<a name="docA001">
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class  　★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005">　Index begin　Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□　Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 stop



<a name="ch05c001">　Index begin　Index this file
2009-11-27-19-33 start
■　Exercise 5.7 problem statement
　　textbook page 84
（Rearrangement, Cyclic Shifts and the AM-GM）

Skillful use of the rearrangement 
inequality often calls for one to
exploit symmetry and to look for
clever specializations of the resulting
bounds. This problem outlines a proof 
of the AM-GM ineqaulity that nicely
illustrates these steps.

<a name="ch05c002">
(a) Show that for positive ck 
    k=1,2,...,n one has

2009-11-27-19-48 here
<a name="ch05c003">
(b) Specialize the result of part (a)
    to show that for all positive xk,
    k=1,2,...,n, one has the rational
    bound

<a name="ch05c004">
(c) Specialize a third time to show that
    for ρ＞0 one also has

and finally indicate how the right choice
of ρ now yields the AM-GM inequality
(2.3)
2009-11-27-20-00 here




<a name="ch05c005">　Index begin　Index this file
2009-11-27-20-02
■　Exercise 5.7 hint
　　textbook page 246
Since the sequence {ck} and {1/ck} are
oppositely ordered, the rearrangement 
inequality (5.12) tells us that for
any permutation σ one has
  n≦c1/cσ(1)+c2/cσ(2)+...+cn/cσ(n) ---eqn.AO004
and part (a) is a special case of 
this observation. 
<a name="ch05c006">
If we set 
  ck=x1x2...xk ---eqn.AO005
in part (a) we get part (b), and if
we then replace xk by ρxk, we get
part (c). Finally, by setting
  ρ=(x1x2...xn)-n ---eqn.AO006
and simplifying, we get the AM-GM
bound.
2009-11-27-20-13 stop



<a name="ch05c007">
2009-11-27-20-51 start
■　Exercise 5.7 solution


First see numerical example. Assume
  {ck}={c1,c2,c3}={1,2,3} ---eqn.AO005
  {1/ck}={1/1,1/2,1/3} ---eqn.AO006
eqn.AO005 is monotone increase.
eqn.AO006 is monotone decrease.
<a name="ch05c008">
Rearrangement Inequality (5.12) tell 
us monotone increase and monotone 
decrease two sequences same index
terms product first, then sum them
this result is smallest possible.
eqn.AO005 and eqn.AO006 are reverse 
order their product sum
 1/1 + 2/2 + 3/3 = 3 ---eqn.AO007
is smallest. Random order (other
order) create greater product sum.
<a name="ch05c009">
For example
keep {1,2,3} as is, but change
{1/1,1/2,1/3} to {1/3,1/1,1/2}. 
that is change reverse order to 
random order. Now 
let  {1 ,  2,  3} and
    {1/3,1/1,1/2} build product sum.
get 1*(1/3) + 2*(1/1) + 3*(1/2)=3.8333
random  order answer 3.8333 is ≧
reverse order answer 3.0

<a name="ch05c010">　Index begin　Index this file
After numerical example, it is easy
to understand that
   {ck} ={ c1 , c2 , c3 } ---eqn.AO005
  {1/ck}={1/c1,1/c2,1/c3} ---eqn.AO006
created product sum
  c1*(1/c1)+c2*(1/c2)+c3*(1/c3)=n=3 ---eqn.AO007
has smallest value. Any randomized
permutation σ creates product sum
must be ≧ the minimum value n
This statement is eqn.AO004

<a name="ch05c011">
Write in clear mathematical equation,
reverse order created minimum product
sum is

<a name="ch05c012">
2009-11-27-21-24 here
eqn.AO004 used "σ(k)" in denominator 1/cσ(k)
represent random order. eqn.AO001 is one
of those random order.
Please pay attention 
to  eqn.AO001 random order
and eqn.AO008 reverse order
They are different.
rearrangement inequality tell us that
  random order product sum
  ≧ 
  reverse order product sum 
then eqn.AO001 is proved.

<a name="ch05c013">
2009-11-27-21-37 here
Next goto part (b). Let
  ck=x1x2...xk ---eqn.AO005
Put ck into eqn.AO001 get

<a name="ch05c014">
In eqn.AO009 cancel numerator and 
denominator same term, result is
part (b) target equation eqn.AO002.

part (c) is easy to observe.
"replace xk by ρxk, we get
 part (c)"

<a name="ch05c015">　Index begin　Index this file
Exercise 5.7 hint say
"Finally, by setting
  ρ=(x1x2...xn)-n ---eqn.AO006
and simplifying, we get the AM-GM 
bound."
Start from eqn.AO003, replace ρ, get

2009-11-27-22-25 tired

<a name="ch05c018">　Index begin　Index this file
[[
Finally, by setting
  ρ=(x1x2...xn)-n ---eqn.AO006
and simplifying, we get the AM-GM 
bound.
]]
<a name="ch05c019"> (9811272245)
2009-11-27-22-45 in
tute0021.htm
change 
from -n to -1/n
[[
Finally, by setting
  ρ=(x1x2...xn)-1/n ---eqn.AO006
and simplifying, we get the AM-GM 
bound.
]]

<a name="ch05c022">
2009-11-27-23-06 here
eqn.AO015 is
  n≦[x1+x2+...+xn]/GM ---eqn.AO016
rewrite as
  GM≦[x1+x2+...+xn]/n ---eqn.AO017
now get
  GM≦AM ---eqn.AO018
2009-11-27-23-08 done !!



<a name="ch05c023">　Index begin　Index this file
2009-11-28-10-19 start
■　Exercise 5.8 problem statement
　　textbook page 85
（Kantorovich's Inequality for Reciprocals）

Show that if
  0＜m＝x1≦x2≦...≦xn＝M＜∞ ---eqn.AO019
then for nonnegative weights with
  p1+p2+...+pn＝1 ---eqn.AO020
one has

<a name="ch05c025">
2009-11-28-10-39 here
where 
  μ=(m+M)/2 ---eqn.AO021
  γ=√(mM)   ---eqn.AO022
This bound provides a natural complement
to the elementary inequality of Exercise
1.2 (page 12), but it also has important
applications in numerical analysis where
<a name="ch05c026">
for example, it has been used to estimate
the rate of convergence of the method of
steepest ascent. To get started with this
proof, one might note that by homogeneity
it suffices to consider the case when 
  γ=1 ---eqn.AO023
the geometry of figure 5.2 then tells a
powerful tale.
2009-11-28-10-46 stop

<a name="ch05c027">
2009-11-29-09-41 start
■　Exercise 5.8 hint
　　textbook page 246

The inequality is unaffected if m, M 
and xj 1≦j≦n are multiplied by a
positive constant, so we can assume
without loss of generality that
  γ=1 ---eqn.AO023
in which case we have
  M=m-1 ---eqn.AO024
and it suffices to show that

<a name="ch05c029">
2009-11-29-09-56 here
where  2μ=m+M=m+m-1 ---eqn.AO025
Now one has xj∈[m,m-1] for all
 1≦j≦n, so we have
  xj+xj-1≦m+m-1≦2μ ---eqn.AO026

<a name="ch05c031">
and these yield the bound (14.53) after
one applies the AM-GM inequality to the
two bracketed terms. There are many
instructive proofs of Kantorovich's 
Inequality; this elegant approach via 
the AM-GM inequality is due to Ptak 
(1995).
2009-11-29-10-19 stop



<a name="ch05c032">　Index begin　Index this file
2009-11-29-10-48 start
■　Exercise 5.8 solution


Inequality has much more freedom than
equality equation. Greater side of an
inequality can be changed/enlarged.
Less value side of an inequality can 
be changed/reduced.
<a name="ch05c033">
Here Exercise 5.8 we rescale whole
input sequence. Let after-scale sequence
has GM=1. What is the scale factor?

<a name="ch05c034">
Assume before-scale sequence has
 GM=gm0
 AM=am0
 minimum element value =m0
 maximum element value =M0

<a name="ch05c035">
Assume after-scale sequence has
 GM=gm1
 AM=am1
 minimum element value =m1
 maximum element value =M1

<a name="ch05c036">
AM/GM calculation base on m0 and M0
 am0=(m0+M0)/2 ---eqn.AO028
 gm0=√(m0*M0)  ---eqn.AO029

<a name="ch05c037">　Index begin　Index this file
Exercise 5.8 hint suggest
"multiplied by a positive constant"
Assume constant is k. Then 
after-scale gm1 by demand is
 gm1=√(m0*k*M0*k)=1  ---eqn.AO030
where '=1' is assigned by us. (eqn.AO023)
<a name="ch05c038">
From eqn.AO030, find k in terms of
gm0 as following
 γ=1=gm1 ---eqn.AO023 
 gm1=√(m0*k*M0*k)=1=k*√(m0*M0)=k*gm0
the net result is
 1=k*gm0
or
 k=1/gm0 ---eqn.AO031
<a name="ch05c039">
Now modify input sequence xj 1≦j≦n 
therefore modify element of smallest
value
  m1=m0*k=m0/gm0 ---eqn.AO032
modify element of greatest value
  M1=M0*k=M0/gm0 ---eqn.AO033
<a name="ch05c040">
Check new GM value as below
  gm1=m1*M1=(m0/gm0)*(M0/gm0)
  gm1=m0*M0/gm0/gm0
  gm1=gm02/gm02 = 1 ---eqn.AO034 (eqn.AO023)
which is same as
  m1*M1=1 ---eqn.AO035
and a new relation
  M1=1/m1 ---eqn.AO036 (see eqn.AO024)

<a name="ch05c041">　Index begin　Index this file
Before scaled sequence element xj0
After  scaled sequence element xj1
related as following
  xj1=xj0*k=xj0/gm0 ---eqn.AO037
<a name="ch05c042">
Next show that xj1≦M1
  xj1/M1=[xj0/gm0]/[1/m1]
  xj1/M1=[xj0/gm0]/[1/m0/gm0]
  xj1/M1=[xj0/gm0]/[gm0/m0]
  xj1/M1=[xj0*m0/gm02]
  xj1/M1=[xj0*m0/(m0*M0)]
  xj1/M1=[xj0/M0] ---eqn.AO038
<a name="ch05c043">
Because we know xj0≦M0, then
eqn.AO038 tell us that 
  xj1≦M1=1/m1 ---eqn.AO039
eqn.AO039 tell us that scaled sequence
elements are all bounded by 1/m1
<a name="ch05c044">
Original sequence element do NOT 
bound by 1/m0, because original
sequence GM NOT=1
eqn.AO039 is important for next
proof.
2009-11-29-11-27 stop

<a name="ch05c045">
2009-11-29-12-00 start
Next prove eqn.AO026

Below write m1 as m, xj1 as x to
simplify equation.

<a name="ch05c046">　Index begin　Index this file
given m≦x≦1/m (see eqn.AO039)
and m≦1 (because gm1=1=√[mM], m＜M)
prove 
x+x^(-1) ?≦? m+m^(-1) ---eqn.AO040
get
[x*x+1]/x ?≦? [m*m+1]/m
get
[x*x+1]*m ?≦? [m*m+1]*x
<a name="ch05c047">
get
x*x*m+m ?≦? m*m*x+x
get
0 ?≦? m*m*x+x-x*x*m-m
get
0 ?≦? x*m*[m-x]+[x-m]
get
0 ?≦? [x-m][1-x*m]

<a name="ch05c048">
since x＜1/m (see eqn.AO039)
x*m＜m/m=1
so [1-x*m]>0
given [x-m]＞0 (xj1＞m1, see ch05c035)
answer
0 ?≦? [x-m][1-x*m] ---eqn.AO041
is true, both [.] are non-negative.
then start line
x+x^(-1) ?≦? m+m^(-1) ---eqn.AO040
is true. That is eqn.AO026 is true.
2009-11-29-12-20 here

<a name="ch05c049">
eqn.AO026 whole line multiply by
probability pj and sum from j=1 to
j=n, the result is eqn.AO027.
eqn.AO027 greater than side still 
has probability pj, they sum to one
and not show up.

<a name="ch05c050">
eqn.AO027 has two terms, simplify as
SUM{p*x} and SUM{p/x}. Apply AM-GM
inequality to these two terms, get
  GM=√[SUM{p*x}*SUM{p/x}]≦
     [SUM{p*x}+SUM{p/x}]/2=AM ---eqn.AO042

<a name="ch05c051">
eqn.AO027 tell us that 
  [SUM{p*x}+SUM{p/x}]≦2μ 
Here use AM as μ 
rewrite eqn.AO042 as
  GM=√[SUM{p*x}*SUM{p/x}]≦
     2μ/2=μ ---eqn.AO043
<a name="ch05c052">
square eqn.AO043 get
  [SUM{p*x}*SUM{p/x}]≦μ2 ---eqn.AO044
eqn.AO044 is final answer.
Remember, scaled sequence gm1=1
that is γ=1 in eqn.5.21.
eqn.AO044 is same as eqn.5.21
Exercise 5.8 solved.
2009-11-29-12-42 stop



<a name="ch05c053">　Index begin　Index this file
2009-11-29-21-15 start
■　Exercise 5.9 problem statement
　　textbook page 85
（Monotonicity Method）

Suppose ak＞0 and bk＞0 for k=1,2,...,n
and for fixed θ∈R, consider the
function

<a name="ch05c055">
2009-11-29-21-32 here
If we set
  θ=1 ---eqn.AO046
we see that f1(0)1/2 gives us the 
left side of Cauchy's inequality 
while f1(1)1/2 gives us the right
side. 
<a name="ch05c056">
Show that fθ(x) is a 
monotone increasing of x on [0,1], 
a fact which gives us a parametric
family of inequalities containing
Cauchy's inequality as a special
case.
2009-11-29-21-39





<a name="ch05c057">　Index begin　Index this file
2009-11-29-21-40 
■　Exercise 5.9 hint
　　textbook page 246
One elegant way to make the mono-
tonicity of fθ evident is to set
  cj=(ajbj)θ ---eqn.AO047
and
  dj=log(aj/bj) ---eqn.AO048
to obtain

<a name="ch05c060">
2009-11-29-22-00 here
where cosh(y)=(ey+e－y)/2 ---eqn.AO051
Since cosh(y) is symmetric about zero
and monotone on [0,∞), the monotonicity
of fθ(.) is now immediate.
This solution follows Steiger (1969)
where a second proof based on Holder's
inequality is also given.
2009-11-29-22-05 stop

<a name="ch05c061">
2009-11-30-14-38 start
■　Exercise 5.9 discussion
Exercise 5.9 problem statement say 
that in eqn.AO045
[[
If we set
  θ=1 ---eqn.AO046
we see that f1(0)1/2 gives us the 
left side of Cauchy's inequality 
while f1(1)1/2 gives us the right
side. 
]]
Is it true? Let us put three equations
side by side as next. Middle one is
Cauchy's inequality.

First one is eqn.AO045 with θ=1 and x=0

<a name="ch05c065">
2009-11-30-15-20 here
After compare above three equations side
by side, we can believe that in eqn.AO045
hold θ=1, then x=0 to x=1 interpolate 
between left side and right side of Cauchy 
inequality. 
2009-11-30-15-30 stop



<a name="ch05c066">　Index begin　Index this file
2009-11-30-16-04 start
■　Exercise 5.9 solution


Problem only as us to show that 
 fθ(x) is a monotone increasing
 of x on [0,1]
fθ(x) has two parameters, θ and x.
But problem require fixed θ∈R
Only variable is x.
<a name="ch05c067">
Review eqn.AO045, we put constant 
terms together, and put variable x 
term together. In eqn.AO045 both
summation have ajθbjθ, it is natural
to define
  cj=(ajbj)θ ---eqn.AO047
and both summation have aj+x*bj-x 
and aj-x*bj+x. We know exp() and log()
are inverse function to each other.
<a name="ch05c068">
If AA and BB are non-negative 
expression [log(m) require m≧0]
  AA = exp(log(AA)) ---eqn.AO055
and
  log(BBt)=t*log(BB) ---eqn.AO056
<a name="ch05c069">
then
  aj+x*bj-x=exp(log(aj+x*bj-x)) ---eqn.AO057
  aj+x*bj-x=exp(x*log(aj+1*bj-1)) ---eqn.AO058
  aj+x*bj-x=exp(x*log(aj/bj)) ---eqn.AO059
Above three equations are same thing.
<a name="ch05c070">
Last one isolate variable x and it
is useful. Based on above observation
we define
  dj=log(aj/bj) ---eqn.AO048
eqn.AO059 become
  aj+x*bj-x=exp(x*dj) ---eqn.AO060
<a name="ch05c071">
eqn.AO045 has another bracket allow
us write
  aj-x*bj+x=exp(-x*dj) ---eqn.AO061
Put all of above together, we find
eqn.AO045 become eqn.AO049

<a name="ch05c072">　Index begin　Index this file
cosh(y) is a well defined function
 cosh(y)=(ey+e－y)/2 ---eqn.AO051
Use eqn.AO051, rewrite eqn.AO049 as
eqn.AO050

<a name="ch05c073">
cj,ck,dj,dk are all constants,
cosh(y) is a monotone increase function. 
from
  fθ(x) = eqn.AO045 = eqn.AO050
we conclude that eqn.AO045 is a monotone
increasing function of x on [0,1].
Drawing suggest that fθ(x) is a monotone
increasing function of x on [0,∞)
2009-11-30-16-48 stop

<a name="ch05c074">
2009-11-30-18-50 start
If a,b two arrays are proportional then
Cauchy's inequality become equality. 
Both side have same value. fθ(x) must
reflect this fact. 
dj is defines as
  dj=log(aj/bj) ---eqn.AO048
<a name="ch05c075">
Since a,b two arrays are proportional
dj is constant for all j=1,...,j=n
See eqn.AO050 for answer.
cosh[(dj－dk)x]=cosh[(0)x]
variable x multiply by zero, whole
equation no variable x effect. Output
is constant as expected.
2009-11-30-19-01 stop



<a name="ch05c076">　Index begin　Index this file
2009-11-30-19-20 start
■　Exercise 5.10 problem statement
　　textbook page 86
（A Proto-Muirhead Inequality）

If the nonnegative real numbers a1,a2
and b1,b2 satisfy
  max{a1,a2} ≧ max{b1,b2} ---eqn.AO062
and
  a1+a2=b1+b2 ---eqn.AO063
<a name="ch05c077">
then for nonnegative x and y, one has
 xb1yb2+xb2yb1 ≦ xa1ya2+xa2ya1  ---eqn.5.22
Prove this assertion by considering 
an appropriate factorization of the
difference of the two sides.
2009-11-30-19-32 here






<a name="ch05c078">
2009-11-30-20-00 start
■　Exercise 5.10 hint
　　textbook page 246
We can assume without loss of 
generality that 
    a1≧a2 ---eqn.AO064
and b1≧b2 ---eqn.AO065
and a1≧b1 ---eqn.AO066 (result of eqn.AO062)
<a name="ch05c079">
Remaining mindful of the relation
  a1+a2=b1+b2 ---eqn.AO063
the proof can be completed by the
factorization
  xa1ya2+xa2ya1 - xb1yb2 - xb2yb1 ---eqn.AO067
 =xa2ya2[xa1-a2+ya1-a2 - xb1-a2yb2-a2 - xb2-a2yb1-a2]
<a name="ch05c080">
  //from eqn.AO063 get a1=b1+b2-a2 ---eqn.AO068
 =xa2ya2[x(b1+b2-a2)-a2+y(b1+b2-a2)-a2
      - xb1-a2yb2-a2 - xb2-a2yb1-a2]
  //above red term become below blue term
 =xa2ya2[xb1-a2*xb2-a2 - xb1-a2*yb2-a2
      - yb1-a2*xb2-a2 + yb1-a2*yb2-a2 ]
 =xa2ya2(xb1-a2 - yb1-a2)(xb2-a2 - yb2-a2)≧0 ---eqn.AO067 end here

Since b1-a2≧b2-a2=a1-b1≧0  ---eqn.AO069
2009-11-30-20-55 here




<a name="ch05c081">　Index begin　Index this file
2009-11-30-21-45 start
■　Exercise 5.10 solution


Exercise 5.10 hint already solved 
this problem. LiuHH add detail in
hint section. Last step is to show
  (xb1-a2 - yb1-a2)(xb2-a2 - yb2-a2)≧0 ---eqn.AO070
<a name="ch05c082">
rewrite eqn.AO070 in simpler form as
  (xm - ym)(xn - yn)≧0 ---eqn.AO071
where m=b1-a2 ---eqn.AO072
and   n=b2-a2 ---eqn.AO073
all of x,y,m,n are non-negative.
We need to show that (xm - ym) and 
(xn - yn) two terms have same sign.

<a name="ch05c083">
Assume  (xm - ym)≧0 ---eqn.AO074
then    xm - ym≧0
        xm ≧ ym
        xm/ym ≧ 1
<a name="ch05c084">
    log(xm/ym)≧log(1)=0
    m*log(x/y)≧0 ---eqn.AO075
similarly
    n*log(x/y)≧0 ---eqn.AO076
eqn.AO071 can be written as
    m*n*log(x/y)*log(x/y) ?≧? 0 ---eqn.AO077

<a name="ch05c085">
from eqn.AO064, eqn.AO065, eqn.AO066
and eqn.AO069, 
we know m and n both ≧ 0
Key point is at 
   log(x/y)*log(x/y)
which is a square of real number,
<a name="ch05c086">
no matter x/y≧1 or x/y＜1
   log(x/y)*log(x/y)
is always non-negative. then
eqn.AO077 is true and eqn.AO067 is 
true. Problem is solved.
2009-11-30-22-04 stop



<a name="ch05c087">　Index begin　Index this file
2009-12-01-10-01 start
■　Exercise 5.11 problem statement
　　textbook page 86
（Chebyshev's Inequality for Tail Probabilities）

One of the most basic properties of
the mathematical expectation E(.)
that one meets in probability theory
<a name="ch05c088">
is that for any random variables X
and Y with finite expectation the
relationship X≦Y implies that
  E(X)≦E(Y) ---eqn.AO078
Use this fact to show that for any
random variable Z with finite mean
  μ=E(Z) ---eqn.AO079
<a name="ch05c089">
one has the inequality
  P(|Z-μ|≧λ)≦E(|Z-μ|2)/λ2 ---eqn.5.23
This bound provides one concrete 
expression of the notation that a
random variable is not likely to be
too far away from the mean, and it
is surely the most used of the several
inequalities that carry Chebyshev's
name.
2009-12-01-10-13 stop





<a name="ch05c090">
2009-12-01-10-15 start
■　Exercise 5.11 hint
　　textbook page 247
Let A denote the event that |Z-μ| is 
at least as large as λ. Now define 
a random variable χA by setting
  χA = 1 ---eqn.AO080
<a name="ch05c091">
if the event A occurs, and setting
  χA = 0 ---eqn.AO081
otherwise. Note that 
  E(χA)=P(A)=P(|Z-μ|≧λ) ---eqn.AO082
Also note that
  χA≦|Z-μ|2/λ2 ---eqn.AO083
<a name="ch05c092">
Since both side are zero if A does 
not occur, and the right side is at
least as large as 1 if the event A
does occur. On taking the expectation
of the last bound one gets Chebyshev's
tail bound(5.23). Admittedly, the
language used in this problem and its 
solution are special to probability
theory, but nevertheless the argument
is completely rigorous.
2009-12-01-10-28 stop





<a name="ch05c093">　Index begin　Index this file
2009-12-01-10-37 start
■　Exercise 5.11 solution


Exercise 5.11 where is sine() ? 
where is cosine()? where is log()?
where is exp()?

<a name="ch05c094">
LiuHH is not familiar with probability
theory and is unable to solve this
problem. Just present few outsider's
question. 

First, eqn.AO082 say
  E(χA)=P(A)=P(|Z-μ|≧λ) ---eqn.AO082
guess E(χA) is expectation value
guess P(A) is the probability that
      event A occurs.
<a name="ch05c095">
If paid five dollars to buy lottery
ticket. If first prize is ten thousand
dollars. If sold one million tickets.
Then expectation value is ten thousand
dollars divide by one million tickets.
=0.01 dollar/ticket
My question is 
How can expectation value 0.01 dollar 
equal to a pure number probability P(A)
Dollar and pure number term dimension
are different.

<a name="ch05c096">
Second LiuHH do not know where the five
dollar fit in this expectation problem.

Again, LiuHH is probability theory
outsider. 
Unable to solve Exercise 5.11 
2009-12-01-10-57 stop

========= Chapter five end here =========

2009-12-01-17-06 done proofread
2009-12-01-18-03 done spelling check

<a name="docB001">
2009-11-28-23-14 start
Today (2009-11-28) can not finish even 
one problem, Exercise 5.8. Drawing and 
programming for Kantorovich's Inequality
take time. But another event also take
much of the day's time.

<a name="docB002">
2009-11-28-15-52 Sister arrive LiuHH room.
2009-11-28-17-07 Sister left Simi Valley.
After 2009-11-28-17-07, it is pile up work
and eat food.
2009-11-28-19-29 sister called, she arrive
Gardena (California) safely.

<a name="docB003">
Sister bring food to LiuHH. Start from 
2009-May or June. Sister come once a
month. Each time bring one month food
supply. 
First ten days, most fresh food.
Second ten days, race with rotten vegetable.
Third ten days, it is 'diet' time, eat less.
Not me buy, It is sister buy food for me.

<a name="docB004">
Sister bring one page, ask me type to 
Chinese text file. This will take tomorrow
three hours time.
2009-11-28-23-28 stop

<a name=20091217>
2009-12-17-11-09 start
Update 2009-12-17 change all tute*.htm
(from tute0007.htm to tute0023.htm)
first:
Correct 'Limit' link from '#docA06'
to '#docA006'
second:
Change Javascript index to read from
jslist1e.js so that update jslist1e.js
then update ALL tute*.htm.
2009-12-17-11-23 stop

<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56