ST 501 Course: Fundamentals of Statistical Inference I. Sujit K. Ghosh.

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1 ST 501 Couse: Fundamentals of Statistical Infeence I Sujit K. Ghosh sujit.ghosh@ncsu.edu Pesented at: 2229 SAS Hall, Depatment of Statistics, NC State Univesity Sujit K. Ghosh ST 501 Couse Slides 1 Textbook: Rice A. J. (2007). Mathematical Statistics and Data Analysis, 3d Edition Outline Chap 1: Pobability Chap 2: Random Vaiables Chap 3: Joint Distibutions Chap 4: Expected Values Chap 5: Limit Theoems Last updated on: August 19, 2018 Sujit K. Ghosh ST 501 Couse Slides 2

2 Why We Need Pobability Theoy? Conside pedicting closing pices of S&P 500 seies fo the next 10 tading days Suppose we ae only inteested in pedicting whethe the pice will go up (+) o down ( ) fom the last tading day In othe wods, we simply want to pedict the sign of log-etun of the pices (i.e. if log. t =log(p t /P t 1 ) we want to pedict sign(log. t )) Suppose fo the last 5 tading days the signs wee +,, +, +, What can you say about the numbe of + signs fo the next 10 tading days? How much confidence do you have in answeing the above question? What assumptions (if any) have you made to answe the above questions? Clealy, nothing can be answeed with cetainty (and we can only povide pobability) Sujit K. Ghosh ST 501 Couse Slides 3 Let s conside a little moe abstact expeiment with an un containing N balls of two colos: RED (R) and WHITE (W) [Go Wolfpack!] Ou goal is to guess how many RED (R) balls ae in the un Suppose we daw n balls with eplacement and ( n) tuns out to be RED How many RED balls does the un contain? Relating to pevious scenaio, if we designate RED ball as negative sign, we have N =15, n =5and =2 If R denotes the unknown numbe of RED balls in the un, a simple minded stategy is to equate the two popotions: R N = n and guess ˆR = n N How accuate is the above answe? Will you answe change if the n balls wee dawn without eplacement? Will you answe change is you knew the sequence of the colos of the balls? Sujit K. Ghosh ST 501 Couse Slides 4

3 Pobability Theoy The concept of pobability dates back to 16th centuy (gambling expeiments) Howeve, the axiomatic development of pobability emeged only in the last centuy Pobability theoy has been applied to a wide vaiety of events: Atmospheic sciences: weathe and climate pedictions Business and Finance: insuances, pices, yields, etc. Cosmology: e.g., likelihood of inflation (Gibbons-Hawking-Stewat measue) Defense: complex eliability systems Electonics: compute opeating system Food sciences: agicultual poducts... Zoology: species abundance Sujit K. Ghosh ST 501 Couse Slides 5 The basic notions of pobability is based on computing long tem fequency of events Sample Space: The set of all possible outcomes of an expeiment; and is typically denoted by Ω (i) Finite sample space: Ω={RR, RW, W R, W W } if two balls ae dawn fom an un containing 4 RED balls and 5 WHITE balls (ii) Countably infinite sample space: Ω={0, 1, 2,...} if we count numbe of defectives in a lage consignment (iii) Uncountable sample space: Ω=(0, ) if we measue time until a ca battey goes dead Events: A subset of outcomes of the sample space; and is typically denoted by E (i) E = {RW, W R, W W } Ω: dawing at most one RED ball (ii) E = {3, 4,...} Ω: at least 3 defectives (iii) E =(0, 20.25] Ω: battey dies within months Sujit K. Ghosh ST 501 Couse Slides 6

4 Set theoetic notions of events (sets): Empty set: ; a set with no elements Intesection of events: E 1 E 2 contains elements of both events E 1 and E 2. Moe geneally n i=1e i = E 1 E 2... E n Union of events: E 1 E 2 contains elements of eithe E 1 and/o E 2.Moe geneally n i=1e i = E 1 E 2... E n Complement of an event: E c contains all elements of Ω not in E Disjoint events: E 1 E 2 = two events have no common elements Popeties of set opeations: Commutative law: E 1 E 2 = E 2 E 1 and E 1 E 2 = E 2 E 1 Associative law: (E 1 E 2 ) E 3 = E 1 (E 2 E 3 ) and (E 1 E 2 ) E 3 = E 1 (E 2 E 3 ) Distibutive law: (E 1 E 2 ) E 3 =(E 1 E 2 ) (E 2 E 3 ) Sujit K. Ghosh ST 501 Couse Slides 7 A pobability measue is a eal-valued function P defined on the subsets of sample space (σ-field of Ω) satisfying the following axioms: (i) P(Ω) = 1 (ii) P(E) 0 fo any E Ω (iii) P( i=1e i )= i=1 P(E i) fo any sequence of mutually disjoint events E 1,E 2,...(i.e., E i E j = fo any i j) Popeties of pobability measues: (i) P(E c )=1 P(E) and P( ) =0 (ii) P(E 1 ) P(E 2 ) if E 1 E 2 Ω (iii) P(E 1 E 2 )=P(E 1 )+P(E 2 ) P(E 1 E 2 ) (iv) P( n i=1e i )= n i=1 P(E i Fi 1) c whee F i 1 = i 1 j=0 E j fo i =1, 2,... with E 0 =. It thus follows that P( n i=1e i ) n i=1 P(E i) (v) P( n i=1e i ) n i=1 P(E i) (n 1) Sujit K. Ghosh ST 501 Couse Slides 8

5 Pobability measue on a finite sample space: Ω={w 1,w 2,...,w m } P(E) = #E #Ω = numbe of elements in E numbe of elements in Ω, whee E Ω An Example: Suppose the NCSU telephone numbe must appea as 513-????. If all sequences of emaining fou digits ae equally likely, what is the pobability that a andomly selected NCSU phone numbe contains seven distinct digits? Hee Ω={fou possible digits among 0, 1, 2,...,9} and hence we have #Ω = = Next let E = {fou distinct digits among 0, 2, 4, 6, 7, 8, 9} and hence #E = b/c the fist digit could be any of seven emaining digits; second could be any of the emaining six, and so on. Hence, we get P(E) = = Thus, we need systematic methods to count without enumeating all possible elements of an event Sujit K. Ghosh ST 501 Couse Slides 9 Some geneal ules of counting using combinatoics If we have successive selections (decisions) with exactly n k choices at the k-th step, a total of k=1 n k = n 1 n 2 n diffeent esults ae possible. If we want to daw a sample of size fom a population of n objects, thee exists (i) n diffeent samples, when sampled with eplacement (ii) (n) = n(n 1)...(n +1)samples, when sampled without eplacement ) = (n) unodeed samples (whee! =( 1)...2.1) (iii) ( n! Examples: Suppose we daw a andom sample of size fom a population of n distinguishable elements: (a) P[no epitition in ou sample] = (n) n (b) P[a fixed element is included] =1 (n 1) n (c) P[a fixed element is included] =1 (n 1) (n) eplacement if sampling with eplacement if sampling with eplacement = n if sampling without Sujit K. Ghosh ST 501 Couse Slides 10

6 The numbe of ways n objects can be gouped into classes with n k in the k-th class, such that n = k=1 n ( ) n k is n 1 n 2 n = n! n 1! n 2! n! ( n The tem n 1 n 2 n ) is called multinomial coefficient, because it can be shown that (x 1 + x x ) n = ( n n 1 n 2 n ) x n 1 1 x n 2 2 x n whee the sum is ove all nonnegative integes n 1,n 2,...,n such that n 1 + n n = n How many tems ae thee in the above summation? The numbe of distinct solutions of n 1 + n n = n ( ) +n 1 is 1 The above expession is also the numbe of ways n indistinguishable objects can be placed into cells P[none of the cells emain empty] = ( ) ( n 1 1 / +n 1 ) 1 This is equivalent to numbe of distinct solutions, such that n k 1 k Sujit K. Ghosh ST 501 Couse Slides Fo any x ( ) x R, = x(x 1) (x +1) (! ) ( ) a 2. Fo any a>0, =( 1) a Fo any x ( ) ( x R, = x+1 1) + ( x 4. Fo any a, b R, n =0 ( a )( b n whee =0, 1, 2,... [( ) ] x =0if <0 +1),fo =0, 1, 2,... ) ( = a+b ) fo n =1, 2, (1 + t) a = =0 ( a ) t fo any a R and t ( 1, 1) 6. Using a = 1 and integating above: log(1 + t) = =0 n ( 1) t Stiling s fomula: n! 2πn n+ 1 2 e n [a n b n lim n a n bn =1 8. Γ(x) 2πx x 1 2 e x whee Γ(x) = 0 t x 1 e t dt is the Gamma function and Γ(x +1)=xΓ(x) fo x>0. In paticula, Γ(n +1)=n! Online Combination and Pemutation calculato: Click ] Sujit K. Ghosh ST 501 Couse Slides 12

7 Example (sign of log-etuns): Conside again dawing n balls fom an un containing N balls with two colos: RED (R) and WHITE (W) Suppose out n balls ae RED. How many RED balls ae thee in the un? Let R = numbe of (unknown) RED balls and so W = N R WHITE balls. (i) Sampling with eplacement: P[ RED balls] = ( n ) R (N R) n N n = ( n R )( N Find R that maximizes the above pobability (ii) Sampling without eplacement: P[ RED balls] = ( n Find R that maximizes the above pobability ) ( ) 1 R n N ) (R) (N R) n (N) n = (R )( N R n ) ( N n) Show that in eithe cases the most likely value of R is given by the ealie simple minded value of ˆR = n N obtained by equating R N = n Notice that the above answe emains the same even if you knew a given ode of the sequence of RED and WHITE balls Sujit K. Ghosh ST 501 Couse Slides 13 sampling with eplacement sampling without eplacement pob.w pob.wo R R Figue 1: Compaing two sampling schemes when N =15,n=5and =2 Sujit K. Ghosh ST 501 Couse Slides 14

8 R codes: N=15; n=5; =2 pob1=function(r){dbinom(,size=n,pob=r/n)} pob2=function(r){dhype(,m=r,n=n-r,k=n)} pa(mfow=c(1,2),cex=0.75) R=0:N; pob.w=pob1(r); pob.wo=pob2(r) plot(r, pob.w, type="b", main="sampling with eplacement") R.hat1=R[which.max(pob.w)] abline(v=r.hat1,col="ed") plot(r, pob.wo, type="b", main="sampling without eplacement") R.hat2=R[which.max(pob.wo)] abline(v=r.hat2,col="ed") What happens when N = 100,n=50and =20? You may use the above lines of codes simply eplacing the fist line with N=100; n=50; =20 Sujit K. Ghosh ST 501 Couse Slides 15 sampling with eplacement sampling without eplacement pob.w pob.wo R R Figue 2: Compaing two sampling schemes when N =100,n=50and =20 Sujit K. Ghosh ST 501 Couse Slides 16

9 Conditional Pobability If D and E ae two events (subsets of the sample space Ω), then the pobability of D given E is P[D E] = P(D E), povided P(E) 0 P(E) Multiplication ule: P(D E) =P(E)P(D E) =P(D)P(E D) Moe geneally, P( n i=1e i )=P(E 1 ) n i=2 P(E i i 1 j=1 E j) If we assume the Makov popety P(E i i 1 j=1 E j)=p(e i E i 1 ) fo i =2, 3,...then P( n i=1e i )=P(E 1 ) n i=2 P(E i E i 1 ) Given any sequence of mutually disjoint events E i s patitioning the sample space Ω= n i=1e i, (whee E i E j = fo i j) wehavethelaw of Total Pobability P(E) = n i=1 P(E E i)= n i=1 P(E E i)p(e i ) The law leads to the Bayes ule: P(E j E) = P(E E j)p(e j ) ni=1 P(E E i )P(E i ) Often the above conditional pobability leads to supising esults when the events E j s ae vey ae Sujit K. Ghosh ST 501 Couse Slides 17 An Example: Suppose that a andomly chosen woman has beast cance is 0.1% and suppose a adiology test is 99.9% accuate in detecting beast cance but povides false positive test 1% of the times. What is the pobability that a woman has beast cance given a positive test? Let E 1 = {woman has beast cance} and E = {adilogy test is positive}. Notice that E 2 = {woman doesn t have beast cance} = E1 c. We ae given, P(E 1 )=0.001 = 1 P(E 2 ). Also we ae told, P(E E 1 )=0.999 and P(E E 2 )=0.01. Thus, by Bayes ule it follows: P(E 1 E) = = ( ) 0.01 Most folks (even those in medical pofessions) get supised by this esult that thee is moe than 90% chance the woman may not have a beast cance given a highly accuate positive test What is the pobability of beast cance given thee successive positive tests? It is almost cetain ( 99.9%)!! Sujit K. Ghosh ST 501 Couse Slides 18

10 The concept of Independent events (not same as disjoint!) Two events E 1 and E 2 ae said to be pobabilistically independent if P(E 1 E 2 )=P(E 1 ) o equivalently, P(E 2 E 1 )=P(E 2 ) Moe succinctly, E 1 and E 2 ae independent if P(E 1 E 2 )=P(E 1 )P(E 2 ) Moe geneally, a collection of n events, E 1,E 2,...,E n ae mutually independent if P( m j=1e ij )= m j=1 P(E i j ) fo all subcollection E i1,e i2,...,e im fo m =2,...,n. Thus, it will equie 2 n n 1 veifications with n events! It is possible that 3 events E 1,E 2,E 3 ae paiwise independent but E 1,E 2,E 3 ae not mutually independent If two events E 1 and E 2 ae disjoint and if P(E 1 )P(E 2 ) > 0, then E 1 and E 2 can t be independent Pactice Poblems fom Chapte 1: 1, 8, 15, 17, 20, 34, 38, 47, 54, 61, 63, 70 & 77 Sujit K. Ghosh ST 501 Couse Slides 19