Intro to Probability Instructor: Alexandre Bouchard

Transcription

1 Intro to Probability Instructor: Aleandre Bouchard

2 Logistics What s new/recent on the website: Webwork: net one will be posted Thursday. Written assignment 2 released during lecture time

3 Plan for today: Continuous random variables, continued Poisson process

4 Review

5 Density f() The function f is a density for X if for any interval A Eample: A = [a, b] area = probability height = density

6 Cumulative distribution function F() (CDF) The probability that a random variable is less or equal to F() = P(X )

7 Def 16 The uniform distribution Density CDF

8 Relation between density and CDF Density CDF c Go from density to CDF by integrating: c The CDF at is the integral (area under the curve) of the density, from - to

9 Relation between density and CDF Density CDF c Go from CDF to density by differentiating: The density at is the derivative (slope) of the CDF c At points where a CDF is differentiable

10 Def 18 The eponential distribution Density CDF f()

11 Functions of continuous random variables

12 E 53 Motivation Suppose I tell you that is the distribution of Richter scales What is the distribution of the amplitudes? For simplicity: Assume Richter scale X is uniform between 0 and 1 What is the distribution of ep(x)?

13 Epectation of continuous random variables

14 E 54 Motivation A earthquake of amplitude costs billion How much money is spent for an average earthquake? Want: E[ep(X)] Recall: not equal to ep(ex)

15 Physical interpretation

16 Def 17 Definition We define the epectation of a continuous random variable X with density f as: X Eample...

17 Epectation of a function of a continuous random variables

18 Prop 14 2 ways As in discrete case, there are actually 2 ways to compute E[g(X)] Brute force : first compute density of g(x), then compute the epectation using definition (what we did so far) Shortcut: Eample... X

19 The Poisson process

20 E 55 Motivation: mutations on a genome Consider a genome G with 3 billion nucleotides (letters A, C, G, T) We epose it to radiation, creating mutations For each letter, there is an independent, 1/(1 billion) chance of a mutation (1e-9) What is the distribution on the total number NG of mutations? ACCGAT 3 billion Eact: NG ~ Bin(p = 1e-9, n = 3e9) Appro: NG ~ Poi(λ = 3)

21 Continuous approimation The genome G is discrete, but the large number letters motivates a continuous approimation, G = [0, 1] Net question: let H = first half of the genome = [0, 1/2] What is the distribution on the number of mutation in the first half of the genome, NH? NH = # points in H = 2 in eample realization below H G NH ~ Poi(3 * length(h) = 1.5)

22 Mutations in subsets, continued What is the distribution on the number of mutation in the first half of the genome, NH? Note: we can do this with as many subsets A, B,.. of the genomes as we want: B A H NH ~ Poi(3 * length(h) = 1.5) NA ~ Poi(3 * length(a) = 1) NB ~ Poi(3 * length(b) = 0.4) NH = # points in H NA = # points in A...

23 Mutations in subsets, continued In general, for any set A, NA ~ Poisson(intensity * length(a)) The intensity is a parameter, abbreviation: i For eample, i = 3 in our eample H B A NH = # points in H NA = # points in A...

24 Relations between numbers of mutations in different subsets Is N H independent of NA? Is N H independent of NB? Is N A independent of NB? Yes No Yes B A H NH = # points in H NA = # points in A...

25 Relations between mutations in subsets In general, for any two sets A and B: if A and B are disjoint, then N A is independent of NB B A H NH = # points in H NA = # points in A...

26 Def 19 Poisson processes B A Consider a random set of points, X Eample of a realization: {0.12, 0.15, 0.57, 0.71} X is called a Poisson process with intensity i if: For all A G, N A ~ Poi(i * length(a)) For all disjoint A, B G, NA is independent of NB NA = # points of X in A NB = # points of X in B

27 Spacings of a Poisson process

28 Spacings So far: for any fied set A, we have the distribution of the # of points in A Net: distribution of the spacings D1, D2, D3: D1 = location of the first point D2 = distance between first and second D3 = distance between second and third,... D1 D2 D3 D4 D5 A

29 Start with first spacing Note: if we set A = [0, 0.2], then these two events are equal: (D1 > 0.2) = (NA = 0) NA = # points of X in A D1 A 0.2

30 Probab. first spacing > 0.2 (D1 > 0.2) = (NA = 0) Hence, P(D1 > 0.2) = P(NA = 0) Get: P(D1 > 0.2) = e k k! Recall: NA ~ Poi(i * length(a)) Here: length(a) = 0.2 λ? k? intensity i * D1 A 0.2 NA = # points of X in A

31 Probab. first spacing > t (D1 > t) = (NA = 0) Hence, P(D1 > t) = P(NA = 0) Recall: NA ~ Poi(i * length(a)) Get: P(D1 > t) = e k k! Here: length(a) = t λ? k? intensity i * t 0 D1 t A NA = # points of X in A

32 Def 18 Eponential distribution, where λ = intensity i Density CDF f()

33 Other spacings? Same story: D1 ~ Ep( i ), D2 ~ Ep( i ), D3 ~ Ep( i ), etc All independent and identically distributed (iid) D1 D2 D3 D4 D5 A

34 Application to earthquake distribution

35 E 56 Earthquake eample Data: 68 earthquake around the Strait of Messina in the period Only magnitude > 4.5 kept Foreshocks and aftershocks removed

36 E 56 Data:... Available under Files in case you are curious: earthquakes.ls 1/average = intensity =

37 E 56

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39 Useful facts Eponential (rate λ) Density: λ ep( -λ ) [0, ) Mean: 1/λ Variance: 1/λ 2 CDF: 1 - ep( -λ) Poisson Process (intensity λ) Let A and B be two disjoint intervals Let their length be LA and LB, Then NA and NB are: independent Poisson r.v.s, with rates λla and λlb Poisson Distribution (mean λ) PMF: λ e - λ /!, {0, 1, 2,...} Mean: λ, Variance: λ

40 E 56a Clicker question The probability that a recurrence interval is shorter than 10 months is given by the epression: A. P(T < 10) where T~Ep( =0.2403) B. P(T < 0.83) where T~Ep( =0.2403) C. P(T < 10) where T~Poisson( l =0.2403*10) D. P(T < 0.83) where T~Poisson( l =0.2403*10)

41 E 56b Clicker question What is the probability that a recurrence interval is shorter than 10 months? A B C D

42 E 56c Clicker question What proportion of the observed recurrence intervals are greater than 5 years? A B C D. 0.38

43 E 56d Clicker question What is the probability that a recurrence interval is greater than 5 years (theoretical value)? A B C D

44 E 56e Clicker question The probability that there are no earthquakes in 5 years is given by the epression: A. P(T < 5) where T~Ep( =0.2403) B. P(T > 5) where T~Ep( =0.2403) C. P(T = 0) where T~Ep( =0.2403) D. P(X = 0) where X~Poisson( l =0.2403*5) E. P(X = 0) where X~Poisson( =0.2403)

45 E 56f Clicker question An earthquake just occurred. What is the probability that one has to wait for more than 30 years to observe the 10th earthquake? A. P(T > 30) where T~Ep( =0.2403) B. P(X = 9) where X~Poisson( l =0.2403*30) C. P(X = 10) where X~Poisson( l =0.2403*30) D. P(X l =0.2403*30) E. P(G > 30) where G~Gamma( =10, =0.2403)