNC - MATH - 548
UNC - MATH - 548
UNC - MATH - 548
UNC - MATH - 548
UNC - MATH - 548
UNC - MATH - 548
UNC - MATH - 548
W 10b/11a( V 3 41/2) Consider the bipartite graph G = (X,Y) in whichX = cfw_1,2,3,8,9, Y = cfw_a,b,c,h,i, and has many edges. In particular,the edge set contains both the matchingM = cfw_(1,a), (3,c), (5,i), (6,f), (7,d), (8,e), (9,h) and the matching
UNC - MATH - 548
C h a p t er s 11, 12, 13:G r a p h T h eor yAt the beginning of the semester it was announced that this course would beprimarily concerned with the oldest and most renowned part of combinatorics,the enumeration of combinatorial objects. But in additi
University of Texas-Tyler - PSYC - 4321
COURSE OUTLINEHISTORY OF PSYCHOLOGY, HERGENHAHN (2001)Chapter 1 - IntroductionA. Why study history?B. The data of history.C. Why science?D. PopperE. KuhnChapter 2 - The Greeks!A. Early GreeksB. Thales (fl 585 B.C.)C. Anaximander (fl 560 B.C.)D
University of Texas-Tyler - PSYC - 4321
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Waterloo - STAT - 340
Result: If f() is a monotone function on (0, 1) then Cov ( f (U ), f (1 U ) ) &lt; 0 for U (0,1)Proof: Suppose U1 , U 2 iid U[0,1].Since f() is monotone f (U1 ) f (U 2 ) f (1 U1 ) f (1 U 2 ) &lt; 0 .Thus f (U1 ) f (1 U1 ) + f (U 2 ) f (1 U 2 ) &lt; f (U1 ) f (1
Waterloo - STAT - 340
DISCRETE UNIFORMRANDOM NUMBERGENERATION (Ross, Ch 3,McLeish, Ch 3.1-3.2)e.g. Stat 340 Class Random NumberGenerator:Write down five digits randomlyselected from cfw_0,1,2,9.e.g.12345Some elementary tests:N=number of digits equalto 7 is binomial
Waterloo - STAT - 340
Generating non-uniformGeneratingrandom variates (RossChapters 4-5) Start with a more general version ofthe inverse transformA General Definition of F-1 Suppose F(x) is any c.d.f. (in red below)Some properties of F-1SomeDefine F 1 ( y) = mincfw_x
Waterloo - STAT - 340
Continuous Distributions:Chapter 523The Cauchy distributiuonCauchy c.d.f. F ( x ) =11xa)+ arctan(2binverse c.d.f. X = F 1 (U ) = a + b tancfw_ (U p .d . f . f ( x ) =1) = a b / tan( U )2b (b 2 + ( x a ) 2 )24What if daily stock returns
Waterloo - STAT - 340
Discrete Event Simulation e.g. Bankqueue ( Ross, Chapter 6)Interested in required number oftellers, throughput rate etc.VariablesTime variable tState variables: these tellthe state of the system attime t: e.g. number of tellersavailable, number o
Waterloo - STAT - 340
Number in system versus time:one simulation, exponentialservice rate =60average time in system 0.20435average number in system 5.0243Varying the service timesCompare exponential service times, constantservice times and uniform service times, allwi
Waterloo - STAT - 340
Example. Simulating inventorymanagement.Basic Problem:when to order (ormanufacture) andhow much.Associated costs:order cost=c(Q)=k+cQk=cost of setupc=cost per item orderedQ=number of itemsorderedCost of holdinginventory (e.g.capital,space, i
Waterloo - STAT - 340
CHAPTER 7The BootstrapBootstrapping and meansquared errorSuppose an estimator is a function of n independentobservationsg ( X 1 ,., X n )This is an estimator of some parameterwhich depends on the c.d.f. F(x) of the random variables( X 1 ,., X n )
Waterloo - STAT - 340
Chapter 8. Variance Reduction. Review: The Call Option Price is theexpected present value (i.e. discountedto present) of future returnsEcfw_e rT max(ST K ,0)where expectation is under the risk neutralmeasure in the Black-Scholes model.1Crude simul
Waterloo - STAT - 340
Stratified SampleConsider choosing one point in interval [0,a] andanother in the interval [a,1]aU1 is in [0, a] and a + (1 a)U 2 is in [a,1].Use a weighted average of the function at these two points1to estimate f (u )du. What should the weights be?
Waterloo - STAT - 340
Comparing estimators for CallOption Pricing Examplescript9 uses a variety of estimators:Best efficiency (74) for stratified-antithetic0.37[ f (.47 + .37U ) + f (.84 .37U )]2.16+[ f (.84 + .16U ) + f (1 .16U )]2This is stratified into [.47,.84]
Waterloo - STAT - 340
Simulation and Bootstrapping for Teaching StatisticsTim C. HesterbergMathSoft, Inc., 1700 Westlake Ave. N., Suite 500Seattle, WA 98109-3044, U.S.A.TimH@statsci.com, http:/www.statsci.com/HesterbergAbstractSome key ideas in statistics and probability
Waterloo - STAT - 340
CHAPTER3Basic Monte Carlo Methodsas an example thesimplewish toC onsiderEuropean call optionfollowing very price \$22problem. Wefunctionprice awith exerciseand payoffV(ST ) = (ST 22)+ . Assume for the present that the interest rate is 0 percenta
Waterloo - STAT - 340
CHAPTER4Variance Reduction TechniquesINTRODUCTIONIn this chapter we discuss techniques for improving on the speed and efciency of a simulation, usually called variance reduction techniques.Much of the simulation literature concerns discrete event sim
Waterloo - STAT - 340
CHAPTER5Simulating the Value of OptionsASIAN OPTIONSAn Asian option, at expiration T, has value determined not by the closing price of the underlying asset as for a European option, but on an average price of the asset over an interval. For example a
Waterloo - STAT - 340
CHAPTER7Estimation and CalibrationINTRODUCTIONVirtually all models have parameters that must be specied in order for themodels to be completely described. Statistical estimation can be used forsome or all of these parameters under two important cond
Waterloo - STAT - 340
SOLUTIONS FOR REVIEW PROBLEMS1.The joint probability density function isf (x, y ) = 2e(x+2y) , 0 &lt; x &lt; and 0 &lt; y &lt; .ThereforeP (XZZ&lt; Y)==Z0=Zcfw_y01.3f (x, y )dxdycfw_(x,y );x&lt;y f (x, y )dxdy2. In this case n = 30 and the observed nu
Waterloo - STAT - 340
02/01/2007STATISTICS 340/CS 437COMPUTER SIMULATIONInstructor:Don McLeishMC 6138TEXT:Simulation, 2nd 4th edition, Sheldon M.Ross. Academic Press(on reserve in libraryQA273.R82)Various notes andtransparencies willbe posted on webpage including
Waterloo - STAT - 340
Covariance: DefinitionIf E(X)=x and E(Y)=y, thenCov(X,Y)= E[(X-x)(Y-y)]Another formula for covariance:Cov(X,Y)=E(XY)-xy51Laws of Covariancefor constants a,b d.Cov(X,X)=var(X)Cov(X,Y)=Cov(Y,X)Cov(aX+b,Y)=a Cov(X,Y)|Cov(X,Y)|SD(X)SD(Y)If X,Y ind
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