Slides 4: Continuous Random Variables

Transcription

1 Slides 4: Continuous Random Variables Continuous RVs Values, numerical, in a continuous range ( real ) Continuous transformations of RAND() Who needs them? What are the long run properties? What transform to use? When the properties are specified. 1

2 Continuous Random Variables First introduction to important theory: CDF: cumulative probability density function. PDF: probability density function. For example, =NORMDIST, Normal distribution. First we ll look constructively through =NORMINV. Later, what are such functions? 2

3 The Normal Distribution Bell shaped curve, most probable value is µ. Most probability is within µ±2σ Symmetric around µ. µ is a location parameter, σ is a scale parameter. NORMDIST(RAND(),µ,σ,TRUE/FALSE) 3

4 Examples of Transformations Y generated by RAND() X = 3Y +2 X = Y 2 What are long run properties of X (shorthand for procedure which generates values for X)? What transformation will yield X with specified properties? 4

5 Back to Non-Uniform Random Numbers Hey! I m having a little problem generating a few random numbers. The thing is this: I have a huge list of people and I need to assign a random number between 0.0 and 1.25 to each one. The problem is I need the following: 70% of people to have a number between 1 and 1.2, 20% of people to have a number greater than 1.2, while 10% of people to have a number below 1. Do you know how to do this??? Thanks Transform RAND() 5

6 One Solution First, split into 3 cases using discrete transform: a) 70%, b) 20%, and c) 10% Use conditional simulation (conditional on each of the three cases). If a) use uniform in (1,1.2) If b) use uniform in (1.2,1.25) If c) use uniform in (0,1) 6

7 General Transforms: Long Run Properties What transformation to use? Specify desired properties by P(R r) = F(r), known as cumulative probability distribution function (cdf). 7

8 Staircase Plot CumProb = Function of Possible Values Poss Values = Inverse Function of CumProb In Excel LOOKUP(U,CumProb,PossVals) Biggest steps catch most U 8

9 Modelling Water Drops Our model is simple and only requires 7 coefficients for generating splashes for head-on impact for a material. A more general model for generating splash for arbitrary impact angles (due to surface inclination or wind) requires 54 coefficients (Material Based Splashing of Water Drops, K. Garg, G. Krishnan and S.K. Nayar, Proceedings of Eurographics Symposium on Rendering, 2007). 9

10 Modelling Details Gaussian means Normal 10

11 Staircase Plot LOOKUP inefficient if very many possible values. Need alternatives if possibilities are continuous. 11

12 Random Cannon Shot Trajectory parabolic, specified maximum m = 1 Lands at R, uniformly random on (0,1) 12

13 Long Run Cannon R = U 13

14 Random Cannon R = 3U +2 Average R is

15 Random Cannon R = U 2 Average R is

16 Random Cannon R =NORMINV(U,µ,σ) Seek target of 0.8 (µ = 0.8) σ = 0.05, so 95% of shots within ±0.1 of target. R=NORMINV(RAND(),0.8,0.05) 16

17 Transforms: Long Run Properties Properties of U Equally likely to be any value in (0,1) Average value 0.5 Properties of 3U +2 Equally likely to be any value in (2,5) Average value 3.5 Properties of U 2 In (0,1), but more likely to be small. Average value? Properties of NORMINV(RAND(),0.8,0.05) Average % within ±0.1 17

18 Linear Transforms: Long Run Properties Properties of U Equally likely to be any value in (0,1) By symmetry, E[U] = 0.5 P(U u) = u for 0 < u < 1 Properties of R = 3U +2 Equally likely to be any value in (2,5) By symmetry, E[U] = 3.5 P(R 3u+2) = u for 0 < u < 1 P(R r) = r 2 3 for 2 < r < 5 18

19 Non-Linear Transforms: Long Run Properties Properties of U Equally likely to be any value in (0,1) By symmetry, E[U] = 0.5 P(U u) = u for 0 < u < 1 Properties of R = U 2 Not equally likely to be any value in (0,1) No symmetry, E[U] =? P(R u 2 ) = u for 0 < u < 1 P(R r) = r for 0 < r < 1 19

20 General Transforms: Long Run Properties Properties of U Equally likely to be any value in (0,1) By symmetry, E[U] = 0.5 P(U u) = u for 0 < u < 1 Properties of R = g(u) Not necessarily equally likely to be any value in (g(0), g(1)) P(R g(u)) = u for 0 < u < 1 P(R r) = function of r; namely g 1 (r) 20

21 General Transforms: Long Run Properties What is the range of R as transforms of U as below? Compute P(R r) for 3 selected values of r in range. Plot/sketch P(R r) as a function of r. R = 3U 2 R = 3U

22 Transforms and P(U u) = u Linear Preserves equally likely Changes the range 10% of U % of 10U Non-linear For example, R can have large values rarely. P(R > r) = P(U > corresponding value of u) P(U u) = u for 0 < u < 1 P(g(U) g(u)) = u for increasing functions g(.) P(R g(u)) = u R obtained by transforming U P(R r) = u where r obtained by g(u) P(R r) = g 1 (r) where r satisfies r = g(u) 22

23 General Transforms: Long Run Properties What transformation to use? Specify desired properties through P(R r) = F(r) Recipe: Form R by F 1 (U) 23

24 Continuous Transformations of RAND 24

25 Bell Shaped Curve Widely used model Symmetric distribution Normal Distribution X N(µ,σ 2 ) f(x) = 1 e (x µ)2 2σ 2 2πσ 2 25

26 Normal Distribution Transform T = Inverse of Normal cdf at U =NORMINV(U,µ,σ)= F 1 (U) Symmetric distribution X N(µ,σ 2 ) f(x) = 1 e (x µ)2 2σ 2 2πσ 2 26

27 Continuous Transformations of U R = g(u) = F 1 (U) and U = g 1 (R) = F(R) 27

28 Modelling Details 28

29 Splash Normal Transform used for N Splash number per second per unit area µ = 73.3, σ = 9.78 N =ROUND(NORMINV(RAND(),73.3,9.78),0) What proportion of N exceeds 100? 29

30 Alternative Solution I need the following: 70% between 1 and % greater than % below 1 30

31 Continuous Random Variables Continuous RVs Values, numerical, in a continuous range ( real ) Continuous transforms of U =RAND() For example, R = 3U +2, R =NORMINV(U) What transform to use? Specify desirable properties by cumulative probability distribution function (cdf) F(r) = P(R r). Use inverse function, i.e. solve F(U) = R, solution U = F 1 (R). Simplest if F(U) is linear Corresponds to R being uniform over a specified range. 31

32 Lab: Systems Marks of 10 students in a class System Lifetime comprised of several components 32

33 Student Marks Ten students of different ability sit an exam. The variation is modelled via the Normal distribution with µ = 58 and σ = 12, i.e., NORMINV(RAND(),µ,σ)). What proportion will get above 40%? Above 80%? What is the probability that the best student gets above 80%? 33

34 System Lifetime A system has 3 components, A, B and C, with redundancy. It is designed such that it will work if either C is working, or both A and B is working. Component Lifetimes of A, B, and C are 10, 15 and 8, respectively. What is the system lifetime? Lifetimes vary around µ = 10,14,13 and normal variation σ = 2. What is the average system lifetime? 34