discrete random variables lecture

Upload: harumoto

Post on 02-Jun-2018

225 views

Category:

Documents


0 download

TRANSCRIPT

  • 8/10/2019 Discrete Random Variables Lecture

    1/13

  • 8/10/2019 Discrete Random Variables Lecture

    2/13

    Random Variables

    A random variable (rv) associates anumber with each outcome in thesample space

    We denote random variables with uppercase letter, XThe observed numerical value once theexperiment is run is denoted by thecorresponding lower case letter, x

    In mathematical terms, a rv is a functionwhose domain is the sample space andthe range is the set or real numbers

    2007 Thomson Brooks/Cole, a part of TheThomson Corporation.

    Random Variables

  • 8/10/2019 Discrete Random Variables Lecture

    3/13

    Random Variable Example

    A tall antenna is built on a mountain topwhere an extreme wind event occurs.Either the antenna fails (F) or survives (S)

    If the rv X is associated withthe outcomes,

    1 indicates that the antennasurvived, 0 indicates that theantenna failed

    S,Fs

    0FX,1SX

    Types of Random Variables

    Discrete Random Variable:takes a finite number of values

    e.g. the number of cars lined up at the MassPike toll plaza #8 (Palmer)

    Continuous Random Variable:takes all values in an interval

    e.g. the time each car mustwait at the toll plaza

  • 8/10/2019 Discrete Random Variables Lecture

    4/13

    Probability Mass Function

    ExampleSuppose you flip two coins, a rv, X, isthe number of heads in the experiment

    What is the sample space?What is the probability of each outcome inthe sample space?

    Probability Mass Function

    A Probability Mass Function (pmf), alsocalled probability distribution , is afunction p(x) that assigns to eachpossible value x that the random variableX can take, its probability

    p(x i) 0 for each possible value x i of X

    xsX:SsallP

    xXPxp

    ixall

    i 1)x(p

  • 8/10/2019 Discrete Random Variables Lecture

    5/13

  • 8/10/2019 Discrete Random Variables Lecture

    6/13

    pmf histogram

    0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 1 2

    p ( x

    )

    pmf:P (X = 0) = 0.5P (X = 1) = 0.3P (X = 2) = 0.2

    Cumulative DistributionFunction Example

    From the previous example, let F(x) denotethe cumulative distribution function (cdf) ofthe rv X

    X pmf, P(X=x) cdf, F(X=x)

    0 0.5 0.5

    1 0.3 0.5+0.3=0.8

    2 0.2 0.5+0.3+0.2=1

  • 8/10/2019 Discrete Random Variables Lecture

    7/13

  • 8/10/2019 Discrete Random Variables Lecture

    8/13

    Cumulative Distribution

    Function Any probability distribution must followthe axioms of probability

    F(- )=0; F( ) = 1F(x) 0 and is weakly increasingIt is continuous in x

    Any function that satisfies these axiomsis a cdf

    Expected Value Example

    Let X = number of working bulldozersafter 6 months. Assume the probabilitythat a bulldozer is working after 6months is 0.8, and there are 3 dozers.

    Find the pmf Find the cdf What is the expected value of the number ofdozers working after 6 months?

  • 8/10/2019 Discrete Random Variables Lecture

    9/13

    Expected Value

    A random variable has probabilities passociated with outcomes x with set ofpossible values D and pmf p(x)

    The expected value is the long runexpected mean, if you were to see Xover and over again

    Dx

    x xpxXE

    Example:Expected ValuesIn previous example, at least 2 dozers areneeded to finish a $100K job. Every dozerthat was brought in after 6 months costs$10K. What is your expected profit if youstart with 3 dozers?

  • 8/10/2019 Discrete Random Variables Lecture

    10/13

    Expected Value of a Function

    Consider that there might be a functionalrelationship with X with a set of possiblevalues D and pmf p(x) such that we havea probability of h(X)

    D

    )x(p)x(h)X(hE

    Properties of Expected Value

    bXEabaXE

    ]X[EXEXXE 2121

  • 8/10/2019 Discrete Random Variables Lecture

    11/13

  • 8/10/2019 Discrete Random Variables Lecture

    12/13

    Example Solution

    E[X2] = 0 .008 + 1 .096 + 4 .384 + 9 .512= 6.24

    E[X] = = 2.4(E[X])2 = 2 = 2.4 2 = 5.76

    2 = 6.24 5.76 = 0.48

    P(x) .008 .096 .384 .512x 0 1 2 3x2 0 1 4 9

    222 ]X[E

    Variance

    The variance is defined as

    22

    D

    2

    2

    2

    ]X[E

    )x(p)x(

    ])X[(E

    )X(V

  • 8/10/2019 Discrete Random Variables Lecture

    13/13

    Properties of Variance &

    Standard Deviation

    Xa)baX(V 22

    XbaX |a|

    Example:Variance Properties

    Your profit is equal to $6000 plus $10 forevery dozer that is working at 6 months.

    What is the variance of your profit?What is the STD of your profit?