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Statistics 510 Notes 7

By James Harper,2014-06-28 19:06
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Statistics 510 Notes 7 ...

Statistics 510: Notes 21

Reading: Sections 7.4-7.5

Schedule: I will e-mail homework 9 by Friday. It will not

    be due in two weeks (the Friday after Thanksgiving).

I. Covariance, Variance of Sums and Correlations (Chapter

    7.4)

The covariance between two random variables is a measure

    of how they are related.

    CovXY(,)The covariance between X and Y, denoted by , is

    defined by

    CovXYEXEXYEY(,)[([])([])]???.

    CovXY(,)0?Interpretation of Covariance: When , higher than expected values of X tend to occur together with

    higher than expected values of Y.

    CovXY(,)0?When , higher than expected values of X

    tend to occur together with lower than expected values of Y.

Example 1: From the Excel file stockdata.xls, we find that

    the correlations of the monthly log stock returns, in

    percentages, of Merck & Company, Johnson & Johnson,

    General Electric, General Motors and Ford Motor

    Company from January 1990 to December 1999 are

     Covariance

    Merck J&J 36.17205

     1

Merck GE 26.85792

    Merck GM 15.92461

    Merck Ford 16.19535

    J&J GE 48.45063

    J&J GM 37.28136

    J&J Ford 40.37623

    GE GM 27.72861

    GE Ford 26.63151

    GM Ford 84.76252

Correlation: The magnitude of the covariance depends on

    the variance of X and the variance of Y. A dimensionless measure of the relationship between X and Y is the

    ?(,)XYcorrelation :

    CovXY(,)

    ??(,)XY. VarXVarY()()

    The correlation is always between -1 and 1. If X and Y are

    ?(,)0XY?independent, then but the converse is not true. Generally, the correlation is a measure of the degree of

    linear dependence between X and Y.

    ab??0,0Note that for ,

    CovaXbYabCovXY(,)(,)(,)(,)aXbYXY????? VaraXVarbYaVarXbVarY()()()()

    (this is what is meant by saying that the correlation is

    dimensionless if X and Y are measured in certain units, and the units are changed so that X becomes aX and Y

    becomes bY, the correlation is not changed).

Example 1 continued: From the Excel file stockdata.xls, we

    find that the correlations of the monthly log stock returns,

    in percentages, of Merck & Company, Johnson & Johnson,

     2

General Electric, General Motors and Ford Motor

    Company from January 1990 to December 1999 are

     Correlation

    MerckJ&J0.419128

    MerckGE0.296728

    Merck GM0.182804

    MerckFord0.182194

    J&JGE0.449641

    J&JGM0.359493

    J&JFord0.381549

    GEGM0.254941

    GEFord0.239957

    GMFord0.793547

Properties of covariance:

By expanding the right side of the definition of the

    covariance, we see that

    CovXYEXYEXYXEYEYEX(,)[[][][][]]????

     [][][][][][][]????EXYEXEYEXEYEXEY

     [][][]??EXYEXEY

Note that if X and Y are independent, then

    ??

    EXYxyfxydxdy[](,)???????

    ??

     ()()?xyfxfydxdyXY??????

    ???? ()()?xfxyfydydxXY??????????

    ? ()[]?xfxEYdxX???

     [][]?EXEY

    Thus, if X and Y are independent,

     3

    CovXYEXYEXEY(,)[][][]0??? (1.1)

The converse of (1.1) is not true. Consider the sample

    space

    S???{(2,4),(1,1),(0,0),(1,1),(2,4)}with each point having equal probability. Define the random variable X to be the first component of the sample point chosen, and Y

    XY(2,4)2,(2,4)4,?????the second. Therefore, and so

    Xon. and Y are dependent, yet their covariance is 0. The

    former is true because

    1122

    ????????PXYPXPY(1,1)(1)*(1) 55525

    To verify the latter, note that

    11EXYEX()[(8)(1)018]0, ()[(2)(1)012)]0 ????????????????

    55

    1and ()[41014]2EY??????

    5

    Thus,

    CovXYEXYEXEY(,)()()()00*20?????.

Proposition 4.2 lists some properties of covariance.

    Proposition 4.2:

    CovXYCovYX(,)(,)?(i)

    CovXXVarX(,)()?(ii)

    CovaXYaCovXY(,)(,)?(iii)

    nmnm??

    CovXYCovXY,(,)???ijij????(iv) . ijij????1111??

Combining properties (ii) and (iv), we have that

     4

    nn??

    ??VarXVarXCovXX()2(,)iiij?????? (1.2) iiij???11??

    XXX,,If are pairwise independent (meaning that and i1n

    Xij? are independent for ), then Equation (1.2) reduces j

    to

    nn??

    ?VarXVarX()ii????. ii??11??

    XExample 2: Let be a hypergeometric

    (,,)nmNm?Xrandom variable i.e., is the number of white balls drawn in n random draws from an urn without

    replacement that originally consists of m white balls and Nm? black balls. Find the variance of X.

    1 if ith ball selected is white?

    I??iLet . 0 if ith ball selected is black?

    XII???Then and 1n

    VarXVarII()()??1n

    n

     ()2(,)??VarICovII iij???iij1??

    m2EI?()II?VarI()iTo calculate , note and so that iiiN

    2mmm222EIVarIEIEI?????(), ()()[()]iiii2. NNN

     5

    CovII(,)To calculate , we use the formula ij

    CovIIEIIEIEI(,)()()()??II?1. Note that if ijijijij

    both the ith and jth balls are white, and 0 otherwise. Thus,

    EII()=P(ith and jth balls are white). By considering the ij

    sequence of experiments look at the ith ball, look at the jth

    stndball, then look at the 1 ball, look at the 2’s ball, ..., look

    at the i-1 ball, look at the i+1 ball, ..., look at the j-1 ball,

    look at the j+1 ball, ..., look at the nth ball, we see that

    P(ith and jth balls are white) =

    mmNNNnmm*(1)*(2)*(3)**(1)*(1)??????

    ?. NNNnNN*(1)**(1)*(1)????

    nn*(1)?

    EII()?ijThus, and NN*(1)?CovIIEIIEIEI(,)()()()??ijijij

    mmmm*(1)?

     *??

    NNNN*(1)?

    mmm?1??

     ????NNN?1??

    ??mmN?

     ???NNN(1)???

    and

     6

VarXVarII()()??1n

    n

     ()2(,)??VarICovIIiij???

    iij1??

    2

    ????mmnnmmN(1)????

     2???n??????NNNNN2(1)??????? ??

    mmn?1????

     11???????NNN?1????

    Note that the variance of a binomial random variable with n

    m

    p?trials and probability of success for each trial is N

    mm??

    1???, so the variance for the hypergeometric is NN??

    n?1??

    1???smaller by a factor of ; this is due to the N?1??

    IInegative covariance between and for the jihypergeometric.

II. Conditional Expectation (Chapter 7.5)

Recall that if X and Y are joint discrete random variables,

    the conditional probability mass function of X, given that

    Yy?PYy()0??, is defined for all y, such that , by

    pxy(,)pxyPXxYy(|)(|)????XY|. py()Y

     7

It is natural to define, in this case, the conditional

    Yy?expectation of X, given that for all values of y such

    py()0?that by Y

    EXYyxPXxYy[|]{|}?????

    x

    . (|)?xpxyXY?|

    x

    Yy?The conditional expectation of X, given that , represents the long run mean value of X in many

    Yy?independent repetitions of experiments in which .

For continuous random variables, the conditional

    Yy?expectation of X, given that , by

    ?

    EXYyxfxydx[|](|)??XY| ???

    fy()0?provided that . Y

Example 3: Suppose that events occur according to a

    ?Poisson process with rate . Let N be the number of

    p?1events occurring in the time period [0,1]. For , let X

    [0,]pbe the number of events occurring in the time period . Find the conditional probability mass function and the

    Nn?conditional expectation of X given that .

     8

9

III. Computing Expectations and Probabilities by

    Conditioning (Section 7.5.2-7.5.3)

    EXY[|]Let us denote by that function of the random

    Yy?EXYy[|]?Yvariable whose value at is . Note that

    EXY[|]is itself a random variable. An important property

    of conditional expectations is the following proposition:

Proposition 7.5.1:

    EXEEXY[][[|]]? (1.3)

If Y is a discrete random variable, then equation (1.3) states

    that

    EXEXYyPYy[][|]{}???? (1.4)

    y

If Y is a continuous random variable, then equation (1.3)

    states that

    ?

    EXEXYyfydy[][|]()??Y ???

One way to understand equation (1.4) is to interpret it as

    EX[]follows: To calculate , we may take a weighted

    average of the conditional expected value of X, given that Yy?EXYy[|]?, each of the terms being weighted by the probability of the event on which it is conditioned.

    Equation (1.4) is a “law of total expectation” that is

    analogous to the law of total probability (Section 3.3, notes

    6).

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