1

6. Mean, Variance, Moments and Characteristic Functions

For a r.v X, its p.d.f represents complete information about it, and for any Borel set B on the x-axis

Note that represents very detailed information, and quite often it is desirable to characterize the r.v in terms of its average behavior. In this context, we will introduce two parameters - mean and variance - that are universally used to represent the overall properties of the r.v and its p.d.f.

B X dxxfBXP .)()( (6-1)

)(xf X

)(xf X

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Mean or the Expected Value of a r.v X is defined as

If X is a discrete-type r.v, then using (3-25) we get

Mean represents the average (mean) value of the r.v in a very large number of trials. For example if then using (3-31) ,

is the midpoint of the interval (a,b).

.)( )( dxxfxXEX XX (6-2)

. )(

)()()(

1

iii

iii

iiii

iiiX

xXPxpx

dxxxpxdxxxpxXEX

(6-3)

),,( baUX

(6-4)

b

a

b

a

ba

ab

abx

abdx

ab

xXE

2)(22

1)(

222

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On the other hand if X is exponential with parameter as in (3-32), then

implying that the parameter in (3-32) represents the mean value of the exponential r.v.

Similarly if X is Poisson with parameter as in (3-45), using (6-3), we get

Thus the parameter in (3-45) also represents the mean of the Poisson r.v.

0

/

0 , )(

dyyedxe

xXE yx (6-5)

.!)!1(

!!)()(

01

100

eei

ek

e

kke

kkekXkPXE

i

i

k

k

k

k

k

k

k

(6-6)

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In a similar manner, if X is binomial as in (3-44), then its mean is given by

Thus np represents the mean of the binomial r.v in (3-44).

For the normal r.v in (3-29),

.)(!)!1(

)!1(

)!1()!(

!

!)!(

!)()(

111

01

100

npqpnpqpiin

nnpqp

kkn

n

qpkkn

nkqp

k

nkkXkPXE

ninin

i

knkn

k

knkn

k

knkn

k

n

k

(6-7)

.2

1

2

1

)(2

1

2

1)(

1

2/

2

0

2/

2

2/

2

2/)(

2

2222

2222

dyedyye

dyeydxxeXE

yy

yx

(6-8)

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Thus the first parameter in is in fact the mean of the Gaussian r.v X. Given suppose defines a new r.v with p.d.f Then from the previous discussion, the new r.v Y has a mean given by (see (6-2))

From (6-9), it appears that to determine we need to determine However this is not the case if only is the quantity of interest. Recall that for any y,

where represent the multiple solutions of the equation But(6-10) can be rewritten as

),( 2NX

),( xfX X )(XgY ).( yfY

Y

.)( )( dyyfyYE YY (6-9)

),(YE

).( yfY )(YE

0y

, i

iii xxXxPyyYyP (6-10)

ix

).( ixgy

,)()( ii

iXY xxfyyf (6-11)

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where the terms form nonoverlapping intervals. Hence

and hence as y covers the entire y-axis, the corresponding x’s are nonoverlapping, and they cover the entire x-axis. Hence, in the limit as integrating both sides of (6-12), we get the useful formula

In the discrete case, (6-13) reduces to

From (6-13)-(6-14), is not required to evaluate for We can use (6-14) to determine the mean of

where X is a Poisson r.v. Using (3-45)

iii xxx ,

,)()()( )( ii

iXiii

iXY xxfxgxxfyyyfy (6-12)

.)()()( )()( dxxfxgdyyfyXgEYE XY (6-13)

).()()( ii

i xXPxgYE (6-14)

)( yfY

,2XY

)(YE

).(XgY

,0y

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.

!)!1(

!!!

!)1(

)!1(

!!)(

2

0

1

1

10 0

0

1

1

1

2

0

2

0

22

eee

em

eei

e

ei

ieii

ie

iie

kke

kke

kekkXPkXE

m

m

i

i

i

i

i i

ii

i

i

k

k

k

k

k

k

k

(6-15)

In general, is known as the kth moment of r.v X. Thus if its second moment is given by (6-15).,)( PX

kXE

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Mean alone will not be able to truly represent the p.d.f of any r.v. To illustrate this, consider the following scenario: Consider two Gaussian r.vs and Both of them have the same mean However, as Fig. 6.1 shows, their p.d.fs are quite different. One is more concentrated around the mean, whereas the other one has a wider spread. Clearly, we need atleast an additional parameter to measure this spread around the mean!

(0,1) 1 NX (0,10). 2 NX

.0

Fig.6.1

)( 11xf X

1x

12 (a)

)( 22xf X

2x

102 (b)

)( 2X

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For a r.v X with mean represents the deviation of the r.v from its mean. Since this deviation can be either positive or negative, consider the quantity and its average value represents the average mean square deviation of X around its mean. Define

With and using (6-13) we get

is known as the variance of the r.v X, and its square root is known as the standard deviation of X. Note that the standard deviation represents the root mean square spread of the r.v X around its mean

X ,

,2 X

][ 2 XE

.0][ 2 2 XEX

(6-16)

2)()( XXg

.0)()(

22

dxxfx XX

(6-17)

2

X

2)( XEX

.

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Expanding (6-17) and using the linearity of the integrals, we get

Alternatively, we can use (6-18) to compute

Thus , for example, returning back to the Poisson r.v in (3-45), using (6-6) and (6-15), we get

Thus for a Poisson r.v, mean and variance are both equal to its parameter

.)(

)( 2)(

)(2)(

22___

2 222

2

2

222

XXXEXEXE

dxxfxdxxfx

dxxfxxXVar

XX

XX

(6-18)

.2

X

.22___

22 2

XXX

(6-19)

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To determine the variance of the normal r.v we can use (6-16). Thus from (3-29)

To simplify (6-20), we can make use of the identity

for a normal p.d.f. This gives

Differentiating both sides of (6-21) with respect to we get

or

.2

1])[()(

2/)(

2

22 22

dxexXEXVar x

(6-20)

),,( 2N

2/)(

2

1

2

1)(

22

dxedxxf xX

2/)( .222

dxe x(6-21)

2/)(3

2

2)( 22

dxex x

,2

1 2

2/)(

2

2 22

dxex x (6-22)

,

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which represents the in (6-20). Thus for a normal r.v as in (3-29)

and the second parameter in infact represents the variance of the Gaussian r.v. As Fig. 6.1 shows the larger the the larger the spread of the p.d.f around its mean. Thus as the variance of a r.v tends to zero, it will begin to concentrate more and more around the mean ultimately behaving like a constant.

Moments: As remarked earlier, in general

are known as the moments of the r.v X, and

),( 2N

)(XVar

2)( XVar (6-23)

,

1 ),( ___

nXEXm nnn

(6-24)

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])[( nn XE (6-25)

are known as the central moments of X. Clearly, the mean and the variance It is easy to relate and Infact

In general, the quantities

are known as the generalized moments of X about a, and

are known as the absolute moments of X.

,1m .22 nm

.n

.)( )(

)(])[(

00

0

knk

n

k

knkn

k

knkn

k

nn

mk

nXE

k

n

Xk

nEXE

(6-26)

])[( naXE (6-27)

]|[| nXE (6-28)

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For example, if then it can be shown that

Direct use of (6-2), (6-13) or (6-14) is often a tedious procedure to compute the mean and variance, and in this context, the notion of the characteristic function can be quite helpful.

Characteristic Function

The characteristic function of a r.v X is defined as

even. ,)1(31

odd, ,0)(

nn

nXE n

n

odd. ),12(,/2!2

even, ,)1(31)|(|

12 knk

nnXE

kk

nn

(6-29)

(6-30)

),,0( 2NX

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Thus and for all

For discrete r.vs the characteristic function reduces to

Thus for example, if as in (3-45), then its characteristic function is given by

Similarly, if X is a binomial r.v as in (3-44), its characteristic function is given by

.)()( dxxfeeE X

jxjXX

(6-31)

,1)0( X 1)( X .

k

jkX kXPe ).()( (6-32)

)( PX

.!

)(

!)( )1(

0 0

jj ee

k k

kjkjk

X eeek

ee

kee (6-33)

.)()()(00

njn

k

knkjn

k

knkjkX qpeqpe

k

nqp

k

ne

(6-34)

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To illustrate the usefulness of the characteristic function of a r.v in computing its moments, first it is necessary to derive the relationship between them. Towards this, from (6-31)

Taking the first derivative of (6-35) with respect to , and letting it to be equal to zero, we get

Similarly, the second derivative of (6-35) gives

.!

)(

!2

)()(1

!

)(

!

)()(

22

2

00

kk

k

k

kk

k

k

kjX

X

k

XEj

XEjXjE

k

XEj

k

XjEeE

(6-35)

.)(1

)(or )()(

00

XX

jXEXjE (6-36)

,)(1

)(0

2

2

22

X

jXE (6-37)

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and repeating this procedure k times, we obtain the kth moment of X to be

We can use (6-36)-(6-38) to compute the mean, variance and other higher order moments of any r.v X. For example, if then from (6-33)

so that from (6-36)

which agrees with (6-6). Differentiating (6-39) one more time, we get

.1 ,)(1

)(0

kj

XEk

Xk

kk

(6-38)

,)(

jeX jeee

j

(6-39)

,)( XE (6-40)

),( PX

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, )()( 22

2

2

jejeX ejejeee

jj

(6-41)

so that from (6-37)

which again agrees with (6-15). Notice that compared to the tedious calculations in (6-6) and (6-15), the efforts involved in (6-39) and (6-41) are very minimal.

We can use the characteristic function of the binomial r.v B(n, p) in (6-34) to obtain its variance. Direct differentiation of (6-34) gives

so that from (6-36), as in (6-7).

,)( 22 XE (6-42)

1)()(

njjX qpejnpe

(6-43)

npXE )(PILLAI

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One more differentiation of (6-43) yields

and using (6-37), we obtain the second moment of the binomial r.v to be

Together with (6-7), (6-18) and (6-45), we obtain the variance of the binomial r.v to be

To obtain the characteristic function of the Gaussian r.v, we can make use of (6-31). Thus if then

22122

2

)()1()()(

njjnjjX qpepenqpeenpj

(6-44)

.)1(1)( 222 npqpnpnnpXE (6-45)

.)()( 22222 22 npqpnnpqpnXEXEX (6-46)

),,( 2NX

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.2

1

2

1

) that so (Let

2

1

2

1

)(Let 2

1)(

)2/(

2/

2

2/

2/))((

2

22

)2(2/

2

2/

2

2/)(

2

222222

222

2222

22

juj

jujuj

jyyjyyjj

xxjX

edueee

duee

juyujy

dyeedyeee

yxdxee

(6-47)

Notice that the characteristic function of a Gaussian r.v itself has the “Gaussian” bell shape. Thus if then

and

),,0( 2NX

,2

1)(

22 2/

2

x

X exf (6-48)

(6-49).)( 2/22 eX

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2/22e

(b)

22 2/ xe

x(a)

Fig. 6.2

From Fig. 6.2, the reverse roles of in and are noteworthy

In some cases, mean and variance may not exist. For example, consider the Cauchy r.v defined in (3-39). With

clearly diverges to infinity. Similarly

2 )(xf X )(X

,)/(

)(22 x

xf X

22

2

22

22 ,1)( dx

xdx

x

xXE

(6-50)

. )1

vs(2

2

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22

.)(

22

dx

x

xXE

To compute (6-51), let us examine its one sided factor

With

indicating that the double sided integral in (6-51) does not converge and is undefined. From (6-50)-(6-52), the mean and variance of a Cauchy r.v are undefined.

We conclude this section with a bound that estimates the dispersion of the r.v beyond a certain interval centered around its mean. Since measures the dispersion of

(6-51)

.

0 22

dx

x

x

tanx

/ 2 / 222 2 2 2 0 0 0

/ 2 / 2

0 0

tan sin sec

sec cos(cos )

log cos log cos ,cos 2

xdx d d

xd

(6-52)

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the r.v X around its mean , we expect this bound to depend on as well.

Chebychev Inequality

Consider an interval of width 2 symmetrically centered around its mean as in Fig. 6.3. What is the probability that X falls outside this interval? We need

2

? || XP (6-53)

2

X

Fig. 6.3

X

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To compute this probability, we can start with the definition of

From (6-54), we obtain the desired probability to be

and (6-55) is known as the chebychev inequality. Interestingly, to compute the above probability bound the knowledge of is not necessary. We only need the variance of the r.v. In particular with in (6-55) we obtain

)(xf X

,|| 2

2

XP

(6-54)

. || )()(

)()()()()(

2

||

2

||

2

||

2

222

XPdxxfdxxf

dxxfxdxxfxXE

x Xx X

x XX

(6-55)

.2

,2 k

.1

|| 2k

kXP (6-56)

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Thus with we get the probability of X being outside the 3 interval around its mean to be 0.111 for any r.v. Obviously this cannot be a tight bound as it includes all r.vs. For example, in the case of a Gaussian r.v, from Table 4.1

which is much tighter than that given by (6-56). Chebychev inequality always underestimates the exact probability.

.0027.03|| XP (6-57)

,3k

)1,0(

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Moment Identities :Suppose X is a discrete random variable that takes

only nonnegative integer values. i.e.,

Then

similarly

,2 ,1 ,0 ,0 )( kpkXP k

0

0 0 1

1

01

)()(

1)()()(

i

k k ki

i

ki

XEiXPi

iXPiXPkXP

2)}1({

)(2

)1()()(

1

1

010

XXEiXP

iikiXPkXPk

i

i

kik

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(6-58)

27

which gives

.)()12()()(1 0

22

i k

kXPkiXPiXE (6-59)

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Birthday Pairing

• A group of n people

• A) the probability that two or more persons will have the same birthday?

• B) the probability that someone in the group will have the birthday that matches yours?

• Solution:– C=“no two persons have the same birthday”

28

29

365)1(

11)1(1)(1)(

)1()1()1(

)(

)

2/)1(/1

1

1

1

1

1

Nxe

eeN

kCPCP

N

k

N

nNNNCP

A

x

NnnNkn

k

n

kn

n

k

For n=23, 5.0)( CP

B) D=“a person not matching your birthday”

Nn

Nnnn

eDP

eNN

NDP

/

/

1)(

)/11()1

()(

30

Equations (6-58) – (6-59) are at times quite useful in simplifying calculations. For example, referring to the Birthday Pairing Problem [Example 2-20., Text], let Xrepresent the minimum number of people in a group fora birthday pair to occur. The probability that “the first n people selected from that group have different birthdays” is given by [P(B) in page 39, Text]

But the event the “the first n people selected have

.)1( 2/)1(1

1

Nnnn

kn eN

kp

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different birthdays” is the same as the event “ X > n.”Hence

Using (6-58), this gives the mean value of X to be

Similarly using (6-59) we get

( 1) / 2( ) .n n NP X n e

2

2 2

( 1) / 2 ( 1/ 4) / 2

1/ 20 0

1/ 2(1/8 ) / 2 (1/8 ) / 2

1/ 2 0

12

( ) ( )

2

1/ 2 24.44.

2

n n N x N

n n

N x N N x N

E X P X n e e dx

e e dx e N e dx

N

(6-60)

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Thus

2

2 2 2

2

0

( 1) / 2 ( 1/ 4) / 2

0 1/ 2

1/ 2(1/8 ) / 2 / 2 ( 1/ 4) / 2

0 0 1/ 2

( ) (2 1) ( )

(2 1) 2 ( 1)

2 2

2 2 12 2 ( )

2 8

1 52 2 1 2 2

4 4779.139.

n

n n N x N

n

N x N x N x N

E X n P X n

n e x e dx

e xe dx xe dx e dx

NN E X

N N N N

82.181))(()()( 22 XEXEXVar

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which gives

Since the standard deviation is quite high compared to the mean value, the actual number of people required for a birthday coincidence could be anywhere from 25 to 40.

Identities similar to (6-58)-(6-59) can be derived in the case of continuous random variables as well. For example,if X is a nonnegative random variable with density functionfX (x) and distribution function FX (X), then

.48.13 X

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0 0 0

0 0 0

0 0

{ } ( ) ( )

( ) ( ) ( )

{1 ( )} ( ) ,

X X

X

X

x

y

E X x f x dx dy f x dx

f x dx dy P X y dy P X x dx

F x dx R x dx

(6-61)

34

where

Similarly

( ) 1 ( ) 0, 0.XR x F x x

2 2

0 0 0

0

0.

{ } ( ) 2 ( )

2 ( )

2 ( )

X X

X

x

y

E X x f x dx ydy f x dx

f x dx ydy

x R x dx