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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

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

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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,

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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 ????????????????

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1and ()[41014]2EY??????

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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

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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

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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

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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

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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 .

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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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