Lecture 3 Continuous Random Variable - 國立中興大學 · Lecture 3 Continuous Random Variable...
Transcript of Lecture 3 Continuous Random Variable - 國立中興大學 · Lecture 3 Continuous Random Variable...
Lecture 3
Continuous Random Variable
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Cumulative Distribution Function
Definition
Theorem 3.1 For any random variable X,
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Continuous Random Variable
Definition
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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?
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[Continued]
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 SY = {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
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Solution (con’t)
=> The is true regardless of the outcome, x. It follows that every outcome has probability ZERO.
imply
How to find P[X = x] ?
This demonstrates that P[X = x] ≤ 0.
P[X = x] ≥0. Therefore, P[X = x] = 0.
Solution (con’t)
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How to find the CDF for X, SX = [0, 1)?
FX(x) = 0 for x < 0 FX(x) = 1 for x ≥ 1.
Solution (con’t)
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x
Quiz
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Probability Density Function
Definition
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slope
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Properties of PDF
Theorem 3.2
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Proof
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Theorem 3.1
Theorem
Proof:
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Example
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Solution
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Quiz
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Expected Value of Continuous RV
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Definition
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:
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Expected Value of g(X)
Properties (Theorem 3.5) For any random variable X,
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Theorem 3.4
Fine the variance and standard deviation of the pointer position X in the wheel example.
Example
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Solution
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Quiz
( )kk E Xµ =
( )
( )-
if is discrete
if is continuous
k
x
k
x p x X
x f x dx X∞
∞
=
∑
∫
The kth moment of X
Families of continuous Random Variables
Uniform
Exponential
Erlang
Gaussian
Standard Normal
Standard Normal Complement
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Uniform Random Variable
Definition
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Properties of Uniform
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Theorem 3.6
Problem
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Solution
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Property of Uniform
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Theorem 3.7
Proof
Definition 2.9 Discrete Uniform
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f
Exponential Random Variable
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Definition
Properties of Exponential
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Theorem 3.8
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?
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Solution (a)
(b)
[Continued]
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Solution (con’t)
(c)
(d)
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Solution (con’t)
(e)
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X is an exponential (λ) random variable
More Properties of Exponential
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Theorem 3.9
Proof
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Example
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Solution
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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
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.
Erlang (n,λ) Random Variable
X is an exponential (λ) random variable
Poisson (α=λT)
Properties of Erlang
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Theorem 3.10
Theorem 3.11
Proof (Theorem 3.11)
Proof (Theorem 3.11) (con’t)
Quiz
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Gaussian Random Variable
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Definition
Property of Gaussian
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Theorem 3.13
Theorem 3.12
If X is a Gaussian (μ,σ) random variable,
Definition
Standard Normal Random Variable
The standard normal random variable Z is the Gaussian (0,1) random variable. The CDF of the standard normal random variable Z is
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Standard Normal Random Variable
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Theorem 3.14
Example
Solution:
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Example
Solution:
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Symmetry property of Gaussian
Symmetry properties of standard normal or Gaussian (0,1) PDF.
Theorem 3.15
Q
Standard Normal Complement CDF
Definition
Example
Solution:
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Quiz
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Definition
Unit Impulse (Delta) Function
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Property of Unit Impulse
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Theorem 3.16 shifting property
Unit Step Function
, where u(x) is the unit step function, defined as
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Theorem 3.17
Definition
Example
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Remark: Using the Dirac delta function we can define the density function for a discrete random variables.
Example (con’t)
PMF CDF PDF
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From the example, we can see that the discrete random variable Y can either represented by a PMF PY(y) with bar at y = 1,2,3, by a CDF with jumps at y = 1,2,3, or by a PDF fY(y) with impulses at y =1,2,3.
The expected value of Y can be caluculate either by summing over the PMF PY(y) or integrating over the PDF fY(y). Using PDF, we have
Equivalent Statements
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Theorem 3.18
Mixed Random Variable
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Definition
Example
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Solution
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Solution (con’t)
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Example
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Solution
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Solution (con’t)
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Example
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Solution
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Solution (con’t)
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Solution (con’t)
Quiz
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Probability Models of g(X)
Proof:
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Theorem 3.19
Example
Solution:
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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 fY(y).
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Solution
The function Y=100X. To find the PDF of Y, we first find the CDF FY(y). Remind that the CDF of X is
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Distributions of Y=g(X)
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Theorem 3.20
Property of Derived RV --Shift
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Theorem 3.21
Proof:
Theorem 3.22
Proof:
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Conditional PDF given an Event
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Definition
Conditional PDF
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Definition
Conditional Expected Value given and Event
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Definition
Problem
Solution:
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Solution (con’t)
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Problem
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Solution
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Quiz
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