) ( (6-1) ( = P X ) B f ( x ) dx . X B Note that - - PowerPoint PPT Presentation

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• 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

dx x f B X P . ) ( ) (ξ

(6-1)

) (x fX ) (x fX

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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).

∫

∞ + ∞ −

= = = . ) ( ) ( dx x f x X E X

X X

η

(6-2)

. ) ( ) ( ) ( ) (

1

∑ ∑ ∑ ∫ ∫ ∑

= = = − = − = = =

i i i i i i i i i i i i i X

x X P x p x dx x x p x dx x x p x X E X

• δ

δ η

(6-3)

), , ( b a U X

(6-4)

∫

+ = − − = − = − =

b a b a

b a a b a b x a b dx a b x X E 2 ) ( 2 2 1 ) (

2 2 2

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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.

∫ ∫

∞ − − ∞

= = =

/

, ) ( λ λ λ

λ

dy ye dx e x X E

y x

(6-5)

λ λ λ

. ! )! 1 ( ! ! ) ( ) (

1 1

λ λ λ λ λ λ λ

λ λ λ λ λ λ

= = = − = = = = =

− ∞ = − ∞ = − ∞ = − ∞ = − ∞ =

∑ ∑ ∑ ∑ ∑

e e i e k e k k e k ke k X kP X E

i i k k k k k k k

(6-6)

λ

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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 ( )! ( ! ! )! ( ! ) ( ) (

1 1 1 1 1

np q p np q p i i n n np q p k k n n q p k k n n k q p k n k k X kP X E

n i n i n i k n k n k k n k n k k n k n k n k

= + = − − − = − − = − =         = = =

− − − − = − = − = − = =

∑ ∑ ∑ ∑ ∑

(6-7)

. 2 1 2 1 ) ( 2 1 2 1 ) (

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

2 2 2 2 2 2 2 2

µ πσ µ πσ µ πσ πσ

σ σ σ σ µ

= ⋅ + = + = =

∫ ∫ ∫ ∫

∞ + ∞ − − ∞ + ∞ − − ∞ + ∞ − − ∞ + ∞ − − −

• dy

e dy ye dy e y dx xe X E

y y y x

(6-8)

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Thus the first parameter in ∼ is infact 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

) , (

2

σ µ N X ), ( x f X

X

) (X g Y = ). (y fY

Y

µ

∫

∞ + ∞ −

= = . ) ( ) ( dy y f y Y E

Y Y

µ

(6-9)

), (Y E ). (y fY ) (Y E > ∆y

( ) ( ) ,

∑

∆ + ≤ < = ∆ + ≤ <

i i i i

x x X x P y y Y y P

(6-10)

i

x ). (

i

x g y =

, ) ( ) (

i i i X Y

x x f y y f ∆ = ∆

∑

(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)

( )

i i i

x x x ∆ + ,

, ) ( ) ( ) ( ) (

i i i X i i i i X Y

x x f x g x x f y y y f y ∆ = ∆ = ∆

∑ ∑

(6-12)

( )

∫ ∫

∞ + ∞ − ∞ + ∞ −

= = = . ) ( ) ( ) ( ) ( ) ( dx x f x g dy y f y X g E Y E

X Y

(6-13)

). ( ) ( ) (

i i i

x X P x g Y E = = ∑

(6-14)

) (y fY ,

2

X Y = ) (Y E ). (X g Y =

, → ∆y

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( ) ( )

. ! )! 1 ( ! ! ! ! ) 1 ( )! 1 ( ! ! ) (

2 1 1 1 1 1 1 2 2 2 2

λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ

λ λ λ λ λ λ λ λ λ λ λ λ λ λ

+ = + =         + =         + − =         + =         + = + = − = = = = =

− ∞ = + − ∞ = − ∞ = − ∞ = ∞ = − ∞ = + − ∞ = − ∞ = − ∞ = − ∞ =

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

e e e e m e e i e e i i e i i i e i i e k k e k k e k e k k X P k X E

m m i i i i i i i i 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).

, ) ( λ P X

( )

k

X E

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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

N X (0,10).

2

N X . = µ

Fig.6.1

) (

1

1 x

f X

1

x 1

2 =

σ (a)

) (

2

2 x

f X

2

x 10

2 =

σ (b)

) (

2

X

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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

µ µ − X ,

( ) ,

2

µ − X

( ) ]

[

2

µ − X E

( )

. ] [

2 2

> − = µ σ X E

X

(6-16)

2

) ( ) ( µ − = X X g . ) ( ) (

2 2

> − = ∫

∞ + ∞ −

dx x f x

X

X

µ σ

(6-17)

2

X

σ

2

) ( µ σ − = X E

X

. µ

∆

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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 ) (

2 2

___ 2 2 2 2 2 2 2 2 2

X X X E X E X E dx x f x dx x f x dx x f x x X Var

X X X X

− = − = − = + − = + − = =

∫ ∫ ∫

∞ + ∞ − ∞ + ∞ − ∞ + ∞ −

µ µ µ µ µ σ

(6-18)

.

2

X

σ

( )

.

2 2 ___ 2 2

2

λ λ λ λ σ = − + = − = X X

X

(6-19)

. λ

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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

• r

( )

. 2 1 ] ) [( ) (

2 / ) ( 2 2 2

2 2

∫

∞ + ∞ − − −

− = − = dx e x X E X Var

x σ µ

πσ µ µ

(6-20)

), , (

2

σ µ N

∫ ∫

∞ + ∞ − − − ∞ + ∞ −

= =

2 / ) ( 2

1 2 1 ) (

2 2

dx e dx x f

x X σ µ

πσ

∫

∞ + ∞ − − −

=

2 / ) (

. 2

2 2

σ π

σ µ

dx e

x

(6-21)

∫

∞ + ∞ − − −

= −

2 / ) ( 3 2

2 ) (

2 2

π σ µ

σ µ

dx e x

x

( )

, 2 1

2 2 / ) ( 2 2

2 2

σ πσ µ

σ µ

= −

∫

∞ + ∞ − − −

dx e x

x

(6-22)

, σ

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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

) , (

2

σ µ N ) ( X Var

2

) ( σ = X Var

(6-23)

, σ 1 ), (

___

≥ = = n X E X m

n n n

(6-24)

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] ) [(

n n

X E µ µ − =

(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.

,

1

m = µ .

2 2

µ σ =

n

m .

n

µ

( )

. ) ( ) ( ) ( ] ) [(

k n k n k k n k n k k n k n k n n

m k n X E k n X k n E X E

− = − = − =

−         = −         =         −         = − =

∑ ∑ ∑

µ µ µ µ µ

(6-26)

] ) [(

n

a X E −

(6-27)

] | [|

n

X E

(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 ( 3 1

• dd,

, ) ( n n n X E

n n

σ

• 

  + = − ⋅ =

+

• dd.

), 1 2 ( , / 2 ! 2 even, , ) 1 ( 3 1 ) | (|

1 2

k n k n n X E

k k n n

π σ σ

• (6-29)

(6-30) ), , (

2

σ N X

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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

( ) ∫

+∞ ∞ −

= = Φ . ) ( ) ( dx x f e e E

X jx jX X ω ω

ω

(6-31)

, 1 ) ( = Φ X 1 ) ( ≤ Φ ω

X

. ω

∑

= = Φ

k jk X

k X P e ). ( ) (

ω

ω

(6-32)

) ( λ P X

. ! ) ( ! ) (

) 1 ( − − ∞ = ∞ = − −

= = = = Φ

∑ ∑

ω ω

λ λ λ ω λ λ ω

λ λ ω

j j

e e k k k j k jk X

e e e k e e k e e

(6-33)

. ) ( ) ( ) (

n j n k k n k j n k k n k jk X

q pe q pe k n q p k n e + =         =         = Φ

∑ ∑

= − = − ω ω ω

ω

(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 ω, and letting it to be equal to zero, we get Similarly, the second derivative of (6-35) gives

( )

. ! ) ( ! 2 ) ( ) ( 1 ! ) ( ! ) ( ) (

2 2 2

• +

+ + + + = =       = = Φ

∑ ∑

∞ = ∞ = k k k k k k k k k jX X

k X E j X E j X jE k X E j k X j E e E ω ω ω ω ω ω

ω

(6-35)

. ) ( 1 ) (

• r

) ( ) (

= =

∂ Φ ∂ = = ∂ Φ ∂

ω ω

ω ω ω ω

X X

j X E X jE

(6-36)

, ) ( 1 ) (

2 2 2 2 =

∂ Φ ∂ =

ω

ω ω

X

j X E

(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 ) ( ≥ ∂ Φ ∂ =

=

k j X E

k X k k k ω

ω ω

(6-38)

, ) (

ω λ λ

λ ω ω

ω

j e X

je e e

j

−

= ∂ Φ ∂

(6-39)

, ) ( λ = X E

(6-40)

), ( λ P X

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( ),

) ( ) (

2 2 2 2 ω λ ω λ λ

λ λ ω ω

ω ω

j e j e X

e j e je e e

j j

+ = ∂ Φ ∂

−

(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

• f (6-34) gives

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

, ) (

2 2

λ λ + = X E

(6-42)

1

) ( ) (

−

+ = ∂ Φ ∂

n j j X

q pe jnpe

ω ω

ω ω

(6-43)

np X E = ) (

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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

( )

2 2 1 2 2 2

) ( ) 1 ( ) ( ) (

− −

+ − + + = ∂ Φ ∂

n j j n j j X

q pe pe n q pe e np j

ω ω ω ω

ω ω

(6-44)

( )

. ) 1 ( 1 ) (

2 2 2

npq p n p n np X E + = − + =

(6-45)

[ ]

. ) ( ) (

2 2 2 2 2 2 2

npq p n npq p n X E X E

X

= − + = − = σ

(6-46)

), , (

2

σ µ N X

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. 2 1 2 1 ) that so (Let 2 1 2 1 ) (Let 2 1 ) (

) 2 / ( 2 / 2 2 / 2 / ) )( ( 2 2 2 ) 2 ( 2 / 2 2 / 2 2 / ) ( 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

ω σ µω σ ω σ µω σ ω σ ω σ µω ω σ σ µω σ ω µω σ µ ω

πσ πσ ω σ ω σ πσ πσ µ πσ ω

− ∞ + ∞ − − − ∞ + ∞ − − + − ∞ + ∞ − − − ∞ + ∞ − − ∞ + ∞ − − −

= = = + = = − = = = − = Φ

∫ ∫ ∫ ∫ ∫

j u j j u j u j j y y j y y j j x x j X

e du e e e du e e j u y u j y dy e e dy e e e y x dx e e

(6-47)

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

), , (

2

σ N X

, 2 1 ) (

2 2

2 / 2 σ

πσ

x X

e x f

−

=

(6-48) (6-49)

. ) (

2 /

2 2ω

σ

ω

−

= Φ e

X

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2 /

2 2ω

σ −

e

ω (b)

2 2 2

/ σ x

e−

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

σ

) (x f X ) (ω

X

Φ

, ) / ( ) (

2 2

x x f X + = α π α

∫ ∫

∞ + ∞ − ∞ + ∞ −

∞ =         + − = + =

2 2 2 2 2 2 2

, 1 ) ( dx x dx x x X E α α π α α π α

(6-50)

. ) 1 vs (

2 2

σ σ

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. ) (

2 2

∫

∞ + ∞ −

+ = dx x x X E α π α

(6-51)

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

.

2 2

∫

∞ +

+ dx x x α

θ α tan = x

/ 2 / 2 2 2 2 2 2 / 2 / 2

tan sin sec sec cos (cos ) log cos log cos , cos 2 x dx d d x d

π π π π

α θ θ α θ θ θ α α θ θ θ π θ θ

+∞

= = + = − = − = − = ∞

∫ ∫ ∫ ∫

(6-52)

2

σ

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the r.v X around its mean µ , we expect this bound to depend on as well. Chebychev Inequality Consider an interval of width 2ε symmetrically centered around its mean µ as in Fig. 6.3. What is the probability that X falls outside this interval? We need

2

σ

( ) ?

| | ε µ ≥ − X P

(6-53)

µ

ε 2

ε µ − ε µ + X

• Fig. 6.3

X

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To compute this probability, we can start with the definition

• f

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

• btain

) (x f X

( )

, | |

2 2

ε σ ε µ ≤ ≥ − X P

(6-54)

[ ]

( ).

| | ) ( ) ( ) ( ) ( ) ( ) ( ) (

2 | | 2 | | 2 | | 2 2 2 2

ε µ ε ε ε µ µ µ σ

ε µ ε µ ε µ

≥ − ≥ ≥ ≥ − ≥ − = − =

∫ ∫ ∫ ∫

≥ − ≥ − ≥ − ∞ + ∞ −

X P dx x f dx x f dx x f x dx x f x X E

x X x X x X X

(6-55)

.

2

σ ,

2

σ

σ ε k =

( )

. 1 | |

2

k k X P ≤ ≥ − σ µ

(6-56)

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Thus with we get the probability of X being outside the 3σ 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 . 3 | | = ≥ σ X P

(6-57) , 3 = k

) 1 , ( = = σ µ

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

• nly nonnegative integer values. i.e.,

Then similarly

• ,

2 , 1 , , ) ( = ≥ = = k p k X P

k

∑ ∑ ∑ ∑ ∑ ∑

∞ = ∞ = ∞ = ∞ + = − = ∞ =

= = = = = = = >

1 1 1

) ( ) ( 1 ) ( ) ( ) (

i k k k i i k i

X E i X P i i X P i X P k X P 2 )} 1 ( { ) ( 2 ) 1 ( ) ( ) (

1 1 1

− = = − = = = >

∑ ∑ ∑ ∑

∞ = − = ∞ = ∞ =

X X E i X P i i k i X P k X P k

i i k i k

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(6-58)

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which gives 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 X represent the minimum number of people in a group for a 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 2 2

∑ ∑

∞ = ∞ =

> + = = =

i k

k X P k i X P i X E (6-59) . ) 1 (

2 / ) 1 ( 1 1 N n n n k n

e

N k

p

− − − =

≈ − ∏ =

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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 N

P X n e−

−

> ≈

{ }

2 2 2

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

1 2

( ) ( ) 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 π π

∞ ∞ ∞ − − − − − = = ∞ − − −

= > ≈ ≈ = = + ≈ + =

∑ ∑ ∫ ∫ ∫

(6-60)

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Thus

2 2 2 2

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

( ) (2 1) ( ) (2 1) 2 ( 1) 2 2 2 2 1 2 2 ( ) 2 8 1 5 2 2 1 2 2 4 4 779.139.

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 N N E X N N N N π π π π

∞ = ∞ ∞ − − − − = − ∞ ∞ − − − − −

= + > = + = +   = + +       = + +     = + + + = + + =

∑ ∑ ∫ ∫ ∫ ∫

82 . 181 )) ( ( ) ( ) (

2 2

= − = X E X E X Var

PILLAI

SLIDE 30

30

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 function fX (x) and distribution function FX (X), then . 48 . 13 ≈

X

σ

PILLAI

( ) ( )

{ } ( ) ( ) ( ) ( ) ( ) {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)

SLIDE 31

31

where Similarly ( ) 1 ( ) 0, 0.

X

R x F x x = − ≥ >

( ) ( )

2 2

.

{ } ( ) 2 ( ) 2 ( ) 2 ( )

X X X

x y

E X x f x dx ydy f x dx f x dx ydy xR x dx

∞ ∞ ∞ ∞ ∞

= = ∫ ∫ ∫ = ∫ ∫ = ∫

SLIDE 32

32

A Baseball Trivia (Pete Rose and Dimaggio): In 1978 Pete Rose set a national league record by hitting a string of 44 games during a 162 game baseball

• season. How unusual was that event?

As we shall see, that indeed was a rare event. In that context, we will answer the following question: What is the probability that someone in major league baseball will repeat that performance and possibly set a new record in the next 50 year period? The answer will put Pete Rose’s accomplishment in the proper perspective. Solution: As example 5-32 (Text) shows consecutive successes in n trials correspond to a run of length r in n

PILLAI

SLIDE 33

33

• trials. From (5-133)-(5-134) text, we get the probability of

r successive hits in n games to be where and p represents the probability of a hit in a game. Pete Rose’s batting average is 0.303, and on the average since a batter shows up about four times/game, we get

r r n r r n n

p p

, ,

1

−

+ − = α α (6-62)

 

k r k r n k k kr n r n

qp ) ( ) 1 (

) 1 ( / ,

− = ∑

+ =       −

α

PILLAI

(6-63)

76399 . 0.303)

• (1
• 1

game) hit / P(no

• 1

game) hit /

• ne

least at (

4 =

= = = P p

(6-64)

SLIDE 34

34

Substituting this value for p into the expressions (6-62)-(6-63) with r = 44 and n = 162, we can compute the desired probability pn. However since n is quite large compared to r, the above formula is hopelessly time consuming in its implementation, and it is preferable to

• btain a good approximation for pn.

Towards this, notice that the corresponding moment generating function for in Eq. (5-130) Text, is rational and hence it can be expanded in partial fraction as where only r roots (out of r +1) are accounted for, since the root z = 1/p is common to both the numerator and the denominator of Here

PILLAI

) (z φ

, 1 1 ) (

1 1

∑

= +

− = + − − =

r k k k r r r r

z z a z qp z z p z φ

). (z φ

(6-65)

n n

p q − =1

SLIDE 35

35 r r k r r r r z z r r k r r z z k

z qp r z z z rp z p z qp z z z z p a

k k

) 1 ( 1 ) ( ) 1 ( lim 1 ) )( 1 ( lim

1 1

+ + − − − − = + − − − =

− → + →

PILLAI

From (6-65) – (6-66) where r k z qp r z p a

r k r r k r k

, , 2 , 1 , ) 1 ( 1 1

• =

+ − − =

(6-66) (6-67)

∑ ∑ ∑ ∑

∞ = ∞ = = + − =

= = − − =

1 ) 1 ( 1

) ( / 1 1 ) ( ) (

n n n n n q r k n k k k r k k k

z q z z A z z z a z

n

• φ

r k r r k r k k

z qp r z p a A ) 1 ( 1 1 + − − = − =

• r

∆

SLIDE 36

36

PILLAI

and However (fortunately), the roots in (6-65)-(6-67) are all not of the same importance (in terms

• f their relative magnitude with respect to unity). Notice

that since for large n, for only the roots nearest to unity contribute to (6-68) as n becomes larger. To examine the nature of the roots of the denominator in (6-65), note that (refer to Fig 6.1) implying that for increases from –1 and reaches a positive maximum at z0 given by . 1

1 ) 1 (

∑

= + −

= − =

r k n k k n n

z A p q

(6-68)

r k zk , 1,2, ,

• =

) 1 (

→

+ − n k

z

, 1 | | >

k

z

1

1 ) (

+

− − =

r rz

qp z z A , 1 ) ( < − = A ) ( , ) / 1 ( ), ( ) 1 ( < ∞ = > − = A p A A qp A

r

) ( , z A z ≥

SLIDE 37

37

which gives There onwards A(z) decreases to Thus there are two positive roots for the equation given by and Since but negative, by continuity has the form (see Fig 6.1)

PILLAI

, ) 1 ( 1 ) ( = + − =

− r r z z

z r qp dz z dA . ) 1 ( 1 + = r qp z

r r

(6-69)

. ∞ −

) ( = z A

1

z z < . 1 / 1

2

> = p z

) 1 ( ≈ − =

r

qp A

1

z . , 1

1

> + = ε ε z

Fig 6.1 A(z) for r odd ) (z A z

1 − 1 z2=1/p z1 z0

SLIDE 38

38

It is possible to obtain a bound for in (6-69). When P varies from 0 to 1, the maximum of is attained for and it equals Thus and hence substituting this into (6-69), we get Hence it follows that the two positive roots of A(z) satisfy Clearly, the remaining roots of are complex if r is z

PILLAI r r

p p qp ) 1 ( − = ) 1 /( + = r r p . ) 1 /(

1 +

+

r r

r r

1

) 1 (

+

+ ≤

r r r

r r qp

(6-70) (6-72) (6-71)

. 1 1 1 r r r z + = + ≥ . 1 1 1 1 1

2 1

> = < + < < p z r z ) (z A

SLIDE 39

39

• dd , and there is one negative root if r is even (see

Fig 6.2). It is easy to show that the absolute value of every such complex or negative root is greater than 1/p >1. α − To show this when r is even, suppose represents the negative root. Then α − ) 1 ( ) (

1 =

− + − = −

+ r r

qp A α α α

PILLAI

Fig 6.2 A(z) for r even ) (z A z

z1 z0 z2 α

−

SLIDE 40

40

so that the function starts positive, for x > 0 and increases till it reaches once again maximum at and then decreases to through the root Since B(1/p) = 2, we get > 1/p > 1, which proves our claim. r z / 1 1 + ≥ ∞ − . 1

0 >

> = z x α 2 ) ( 1 ) (

1

+ = − + =

+

x A x qp x x B

r r PILLAI

(6-73)

α

Fig 6.3 Negative root

α

1

) (x B ) ( = α B

z0

1/p

SLIDE 41

41

Finally if is a complex root of A(z), then so that

• r

Thus from (6-72), belongs to either the interval (0, z1)

• r the interval in Fig 6.1. Moreover , by equating

the imaginary parts in (6-74) we get

θ

ρ

j

e z =

1 ) (

) 1 ( 1

= − − =

+ + θ θ θ

ρ ρ ρ

r j r r j j

e qp e e A (6-74)

1 ) 1 ( 1

1 | 1 |

+ + +

+ ≤ + =

r r r j r r

qp e qp ρ ρ ρ

θ

. 1 ) 1 (

sin sin

= + θ θ ρ r qp

r r

) , ( 1 ∞

p

ρ

. 1 ) (

1 <

− − =

+ r r

qp A ρ ρ ρ

PILLAI

(6-75)

SLIDE 42

42

But equality being excluded if Hence from (6-75)-(6-76) and (6-70)

• r

But As a result lies in the interval only. Thus ρ .

1

z z < ) , ( 1 ∞

p

, 1 ) 1 (

sin sin

+ ≤ + r r θ θ . ≠ θ

r r r r r r

r r z qp r qp r       + > = + > ⇒ > + 1 ) 1 ( 1 1 ) 1 ( ρ ρ . 1 1 r z + ≥ > ρ . 1 1 > > p ρ

PILLAI

(6-76) (6-77)

SLIDE 43

43

To summarize the two real roots of the polynomial A(z) are given by and all other roots are (negative or complex) of the form Hence except for the first root z1 (which is very close to unity), for all other roots As a result, the most dominant term in (6-68) is the first term, and the contributions from all other terms to qn in (6-68) can be bounded by , 1 1 ; , 1

2 1

> = > + = p z z ε ε . 1 1 where > > = p e z

j k

ρ ρ

θ

. all for rapidly

) 1 (

k z

n k

→

+ − PILLAI

(6-78) (6-79)

SLIDE 44

44

Thus from (6-68), to an excellent approximation This gives the desired probability to be .

) 1 ( 1 1 + −

=

n n

z A q

(6-81)

0. 1 1 |) | ( ) 1 ( |) | ( |) | ( ) 1 ( 1 |) | ( 1 | || |

1 1 1 2 1 2 2 ) 1 ( 2 ) 1 (

→ ≤ + − = + ≤ + − − ≤ ≤

+ + + = + = = + − = + −

∑ ∑ ∑ ∑

q p q p r r p z p q r z p p z p q r z p z A z A

n n n r k r k r k n r k r k r k r k n k k r k n k k

(6-80)

PILLAI

SLIDE 45

45

Notice that since the dominant root z1 is very close to unity, an excellent closed form approximation for z1 can be obtained by considering the first order Taylor series expansion for A(z). In the immediate neighborhood of z =1 we get so that gives

• r

. ) ( ) 1 ( 1 ) ( 1 1 1

) 1 ( 1 1 1 + −

      + − − − = − =

n r r n n

z pz q r pz q p

(6-82)

PILLAI

ε ε ε ) ) 1 ( 1 ( ) 1 ( ) 1 ( ) 1 (

r r

qp r qp A A A + − + − = ′ + = + ) 1 ( ) ( 1 = + = ε A z A , ) 1 ( 1

r r

qp r qp + − = ε

SLIDE 46

46

Returning back to Pete Rose’s case, p = 0.763989, r = 44 gives the smallest positive root of the denominator polynomial to be (The approximation (6-83) gives ). Thus with n = 162 in (6-82) we get to be the probability for scoring 44 or more consecutive

45 44

1 ) ( z qp z z A − − = . 0549 0000016936 . 1

1 =

z 0548 0000016936 . 1

1 =

z 0002069970 .

162 =

p

(6-84)

. ) 1 ( 1 1

1 r r

qp r qp z + − + ≈

(6-83)

SLIDE 47

47

hits in 162 games for a player of Pete Rose’s caliber a very small probability indeed! In that sense it is a very rare event. Assuming that during any baseball season there are

• n the average about (?) such players over

all major league baseball teams, we obtain [use Lecture #2, Eqs.(2-3)-(2-6) for the independence of 50 players] to be the probability that one of those players will hit the desired event. If we consider a period of 50 years, then the probability of some player hitting 44 or more consecutive games during one of these game seasons turns out to be −

PILLAI

50 25 2 = × . 40401874 . ) 1 ( 1

50 1

= − − P 0102975349 . ) 1 ( 1

50 162 1

= − − = p P

(6-85)

SLIDE 48

48

(We have once again used the independence of the 50 seasons.) Thus Pete Rose’s 44 hit performance has a 60-40 chance of survival for about 50 years.From (6-85), rare events do indeed occur. In other words, some unlikely event is likely to happen. However, as (6-84) shows a particular unlikely event such as Pete Rose hitting 44 games in a sequence is indeed rare. Table 6.1 lists p162 for various values of r. From there, every reasonable batter should be able to hit at least 10 to 12 consecutive games during every season! − −

SLIDE 49

49

0.95257 10 0.48933 15 0.14937 20 0.03928 25 0.000207 44 pn ; n = 162 r Table 6.1 Probability of r runs in n trials for p=0.76399.

As baseball fans well know, Dimaggio holds the record of consecutive game hitting streak at 56 games (1941). With a lifetime batting average of 0.325 for Dimaggio, the above calculations yield [use (6-64), (6-82)-(6-83)] the probability for that event to be

SLIDE 50

50

Even over a 100 year period, with an average of 50 excellent hitters / season, the probability is only (where ) that someone will repeat or outdo Dimaggio’s performance.Remember, 60 years have already passed by, and no one has done it yet! . 0000504532 . =

n

p

PILLAI

(6-86) (6-87)

2229669 . ) 1 ( 1

100

= − − P 00251954 . ) 1 ( 1

50

= − − =

n

p P