Lecture 3 Continuous Random Variable 1
Cumulative Distribution Function Definition Theorem 3.1 For any random variable X, 2
Continuous Random Variable Definition 3
Example Suppose we have a wheel of circumference one meter and we mark a point on the perimeter at the top of the wheel. In the center of the wheel is a radial pointer that we spin. After spinning the pointer, we measure the distance, X meters, around the circumference of the wheel going clockwise from the marked point to the pointer position as shown in the following figure. Clearly, 0 X < 1. Also, it is reasonable to believe that if the spin is hard enough, the pointer is just as likely to arrive at any part of the circle as at any other. (Note that the random pointer on disk of circumference 1) (a) For a given x, what is the probability P[X = x]? (b) What is the CDF of X? 4
Solution a) A reasonable approach is to find a discrete approximation to X. Marking the perimeter with n equal-length arcs numbered 1 to n. Let Y denote the number of the arc in which the pointer stops. Denote the range of Y by S Y = {1, 2,, n}. The PMF of Y is From the wheel on the right side of the figure, we can deduce that if X = x, then Y =, where the notation is defined as s ceiling integer [Continued] 5
Solution (con t) Note that the event, which implies that We observe this is true no matter how finely we divide up the wheel. To find P[X = x], we consider larger and larger values of n. As n increases, the arcs on the circle decrease in size, approaching a single point. The probability of the pointer arriving in any particular arc decrease until we have in the limit, This demonstrates that P[X = x] 0. Note that the first axiom of probability states that P[X = x] 0. Therefore, P[X = x] = 0. => The is true regardless of the outcome, x. It follows that every outcome has probability ZERO. 6
Solution (con t) imply How to find P[X = x]? This demonstrates that P[X = x] 0. P[X = x] 0. Therefore, P[X = x] = 0. => The is true regardless of the outcome, x. It follows that every outcome has probability ZERO. 7
Solution (con t) b) Observing that any outcome x S X = [0, 1). We have F X (x) = 0 for x < 0, and F X (x) = 1 for x 1. To find the CDF for x between 0 and 1 we consider the event {X x} with x growing from 0 to 1. Each event corresponds to an arc on the circle. The arc is small when x 0 and it includes nearly the whole circle when x 1. F X (x) = P[X x] is the probability that the pointer stops somewhere in the arc. This probability grows from 0 to 1 as the arc increases to include the whole circle. Given our assumption that the pointer has no preferred stopping places, it is reasonable to expect the probability to grow in proportion to the fraction of the circle occupied by the arc X x. This fraction is simply X. To be more formal, we can refer to the figure and note that with the circle divided into n arcs, Therefore, the probabilities of the three events satisfy 8
Solution (con t) How to find the CDF for X, S X = [0, 1)? F X (x) = 0 for x < 0 F X (x) = 1 for x 1. 9
Solution (con t) x 10
Quiz 11
Probability Density Function Definition 13
slope 14
Properties of PDF Theorem 3.2 15
Proof Theorem 3.1 16
Theorem Proof: 17
Example 18
Solution 19
Quiz 20
Expected Value of Continuous RV Definition 23
Problem We found that the stopping point X of the spinning wheel experiment was a uniform random variable with PDF Find the expected stopping point E[X] of the pointer. Solution: 24
Expected Value of g(x) Theorem 3.4 Properties (Theorem 3.5) For any random variable X, 25
Example Fine the variance and standard deviation of the pointer position X in the wheel example. 26
Solution 27
Quiz The k th moment of X k E X k - x k x p x k x f x dx if X if X is discrete is continuous 28
Families of continuous Random Variables Uniform Exponential Erlang Gaussian Standard Normal Standard Normal Complement 30
Uniform Random Variable Definition 31
Properties of Uniform Theorem 3.6 32
Problem 33
Solution 34
Property of Uniform Theorem 3.7 35
Proof f Definition 2.9 Discrete Uniform 36
Exponential Random Variable Definition 37
Properties of Exponential Theorem 3.8 38
Example (a) What is the PDF of the duration in minutes of a telephone conversation? (b) What is the probability that a conversation will last between 2 and 4 minutes? (c) What is the expected duration of a telephone call? (d) What are the variance and standard deviation of T? (e) What is the probability that a call duration is within 1 standard deviation of the expected call duration? 39
Solution (a) (b) [Continued] 40
Solution (con t) (c) (d) 41
Solution (con t) (e) 42
More Properties of Exponential Theorem 3.9 X is an exponential (λ) random variable 43
Proof 2013/10/1 2013 Random Process @EE.NCHU 44
Example 45
Solution Since E [T ] = 1/λ = 3, for company B, which charges for the exact duration of a call, Because is a geometric random variable with, therefore, the expected revenue for Company A is 46
Erlang (n,λ) Random Variable Definition order Note 1: Erlang distribution is the distribution of the sum of k iid exponential random variables. The rate of Erlang distribution is the rate of this exponential distribution. Erlang(1, λ) is identical to Exponential(λ). Note 2: Erlang is also related to Pascal. X is an exponential (λ) random variable 48
Properties of Erlang Theorem 3.10 Theorem 3.11 Poisson (α=λt) 50
Proof (Theorem 3.11) 51
Proof (Theorem 3.11) (con t) 52
Quiz 53
Gaussian Random Variable Definition 55
Property of Gaussian Theorem 3.12 If X is a Gaussian (μ,σ) random variable, Theorem 3.13 56
Standard Normal Random Variable Definition The standard normal random variable Z is the Gaussian (0,1) random variable. The CDF of the standard normal random variable Z is 57
Standard Normal Random Variable Theorem 3.14 58
Example Solution: 59
60
Example Solution: 61
Symmetry property of Gaussian Theorem 3.15 Symmetry properties of standard normal or Gaussian (0,1) PDF. Q 62
Standard Normal Complement CDF Definition 63
Example Solution: 64
Quiz 65
68
Unit Impulse (Delta) Function Definition 69
70
Property of Unit Impulse Theorem 3.16 shifting property 71
Unit Step Function Definition, where u(x) is the unit step function, defined as Theorem 3.17 72
Example Remark: Using the Dirac delta function we can define the density function for a discrete random variables. 73
Example (con t) From the example, we can see that the discrete random variable Y can either represented by a PMF P Y (y) with bar at y = 1,2,3, by a CDF with jumps at y = 1,2,3, or by a PDF f Y (y) with impulses at y =1,2,3. PMF CDF PDF The expected value of Y can be caluculate either by summing over the PMF P Y (y) or integrating over the PDF f Y (y). Using PDF, we have 74
Equivalent Statements Theorem 3.18 75
Mixed Random Variable Definition 76
Example 77
Solution 78
Solution (con t) 79
Example 80
Solution 81
Solution (con t) 82
Example 83
Solution 84
Solution (con t) 85
Solution (con t) 86
Quiz 87
Probability Models of g(x) Theorem 3.19 Proof: 89
Example Solution: 90
Example Y centimeters is the location of the pointer on the 1-meter circumference of the circle. Note that X is the location of the pointer in a unit of meter. Use X to derive f Y (y). 91
Solution The function Y=100X. To find the PDF of Y, we first find the CDF F Y (y). Remind that the CDF of X is 92
Distributions of Y=g(X) Theorem 3.20 93
Property of Derived RV --Shift Theorem 3.21 Proof: 94
Theorem 3.22 Proof: 95
Conditional PDF given an Event Definition 96
Conditional PDF Definition 97
Conditional Expected Value given an Event Definition 98
Problem Solution: 99
Solution (con t) 100
Problem 101
Solution 102
Quiz 103