lect6a
TRANSCRIPT
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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 ofits 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 f X
)( x f 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 avery 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
ii
i
ii
iiii
iii X
x X P x p x
dx x x p xdx x x p x X E X δδη
(6-3)
),,( baU X
(6-4)∫ +
=−
−=
−=
−=
b
a
b
a
ba
ab
ab x
abdx
ab
x X E
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 meanvalue 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,)( λλ
λλ dy yedxe
x X E y x (6-5)
λ
λ
λ
.!)!1(
!!)()(
01
100
λλλλλ
λλ
λλλλ
λλ
===−
=
====
−∞
=
−∞
=
−
∞
=
−∞
=
−∞
=
∑∑
∑∑∑
eei
ek
e
k k e
k kek 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()!(
!
!)!(
!)()(
111
01
100
npq pnpq piin
n
npq pk k n
n
q pk k n
nk q p
k
nk k X kP X E
ninin
i
k nk n
k
k nk n
k
k nk n
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
µπσ
µπσ
µπσπσ
σσ
σσµ
=⋅+=
+==
∫∫
∫∫∞+
∞−
−∞+
∞−
−
∞+
∞−
−∞+
∞−
−−
dyedy ye
dye ydx 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 isthe quantity of interest. Recall that for any y,
where represent the multiple solutions of the equation
But(6-10) can be rewritten as
),( 2σµ N X
),( x f X X )( X g Y =).( y f Y
Y µ
∫ ∞+
∞−==
.)()( dy y f yY E Y Y µ (6-9)
),(Y E
).( y f Y )(Y E 0>∆ y
( ) ( ) ,∑ ∆+≤
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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 x x x ∆+ ,
,)()()()( ii
i X ii
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)
).()()( ii
i x X P x g Y E == ∑ (6-14))( y f Y
,2 X Y =
)(Y E
).( X g Y =
,0→∆ y
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( )
( ) . !)!1(
!!!
!)1(
)!1(
!!)(
2
0
1
1
10 0
0
1
1
1
2
0
2
0
22
λλλλ
λλ
λλ
λλλλλ
λλ
λλ
λλλ
λλλλ
λλλ
λλ
λλ
+=+=
+=
+
−=
+=
+=
+=−
=
====
−
∞
=
+−
∞
=
−
∞
=
−
∞
=
∞
=
−
∞
=
+−
∞
=
−
∞
=
−∞
=
−∞
=
∑∑
∑∑ ∑
∑∑
∑∑∑
eee
em
eei
e
ei
ieii
ie
iie
k k e
k k e
k ek k X P k X E
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 k th 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 moreconcentrated 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
.0=µ
Fig.6.1
)( 11 x f X
1 x
1
2
=σ(a)
)( 22 x f X
2 x
102 =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 itsaverage 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 squareroot 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
µµ − X ,
( ) ,2µ− X
( ) ][ 2µ− X E
( ) .0][ 22 >−= µσ X E X
(6-16)
2)()( µ−= X X g
.0)()(
22 >−= ∫ ∞+
∞−dx x f x X X µσ
(6-17)
2
X
σ2)( µσ −= 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 computeThus , 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 ___
2222
2
2
222
X X X E X E X E
dx x f xdx x f x
dx x f x x X Var
X X
X X
−=−=−=
+−=
+−==
∫∫
∫∞+
∞−
∞+
∞−
∞+
∞−
µ
µµ
µµσ
(6-18)
.2 X
σ
( ) .22 ___
22 2 λλλλσ =−+=−= X X X
(6-19)
.λPILLAI
T d t i th i f th l )( 2N
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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
( ) .21])[()(
2/)(
222
22
∫ ∞+
∞−−−−=−= dxe x X E X Var x σµπσµµ (6-20)
),,( 2σµ N
∫∫ ∞+∞−−−∞+
∞−== 2/)(
2
1
2
1)(22
dxedx x f x X σµ
πσ
∫ ∞+
∞−
−− =
2/)( .222
σπσµ dxe x (6-21)
∫ ∞+
∞−
−− =−
2/)(
3
2
2)( 22
πσ
µ σµ dxe x x
( ) ,2
1 2
2/)(
2
2 22 σπσ
µ σµ =−∫ ∞+
∞−
−− dxe 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
),( 2σµ N
)( X Var
2)( σ= X Var (6-23)
,σ
1 ),( ___
≥== n X E X m nnn(6-24)
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])[( nn X E µµ −= (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=µ .22 µσ = nm
.nµ
( ) .)( )(
)(])[(
00
0
k n
k
n
k
k nk n
k
k nk
n
k
nn
m
k
n X E
k
n
X k n E X E
−
=
−
=
−
=
−
=−
=
−
=−=
∑∑
∑
µµ
µµµ
(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(31
odd, ,0)(
nn
n X E
n
n
σ
+=−⋅=
+ odd. ),12(,/2!2
even, ,)1(31)|(|
12 k nk
nn X E
k k
nn
πσσ
(6-29)
(6-30)
),,0( 2σ 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), itscharacteristic function is given by
( ) ∫+∞
∞−==Φ .)()( dx x f ee E X
jx jX
X
ωωω (6-31)
,1)0( =Φ X 1)( ≤Φ ω X .ω
∑ ==Φk
jk
X k X P e ).()( ωω (6-32)
)( λ P X
.!
)(
!)( )1(
0 0
−−∞
=
∞
=
−− ====Φ ∑ ∑ ωω λλλ
ωλλω λλω
j j ee
k k
k jk jk
X eeek
ee
k ee (6-33)
.)()()(00
n j
n
k
k nk j
n
k
k nk jk X q peq pe
k nq p
k ne +=
=
=Φ ∑∑ =
−
=
− ωωωω (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
!
)(
!
)()(
22
2
00
+++++=
=
==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
ωωω
ωω
ω ω
(6-35)
.)(1
)(or)()(
00 == ∂Φ∂
==∂Φ∂
ωω ωω
ωω
X X
j X E X jE (6-36)
,)(1)(0
2
2
2
2
=∂Φ∂=
ωω
ω X j
X E (6-37)
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and repeating this procedure k times, we obtain the k th
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
≥∂Φ∂=
=
k j
X E k
X k
k
k
ωω
ω (6-38)
,)( ωλλ λωω ω
je X jeee j
−=∂Φ∂ (6-39)
,)( λ= X E (6-40)
),( λ P X
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( ),)()( 222
2ωλωλλ λλ
ωω ωω je je X e je jeee
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
of (6-34) gives
so that from (6-36), as in (6-7).
,)( 22 λλ += X E (6-42)
1)()( −+=
∂Φ∂ n j j X q pe jnpe ωω
ωω
(6-43)
np X E =)(PILLAI
O diff i i f (6 43) i ld
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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
( )
2212
2
2
)()1()()( −− +−++=
∂
Φ∂ n j jn j j X q pe penq peenp j ωωωω
ω
ω
(6-44)
( ) .)1(1)( 222 npq pn pnnp X E +=−+= (6-45)
[ ] .)()( 2222222 npq pnnpq pn X E X E X =−+=−=σ (6-46)
),,( 2σµ N X
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1 22∞+∫
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.2
1
2
1
)thatso (Let
2
1
2
1
)(Let2
1)(
)2/(
2/
2
2/
2/))((
2
22
)2(2/
2
2/
2
2/)(
2
222222
222
2222
22
ωσµωσωσµω
σωσωσµω
ωσσµωσωµω
σµω
πσ
πσ
ωσωσ
πσπσ
µπσ
ω
−∞+
∞−
−−
∞+
∞−
−+−
∞+
∞−
−−∞+
∞−
−
∞+
∞−
−−
==
=
+==−
==
=−=Φ
∫
∫
∫∫
∫
ju j
ju ju j
j y y j y y j j
x x j
X
edueee
duee
ju yu j y
dyeedyeee
y xdxee
(6-47)
Notice that the characteristic function of a Gaussian r.v itself
has the “Gaussian” bell shape. Thus if ∼ then
and
),,0( 2σ N X
,2
1)(
22 2/
2
σ
πσ x
X e x f −= (6-48)
(6-49).)( 2/22ωσω −=Φ e X
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2/22ωσ−e
ω(b)
2/ σ xe−
x
(a)
Fig. 6.2
From Fig. 6.2, the reverse roles of in and arenoteworthy
In some cases, mean and variance may not exist. Forexample, consider the Cauchy r.v defined in (3-39). With
clearly diverges to infinity. Similarly
2σ )( x f X )(ω X Φ
,
)/(
)( 22 x x f X += α
πα
∫∫ ∞+
∞−
∞+
∞−∞=
+−=
+=
22
2
22
22 ,1)( dx
xdx
x
x X E
αα
πα
απα
(6-50)
.)1
vs(2
2
σσ
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.)(
22∫ ∞+
∞− += dx
x
x X E
απα
(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
.
0 22∫ ∞+
+ dx
x
x
α θα tan= x
/ 2 / 22
2 2 2 2 0 0 0
/ 2 / 2
0 0
tan sin secsec cos
(cos ) log cos log cos ,
cos 2
x dx d d x
d
π π
π π
α θ θα θ θ θα α θ θ
θ πθ
θ
+ ∞= =
+
= − = − = − = ∞
∫ ∫ ∫
∫ (6-52)
2σPILLAI
h d i hi b d
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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
2σ
( ) ? || εµ ≥− X P (6-53)
µε2
εµ − εµ + X
Fig. 6.3
X
PILLAI
To compute this probability we can start with the definition
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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 thevariance of the r.v. In particular with in (6-55) we
obtain
)( x f X
( ) ,||2
2
ε
σεµ ≤≥− X P
(6-54)
[ ]
( ) .||)()(
)()()()()(
2
||
2
||
2
||
2
222
εµεεε
µµµσ
εµεµ
εµ
≥−≥≥≥
−≥−=−=
∫∫∫∫
≥−≥−
≥−∞+
∞−
X P dx x f dx x f
dx x f xdx x f x X E
x X
x X
x X X
(6-55)
.2σ
,2
σσε k =
( ) .1|| 2k k X P ≤≥− σµ (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). Chebychevinequality always underestimates the exact probability.
( ) .0027.03|| =≥ σ X P (6-57)
,3=k
)1,0( == σµ
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Moment Identities :
Suppose X is a discrete random variable that takesonly nonnegative integer values. i.e.,
Then
similarly
,2,1,0 ,0)( =≥== k pk X P k
∑∑ ∑ ∑ ∑∑∞
=
∞
=
∞
=
∞
+=
−
=
∞
=
===
====>
0
0 0 1
1
01
)()(
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
010
−==
−===>
∑∑∑∑
∞
=
−
=
∞
=
∞
=
X X E i X P
iik i X P k X P k
i
i
k ik
PILLAI
(6-58)
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which gives
Equations (6-58) – (6-59) are at times quite useful insimplifying 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
.)()12()()(1 0
22
∑ ∑∞
=
∞
=>+===
i k
k X P k i X P i X E (6-59)
.)1( 2/)1(1
1
N nnn
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
0 0
1/ 2(1/8 ) / 2 (1/ 8 ) / 2
1/ 2 0
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)
PILLAI
∞
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29
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 1
2 2 ( )2 8
1 52 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))(()()( 22 =−= X E X E X Var
PILLAI
which gives
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30
g
Since the standard deviation is quite high compared to themean 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
f X ( x) and distribution function F X ( X ), then
.48.13 ≈ X
σ
PILLAI
( )( )
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)
where
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31
where
Similarly
( ) 1 ( ) 0, 0. X
R 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
∞ ∞
∞ ∞
∞
= =∫ ∫ ∫= ∫ ∫
= ∫
A B b ll T i i (P t R d Di i )
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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 willrepeat 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
t i l F (5 133) (5 134) t t t th b bilit f
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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. PeteRose’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 nn p p ,,1 −+−= αα (6-62)
k r k
r n
k
k kr n
r n qp )()1()1(/
0
, −= ∑+
=
−α
PILLAI
(6-63)
76399.00.303)-(1-1
game)hit /P(no-1
game)hit /oneleastat(
4
==
== P p
(6-64)
Substituting this value for p into the expressions
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34
Subs u g s v ue o p o e e p ess o s
(6-62)-(6-63) with r = 44 and n = 162, we can compute the
desired probability pn. However since n is quite largecompared to r , the above formula is hopelessly time
consuming in its implementation, and it is preferable to
obtain 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 thedenominator of Here
PILLAI
)( z φ
,
1
1)(
11 ∑
=+
−=
+−
−=
r
k k
k
r r
r r
z z
a
z qp z
z p z φ
).( z φ
(6-65)
nn pq −=1
r r zzzp ))(1(
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35
r r
k
r r r r
z z
r r
k
z z k
z qpr
z z z rp z p
z qp z
z z z pa
k
k
)1(1
)()1(lim
1
))(1( lim
1
1
++−−−−
=
+−−−
=
−
→
+→
PILLAI
From (6-65) – (6-66)
where
r k z qpr
z pa
r
k
r
r
k
r
k ,,2,1 ,)1(1
1=
+−−=
(6-66)
(6-67)∑∑ ∑∑ ∞
=
∞
= =
+−
=
==
−−
=00 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 qpr
z p
a A )1(1
1
+−
−
=−=
or
∆
and
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36
PILLAI
and
However (fortunately), the roots in
(6-65)-(6-67) are all not of the same importance (in terms
of 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 positivemaximum at z 0 given by
.11
)1(∑=
+−=−= r
k
n
k k nn z A pq (6-68)
r k z k
,1,2, , =
0)1( →+− nk
z ,1|| >k z
11)( +−−= r r z qp z z A
,01)0(
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37
which gives
There onwards A( z ) decreases to Thus there are two
positive roots for the equation given by
and Since but negative, bycontinuity has the form (see Fig 6.1)
PILLAI
,0)1(1)(
0
0
=+−=− z z
z r qpdz
.)1(
10 +
=r qp
z r
r
(6-69)
.∞−0)( = z A
01 z z <
.1/12 >= p z 0)1( ≈−= r
qp A1
z .0 ,11
>+= εε z
Fig 6.1 A( z ) for r odd
)( z A
z 1−
1 z 2=1/ p
z 1 z 0
It is possible to obtain a bound for in (6 69) Whenz
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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
0 z
PILLAI
r r p pqp )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).1110r r
r z +=+≥
.11
1
11 21 >=
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39
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.
To show this when r is even, suppose represents the
negative root. Then
α−
0)1()(1
=−+−=− +r r
qp A αααPILLAI
Fig 6.2 A( z ) for r even
)( z A
z z 1 z 0 z 2
α−
so that the function
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40
so a e u c o
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 /110 +≥∞− .10 >>= z x α
2)(1)(1
+=−+= +
x A xqp x x B r r
PILLAI
(6-73)
α
Fig 6.3 Negative root
α
1
)( x B
0)( =α B
z 0 1/ p
Finally if is a complex root of A(z) thenθρ jez =
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41
Finally if is a complex root of A( z ), then
so that
or
Thus from (6-72), belongs to either the interval (0, z 1)
or the interval in Fig 6.1. Moreover , by equating
the imaginary parts in (6-74) we get
ρ e z =
01)( )1(1 =−−= ++ θθθ ρρρ r jr r j j eqpee A (6-74)
1)1(1
1 |1|+++
+≤+= r r r jr r
qpeqp ρρρ θ
.1)1(
sin
sin=
+
θ
θρ
r qp r r
),( 1 ∞ p
ρ
.01)( 1
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42
equality being excluded if Hence from (6-75)-(6-76)
and (6-70)
or
But As a result lies in the interval only.
Thus
ρ.01
z z < ),( 1 ∞ p
,1
)1(
sin
sin
+≤
+r
r
θ
θ
.0≠
r
r
r
r r r
r
r z
qpr
qpr
+>=
+
>⇒>+1
)1(
1 1)1( 0ρρ
.1
10r
z +≥>ρ
.11
>> pρPILLAI
(6-76)
(6-77)
T i th t l t f th l i l
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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 z 1 (which is very close tounity), 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
,11
;0 ,1 21 >=>+= p
z z εε
.11
where >>= p
e z jk ρρ θ
.allforrapidly0
)1(
k z
n
k →
+−
PILLAI
(6-78)
(6-79)
|||| )1()1( ≤ +−+− ∑∑ zAzAr
nr
n
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44
Thus from (6-68), to an excellent approximation
This gives the desired probability to be
.)1(11
+−= nn
z Aq (6-81)
0. 11
|)|()1(
|)|(
|)|()1(1
|)|(1
||||
11
1
2
1
2
2
)(
2
)(
→≤+−=
+≤
+−−
≤
≤
++
+
=
+
=
==
∑
∑
∑∑
q p
q p
r r
p z pqr
z p
p z pqr
z p
z A z A
nn
nr
k r
k
r k
nr
k r
k
r
k
k k k
k k k
(6-80)
PILLAI
)(1 r pz
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45
Notice that since the dominant root z 1 is very close to
unity, an excellent closed form approximation for z 1 can
be obtained by considering the first order Taylor seriesexpansion for A( z ). In the immediate neighborhood of z =1
we get
so that gives
or
.
)()1(1
)(111 )1(1
1
1 +−
+−
−−=−= n
r nn z
pz qr
pz q p (6-82)
PILLAI
εεε ))1(1()1()1()1( r r qpr qp A A A +−+−=′+=+
0)1()(1
=+= ε A z A
,
)1(1
r
r
qpr
qp
+−
=ε
.)(
11 r
r qp z +≈ (6-83)
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46
Returning back to Pete Rose’s case, p = 0.763989, r = 44gives 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
45441)( z qp z z A −−=
.05490000016936.11 = z 05480000016936.11 = z
0002069970.0162
= p (6-84)
)1(11 r qpr +−
(6 83)
hits in 162 games for a player of Pete Rose’s caliber a
ll b bili i d d! I h i i
−
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47
very small probability indeed! In that sense it is a very
rare event.Assuming that during any baseball season there are
on 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
50252 =×
.40401874.0)1(1 501 =−− P
0102975349.0)1(150
1621 =−−= p P
(6-85)
(We have once again used the independence of the 50
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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), rareevents do indeed occur. In other words, some unlikely
event is likely to happen.
However, as (6-84) shows a particular unlikely eventsuch 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!
−−
p ; n = 162r
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49
0.9525710
0.4893315
0.1493720
0.03928250.00020744
pn ; n 162r
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
.0000504532.0=p (6-86)
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50
Even over a 100 year period, with an average of 50excellent 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!
.0000504532.0n
p
PILLAI
(6-86)
(6-87)2229669.0)1(1100
0 =−− P 00251954.0)1(1 500 =−−= n p P