Conditional Expectation for discrete random abhisheksaha/sta4321/ Expectation for discrete random variables ... Conditional Expectation for discrete random variables ... Let X denote the proportion of im-

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  • Lecture 28

    Agenda

    1. Conditional Expectation for discrete random variables

    2. Joint Distribution of Continuous Random Variables

    Conditional Expectation for discrete random

    variables

    Let X and Y be two discrete random variables. For X = x we know thatinstead of using {pY (y) : y Range(Y )} its better to use {pY |X=x(y) : y Range(Y )}. We defined this conditional distribution as,

    pY |X=x(y) =pX,Y (x, y)

    pX(x)=

    P (X = x, Y = y)

    P (X = x)

    for x Range(X) and y Range(Y ). Note that this definition is con-ceptually nothing new, because we have already learned the definition ofconditional probability. Now we define conditional expectation of Y givenX = x.

    Definition 1. For two discrete random variables X and Y , for some x Range(X), we define the conditional expectation of Y given X = x as,

    E(Y |X = x) =

    yRange(Y )

    y pY |X=x(y)

    Please note that E(X|Y = y) can be defined similarly.

    Example

    The joint distribution of X and Y is given in the following table

    X=0 X=1 X=3Y=-1 0.11 0.03 0.00Y=2.5 0.03 0.09 0.16Y=3 0.15 0.15 0.06

    Y=4.7 0.04 0.16 0.02

    1

  • Lets find E(Y |X = 1).

    P (X = 1) = 0.03 + 0.09 + 0.15 + 0.16 = 0.43

    Hence,

    E(Y |X = 1) = 1 0.030.43

    + 2.5 0.090.43

    + 3 0.150.43

    + 4.7 0.160.43

    = 3.2488

    Joint Distribution of Continuous Random Vari-

    ables

    We previously studied the joint probability mass function of two discreterandom variables, which described their joint behaviour. Today, we repeatthe same procedure for continuous random variables.

    Lets recollect that for a continuous random variable X, there has to bea density function fX such that,

    fX(x) 0 x R

    a < b, P (a < X < b) = bafX(x)dx

    Suppose we have two continuous random variables X and Y . The jointprobability behaviour of X and Y is described by the joint probabilitydensity function fX,Y of X and Y which has the following properties

    fX,Y (x, y) 0 (x, y) R2

    P (a X b, c Y d) = ba

    dcfX,Y (x, y)dxdy for all a < b and

    c < d.

    It follows that the joint probability distribution function FX,Y isdefined as,

    FX,Y (a, b) = P (X a, Y b) = a

    b

    fX,Y (x, y)dxdy

    Example

    A certain process for producing an industrial chemical yeilds a product thatcontains two main types of impurities. Let X denote the proportion of im-purities of Type I and Y denote the proportion of impurities of Type II.

    2

  • Suppose that the joint density of X and Y can be modelled as,

    fX,Y (x, y) = 2(1 x) if 0 x 1, 0 y 1= 0 otherwise

    Compute P (0 X 0.5, 0.4 Y 0.7).

    P (0 X 0.5, 0.4 Y 0.7) = 0.50

    0.70.4

    2(1 x)dydx

    =

    0.50

    (0.7 0.4) 2(1 x)dx

    = 0.3 0.50

    2(1 x)dx

    = 0.3[(1 x)2

    ]0.50

    = 0.3[(1 0.5)2 + (1 0)2

    ]= 0.3 [1 0.25]= 0.3 0.75= 0.225

    Lemma 1. If X and Y are continuous random variables, with joint pdf fX,Ythen the individual or marginal pdf s fX and fY are given by,

    fX(x) =

    fX,Y (x, y)dy

    fY (y) =

    fX,Y (x, y)dx

    For the industrial production example considered previously, we compute

    3

  • fX and fY .

    fX(x) =

    fX,Y (x, y)dy

    =

    2(1 x)1{0x1,0y1}dy

    =

    10

    2(1 x)1{0x1}dy

    = 2(1 x)1{0x1}= 2(1 x) if 0 x 1= 0 otherwise

    fY (y) =

    fX,Y (x, y)dx

    =

    2(1 x)1{0x1,0y1}dx

    =

    10

    2(1 x)1{0y1}dx

    = 1{0y1}[(1 x)2

    ]10

    = 1{0y1} 1= 1 if 0 y 1= 0 otherwise

    Please note that by using the indicator function 1{...} we are consideringthe various cases at once and making the calculations easier.

    Suppose we are interseted in the behaviour of the random variable Ygiven X = x. To develop a framework for expressing this behaviour weneed the notion of conditional probability density function (or conditionalpdf in short). But before we go on lets note one thing. When we defined

    P (A|B) = P (AB)P (B)

    , we inherently assumed that P (B) > 0. But if we want to

    calculate P (Y 0.5|X = 0.3), we cant assume P (X = 0.3) > 0 because Xis continuous. So we need another strategy.

    Definition 2. Let X and Y be continuous random variables with joint pdffX,Y and marginal pdf s fX and fY . Then for any x such that fX(x) > 0,

    4

  • we define the conditional pdf of Y given X = x as,

    fY |X=x(y) =fX,Y (x, y)

    fX(x)

    for all y R.

    Note that when fX(x) = 0 we dont define the above quantity. Now wecan calculate

    P (Y 0.5|X = 0.3) = 0.5

    fY |X=0.3(y)dy

    Homework::

    1. For homework from Lecture 27, find E(Y |X = 1) for Problem1 and E(X|Y = 1) for Problem 2.

    2. 5.27,5.31,5.32,5.34

    5