Probability and Statistics I

Transcription

1 Constantn Rothkopf 1 Probablty and Statstcs I FIAS Summer School on Theoretcal Neuroscence and Complex Systems Frankfurt, August 2-24, 2008

2 from Kersten and Yulle () Constantn Rothkopf 2

3 Constantn Rothkopf 3

4 Constantn Rothkopf 4 Part 1: Basc Probablty sample space and events basc probablty laws condtonal probablty ndependence condtonal ndependence Bayes rule

5 Constantn Rothkopf 5 Sample Space & Events The sample space conssts of all possble outcomes of an experment: Flp a con (dscrete, bnary) S={H,T} Roll a de (dscrete) S={1,2,3,4,5,6} Measure the lfetme of an anmal (contnuous) S=[0, ) One can defne an event as a set of outcomes: Flp a con and observe heads E={H} Roll a de and observe an even number E={2,4,6} Measure the lfetme of an anmal and observe a lfetme smaller than 1 day E=[0,24h)

6 Probablty defned on events Probablty Axoms: 1. probablty of event A: A) s between zero and one: 0 A) 1 2. probablty of certan event S s one: S)=1 3. f events A and B are mutually exclusve, then A v B) = A)+B) ( A or B occurred ) One nterpretaton of probablty: relatve frequency. Repeat the experment N tmes and count the number of tmes that event A occurred N A : A) = lm N The other nterpretaton s that of a degree of uncertanty. (Some experments can only be done once!) Example: de roll, S={1,2,3,4,5,6}. D=1) = D=2) = = 1/6 far de shorthand notaton: D=1) = 1) Constantn Rothkopf 6 N N A

7 Constantn Rothkopf 7 Example: rollng two dce The sample space consst of all possble outcomes of the experment Probabltes of outcomes: from Ross (1997)

8 Constantn Rothkopf 8 Example: rollng two dce There are N=36 dfferent outcomes. Assume all outcomes o are dstnct and have the same probablty (far dce): o) = 1/N = 1/36. Queston: consder the event A: sum s less than 6, that s composed of N A =10 outcomes. What s A s probablty A)? Answer: (usng the thrd axom) N ( ) = A P A o) = N A o) = = 10 / 36 = N o A

9 Constantn Rothkopf 9 Venn dagrams fgure from Ptman, 1999 rectangle corresponds to unversal set, other events represented as ellpses, wth ther probablty gven by ther area

10 Constantn Rothkopf 10 Elementary Rules = not v = or ^ = and, A^B) = A,B) A) + A) = 1 A v B) = A) + B) A,B) A) = A,B) + A, B)

11 Constantn Rothkopf 11 Sum Rule Let A be a partton of all possble outcomes,.e. the A are mutually exclusve and ther unon s S. Then the probablty of any event B can be wrtten as: Illustraton wth Venn dagram: P ( B) = B, A ) B A

12 Constantn Rothkopf 12 Condtonal Probablty - Product Rule Idea: nteracton of events, how does knowledge that one event occurred change our belef that the other event also has occurred? Defnton: condtonal probablty ( probablty of A gven B ): A B) =A,B)/B) [or equvalently: A,B) = A B)B)] Product rule: A,B) = A B) B) Illustraton wth Venn dagram: A A^B B

13 Constantn Rothkopf 13 Chan Rule Repeatedly apply the product rule, whch follows drectly from the defnton of condtonal probablty. Product rule: A,B) = A B) B) Now apply ths to: A,B,C) = A,B C) C) = A B,C) B C) C) and n general ths leads to the Chan rule: X 1, X 2,, X N ) = X 1 )X 2 X 1 )X 3 X 2,X 1 ) X N X N-1,,X 1 ) = X N )X N-1 X N )X N-2 X N,X N-1 ) X 1 X N, X N-1,, X 2 )

14 Constantn Rothkopf 14 Condtonal Probablty example Consder the roll of two dce and the followng two events: Event A: D 1 = 2 Event B: D 1 +D 2 < 6 Queston: what s A B)? Answer: A,B)/B) = (3/36)/(10/36) = 3/10 = 0.3

15 Constantn Rothkopf 15 Independence of two Events Defnton: two events A,B are ndependent f A,B) = A)B). equvalently: A B) = A) or: B A) = B) Quz: are these two dce world events ndependent? A = sum s even, B = D 2 >3 A) = 0.5 B) = 0.5 Answer: Yes, snce A,B) = 9/36 = 0.25 = A)B)

16 Constantn Rothkopf 16 Condtonal Independence Defnton: two events A,B are condtonally ndependent gven a thrd event C f A,B C) = A C)B C). Note: A,B C) = A,B,C)/C) Illustraton wth Venn dagram: C A A^B^C B

17 Constantn Rothkopf 17 Condtonal Independence example Example: A: D 1 =2, B: D 2 >3, C: D 1 +D 2 =7 are A and B condtonally ndependent gven C? A,B C) = 1/6 A C) = 1/6 B C) = 1/2 Thus: A,B C) A C) B C) (no condtonal ndependence)

18 Bayes rule Bayes Theorem [Bayes, 1763]: P ( A B) = B A) A) B) Reverend Thomas Bayes Proof: We can decompose A,B) n two ways usng condtonal probabltes: A,B) = A B)B), but also A,B) = B A)A). Hence we have: A B)B) = B A)A), from whch Bayes rule follows mmedately. Constantn Rothkopf 18

19 Termnology Bayes Theorem: P ( A B) = B A) A) B) Note: lkelhood and pror probablty are often much easer to measure than the posteror probablty, so t makes sense to express the latter as a functon of the former. We can also express the evdence B) as a functon of lkelhood and pror usng the law of total probablty assumng a set of A formng a partton: Thus: posteror probablty lkelhood = B, A = ) pror probablty P ( B) B A ) A ) A B) = B A) A) B A ) evdence A Constantn Rothkopf 19 )

20 Constantn Rothkopf 20 Bayes rule example: blood test Consder a blood test to detect a dsease D: the test can be postve + or negatve -. For people havng the dsease, the test s postve n 95% of cases. For people wthout the dsease the test s postve n 2% of the cases. Let s assume we also know that 1% of the populaton has ths dsease. Queston: What s the probablty that a person wth a postve test result has the dsease? + D)=0.95, + D)=0.02, D)=0.01, D)=0.99. D +) = + D)D)/+) ; +)=+ D)D) + + D) D) = 0.95*0.01 / (0.95* *0.99) = 32% (lower than you may have expected)

21 Bayes rule n Percepton Example: consder the followng vson problem. You have to recognze an anmal, the possble events beng A (mouse, cat, dog, ), n an mage I from a possble set of mages. Applcaton of Bayes rule gves: what mages are generated by a mouse and how probable s each of them? how frequently do I encounter a mouse? A = mouse I = ) = I = A = mouse) A I = ) = mouse) s ths a mouse n the mage? how frequently do I encounter any possble mage? Constantn Rothkopf 21

22 Bayes rule n decodng Example: consder the followng decodng problem. You have to recognze a drecton of moton, the event beng S (dots movng to the left or to the rght), from a recordng of neuronal frng rate, where the event s that the frng rate R s 6oHz. Applcaton of Bayes rule gves: If the dots are movng left, how lkely s t to observe a rate of 60Hz? How lkely s a moton drecton to the left across trals? A = mouse I = ) S=left R=60Hz = I = A = mouse) A = (R=60Hz) I = ) R=60Hz S=left mouse) S=left What s the probablty that the dots are movng left, f we measure a rate of 60Hz? How lkely s t to observe a frng rate of 60Hz? Constantn Rothkopf 22

23 Constantn Rothkopf 23 Part 2: Random Varables defnton of random varables probablty mass functons probablty densty functon expected values, mean, and varance example dstrbutons: bnomal, exponental, Posson, Gaussan

24 Constantn Rothkopf 24 Random Varables So far, we have consdered events on sample spaces. But often we are nterested n functons of events, whch leads to the defnton: A random varable s a real valued functon defned on the sample space. Examples: X: the sum of two dce Y: the square of the sum of two dce Z: the number of rolls of two dce untl the total sum s larger then 10 Repeat the experment over and over agan: dstrbuton. The probablty of a result expresses the chance of that result happenng.

25 Constantn Rothkopf 25 From events to random varables The elementary probablty rules hold wth regard to random varables Probablty of an event E={x k } on the sample space S={x 1, x 2,, x,, x k, x N }: defne random varable X as the outcome and then x k ) = X=x k ) X=x Y=y) = X=x, Y=y ) / Y=y)

26 Constantn Rothkopf 26 Dscrete vs.. contnuous from Dayan & Abbott (2001)

27 Constantn Rothkopf 27 Probablty Mass functon (pmf( pmf) A dscrete random varable D has an assocated probablty mass functon (pmf) P, defned over the set of outcomes U: D = a) = a), a U, wth the followng propertes: D 0 a D = a) = a) = a = a) a 1 1 P (a ) unform dstrbuton Example: de roll S = {1, 2, 3, 4, 5, 6} What s the pmf of ths random varable? a

28 The Bnomal Dstrbuton Consder makng a sequence of n bnary experments, e.g., con tosses: Bernoull r.v. X~B(p). Assume: X=1) = p, X=0) = 1 p = q. What s the probablty of gettng a postve result k tmes? Answer: the bnomal dstrbuton n, k) = n k p k q n k, where n k = n! k!( n k)! s the number of k-element subsets of an n-element set ( n choose k ). Example: consder a sequence of three far con tosses. What s the probablty of observng heads exactly once? count outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT use formula: (3! / (1! x 2!)) x (0.5) 1 x (0.5) 2 = 6/(2x2x4) = 3/8 Constantn Rothkopf 28

29 Posson Dstrbuton The Posson dstrbuton can be used f there s a small probablty of success per unt tme that does not depend on prevous successes. If the mean number of successes n a certan tme nterval s gven by µ, then the probablty of observng k successes n a tme nterval s gven by: k) = e µ k µ k! Note: The Posson dstrbuton s also a good approxmaton for the bnomal dstrbuton f N s large and p s small (rare successes). Ths s useful snce bnomal dstrbuton s awkward to compute for large N,k. Constantn Rothkopf 29

30 Constantn Rothkopf 30 Cortcal neurons are (approx.) Posson Example: spkng of neurons over tme (spke tran) s often modeled as a Posson process: every mllsecond there s a certan probablty of the neuron sendng a spke ndependent of what happened earler. (Ths smple model gnores thngs lke the refractory perod.) tme Assume a neuron that fres on average 1 spke per 100ms. What s the probablty of observng exactly 3 spkes durng a tme nterval of 250 ms? k) = If there s one spke n 100ms we can expect k =2.5 spkes n 250ms. 3) = e -2.5 (2.5) 3 / 3! 0.15 e µ k µ k!

31 Constantn Rothkopf 31 Probablty Densty functon (pdf( pdf) probablty densty functon (pdf) from D. Ballard

32 Constantn Rothkopf 32 Non-negatvty and Normalzaton The probablty densty s non-negatve; t needs to ntegrate to one. A ) A 0 ) = 1 p( x) 0 p( x) dx = 1 Some made-up examples: p(x) p(x) p(x) x x x

33 Constantn Rothkopf 33 Example: unform dstrbuton p(x) 0.5 What s the probablty that X = 1.3 exactly: X = 1.3) =? x Fnte probablty corresponds to area under the pdf. 1 < X < 1.5) =? p( x) dx = The probablty at a partcular value x may exceed 1 (compare to pmf)

34 Constantn Rothkopf 34 Gaussan Dstrbuton The Gaussan s the most mportant contnuous dstrbuton: p( x) 1 exp 2πσ ( x µ ) = 2 2σ 2 from Duda et.al. (2001) Mean and varance of Gaussan are µ and σ 2, respectvely X ~ N(µ, σ 2 ) Note: many dstrbutons observed n nature look lke ths. Note: Gaussan can also be used to approxmate bnomal dstrbuton. Note: The convoluton of two Gaussans s agan a Gaussan.

35 Constantn Rothkopf 35 Gaussan and tunng curves from Dayan & Abbott (2001), orgnally from Hubel & Wesel, 1968

36 Constantn Rothkopf 36 Approxmatng the bnomal pmf Example bnomal dstrbutons for varyng numbers of far con tosses,.e. p=1/2 from Ptman (1999)

37 Constantn Rothkopf 37 Exponental Dstrbuton for x 0 wth parameter µ (often λ = 1/ µ): p( x) 1 x = exp µ µ mean: µ varance: µ 2 The exponental random varable X s memoryless, because f t represents the lfetme of a process, then the dstrbuton of the remanng lfetme s ndependent on the tme already passed.

38 Constantn Rothkopf 38 Expected values Consder a numercal random varable X wth an assocated pmf X=x) = x). The expected value or mean s defned as: E ( X ) = µ X = x X = x ) We can also consder the expected value of any functon f(x) of a random varable. Ths s defned as: ( f X )) f ( x ) X x ) E = = ( Note: expectaton s a lnear operaton! We wll use ths when calculatng the mean of the bnomal random varable.

39 Expected values for contnuous Constantn Rothkopf 39 random varables Straghtforward generalzaton wth sums turnng nto ntegrals. In the dscrete case: ( f X )) f ( x ) X x ) E = = ( Now the contnuous case: [ f ( X )] E f ( x) p( x) dx

40 Constantn Rothkopf 40 Mean, Varance, and hgher moments Mean and varance are useful for summarzng the most essental features of a pmf x). The mean s defned as (see above): E ( X ) = µ X = x X = x ) The varance (about the mean) s defned as: E (( X µ ) ) = σ = ( x µ ) X = X Hgher order moments (about the mean) are defned accordngly: X E (( X µ ) ) = σ = ( x µ ) X = X X related to the skewness of a dstrbuton X X 2 2 related σ X = to the kurtoss ( x of a X ) dstrbuton X (peakyness) x ) 24 E (( X µ ) ) = µ = X x x ) )

41 Constantn Rothkopf 41 Calculatng mean and varance Calculatng mean and varance Mean: Varance useful relaton: = = = dx x xp X E x P x X E ) ( ] [ ) ( ] [ µ ( ) ( ) dx x p x X E x P x X E ) ( ] ) [( ) ( ] ) [( = = = µ µ µ µ σ ( ) ] [ ] [ X E X = E σ

42 Calculaton for bnomal dstrbuton Accordng to the defnton of the expected value: But expectaton s a lnear operaton. Each toss of the con s a Bernoull experment. Assumng ndependent tosses: E[X] = E[X 1 +X 2 +X 3 + +X n ]= E[X ] = p = np Constantn Rothkopf 42

43 Constantn Rothkopf 43 Some examples Bernoull random varable wth parameter 0 p 1: mean: µ =p varance: σ 2 =p(1-p) Bnomal random varable wth parameters 0 n, 0 p 1: mean: µ =np varance: σ 2 =np(1-p) Posson random varable wth parameter λ : mean: µ = λ varance: σ 2 = λ Exponental random varable wth parameter λ : mean: µ = 1/ λ varance: σ 2 = 1 / λ 2 Gaussan random varable wth parameters µ, σ 2 : mean: µ varance: σ 2

44 Constantn Rothkopf 44 Cumulatve dstrbuton functon pdf: cdf: p( x) Φ( x ) x < X x + Δx) lm Δ x 0 Δx = = x < X ) = x p( x') dx' cdf gves probablty of X beng smaller than a specfc x Notes: cdf s monotoncally ncreasng (probabltes are non-negatve) cdf takes values between zero and one: 0 Φ( x) 1 pdf s dervatve of cdf probablty of X beng between A and B s gven by: P ( A < X B) = Φ( B) Φ( A) = p( x) dx B A

45 Constantn Rothkopf 45 Graphcal Cumulatve representatons dstrbuton functon of pdf & cdf unform pdf cdf 1 A B A B bmodal pdf cdf 1

46 Constantn Rothkopf 46 Central Lmt Theorem Another reason why the Gaussan s fundamentally mportant: from Ptman (1999)

47 Constantn Rothkopf 47 Pdf of sum of N unform dstrbutons What s the dstrbuton of the sum of N unform random varables? N=1 N=2 N=3 N=4 Note: one can show that the pdf of the sum of two random varables X 1 and X 2 s the convoluton of the two pdfs.

48 Constantn Rothkopf 48 Example: Random Walk A smple model that shows up n many contexts: Brownan moton of partcle n flud neuron s membrane potental under nfluence of exctatory and nhbtory synaptc nputs and many more consder partcle at poston X t {0,±1,±2, } at tme t at each tme step the partcle makes a step to the left or rght wth probablty 0.5: X t+1 =x+1 X t =x) = X t+1 =x-1 X t =x) = 0.5.

49 Constantn Rothkopf 49 Random Walk Queston: Assume partcle starts at X 0 =0, what s probablty of fndng t at a partcular locaton X after T steps? Soluton: (wth central lmt theorem) denote dstance of sngle step by d mean dstance of sngle step: µ = E[d] = 0.5*(-1) + 0.5*1 = 0 varance of dstance of sngle step: σ 2 = E[(d- µ) 2 ] = 0.5* *1 = 1 Accordng to central lmt theorem: dstrbuton of dstance traveled after T steps s approxmately Gaussan wth mean µ * T = 0 and varance: σ 2 = T, or equvalently standard devaton of T. The longer you wat the broader the dstrbuton of X.

50 Random walk and neurons from Shadlen (1999) Constantn Rothkopf 50

51 Constantn Rothkopf 51 Questons? Thanks to Jochen Tresch for large part of the materal for ths lecture