Fundamentals of Probability and Stochastic Processes with Applications to Communications ( PDFDrive.com )

196 7 Gaussian Distributions

Xn σ2X ¼ Xn σ2i ð7:46Þ
μX ¼ μi ð7:47Þ

i¼1 ( Xn Xn )i¼1

X$N μi; σi2

i¼1 i¼1

Proof Consider the case of n ¼ 3 as follows:

X ¼ X1 þ X2 þ X3 ð7:48Þ

Let

Y ¼ X1 þ X2 ð7:49Þ

By (7.34),

Y $ À þ μ2; σ21 þ σ22Á
N μ1

Substituting (7.49) into (7.48), we have

X ¼ Y þ X3

By (7.34),

X $ À þ μ3; σ2Y þ σ23Á ð7:50Þ
N μY

By (5.21),

μY ¼ EðYÞ ¼ EðX1 þ X2Þ ¼ EðX1Þ þ EðX2Þ ¼ μ1 þ μ2 ð7:51Þ

By (5.52), and, since Cov(X1, X2) ¼ 0 because X1 and X2 are independent, we
have

σY2 ¼ VarðYÞ ¼ VarðX1Þ þ VarðX2Þ ¼ σ12 þ σ22 ð7:52Þ

Substituting (7.51) and (7.52) into (7.50), we have

X $ À þ μ2 þ μ3; σ12 þ σ22 þ σ32Á
N μ1

This proves (7.47) for n ¼ 3. By mathematical induction, (7.48) is proven for an
arbitrary n.

Q.E.D.

7.3 Two Jointly Gaussian RVs 197

7.3 Two Jointly Gaussian RVs

Two RVs X and Y are said to be jointly normal, if its joint pdf is given by the
following equation:

& 2 À2ρðxÀμσXXÞσðYyÀμY Þþ 2'

f ðx; yÞ ¼ p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi eÀ2ð1À1 ρ2 Þ xÀμX yÀμY ð7:53Þ
1 À ρ2 σX σY

XY 2π σ X σ Y

where μX, μY, σ2X, σ2Y, and À 1 ρ 1 are constants. When X and Y are jointly
normal, we write

ðX; YÞ $ À μY ; σX2 ; σY2 ; Á ð7:54Þ
N μX; ρ

The above pdf is called the bivariate normal pdf. The bivariate normal pdf
satisfies the following conditions.

Theorem 7.3.1 A bivariate normal pdf given by (7.53) satisfies the following
condition:

Z þ1 Z þ1 ð7:55Þ
f XYðx; yÞdxdy ¼ 1

À1 À1

This is a condition that any pdf must satisfy and is not a special condition for a
bivariate normal pdf. In other words, by the definition of a pdf, the integral of a pdf
over the entire real line(s) must be one.

Proof Rewrite (7.53) as follows:

f XY ðx; yÞ ¼ 2π σ X σ Y p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi eEðx;yÞ ð7:56Þ
1 À ρ2

where the exponent is given by the following equation:

1 ( À μX 2 2ρðx À μX Þðy À μY Þ À μY 2)
À x σX σXσY y σY
Eðx; yÞ ¼ À2ð1 À þ
ρ2Þ

1 ( À μX 2 ðx À μXÞ ρðy À μY Þ À μY 2 À μY 2 À μY 2)
À x σX σX σY ρ2 y σY ρ2 y σY y σY
¼ À2ð1 À 2 þ À þ
ρ2Þ

¼ À2ð1 1 "& À À μY '2 þ À À ρ2Á y À μY 2#
À x À μX ρy σY 1 σY
σX
ρ2Þ

ð7:57Þ

Substituting (7.57) into (7.56), we obtain the following equation:

198 7 Gaussian Distributions

n o2 2 !

p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ2ð1À1 ρ2Þ xÀμX À ρ yÀμY þ ð1 À ρ2 Þ yÀμY
σX σY σY
f XYðx; yÞ ¼
2πσXσY 1 À ρ2
n o2 2
xÀμX yÀμY yÀμY
¼ pffiffiffiffiffi p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ2ð1À1 ρ2Þ σX À ρ σY p1 ffiffiffiffiffieÀ21 σY

2πσX 1 À ρ2 σY 2π 2
n o2
¼ pffiffiffiffiffi p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ2ð1À1ρ2Þσ2X ðx À μX Þ À ρσσXY ðy À μY Þ p1 ffiffiffiffiffi eÀ12 yÀμY
σY

2πσX 1 À ρ2 σY 2π
n ρσσXY μY o2 2
pffiffiffiffiffiffiffiffiffiffip1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ2ð1À1ρ2Þσ2X x ρσσXY y yÀμY
¼ À þ μX À p1 ffiffiffiffiffi eÀ12 σY

2πσX 1 À ρ2 2 σY 2π

¼ pffiffiffiffiffi p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ2ð1À1ρ2Þσ2Xfx À Ag2 p1 ffiffiffiffiffi eÀ12 yÀμY
σY

2πσX 1 À ρ2 & '2 σY 2π

p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ21 pxÀffiAffiffiffiffiffiffi yÀμY 2
σY
¼ pffiffiffiffiffi σX 1Àρ2 p1 ffiffiffiffiffi eÀ12

2πσX 1 À ρ2 σY 2π

ð7:58Þ

where

A ¼ ρσσXY y þ μX À ρσσXY μY ð7:59Þ

Substituting (7.58) into (7.55), we obtain the following equation:

Z þ1 8<Z þ1 n o2 9 2
=
Z þ1 Z þ1 p1 ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ21 pxÀffiAffiffiffiffiffi dx; eÀ12 yÀμY
σY
À1 À1 f XY ðx; yÞdxdy ¼ À1 : À1 pffiffiffiffiffi σX 1Àρ2 dy
pffiffiffiffiffi
2πσX 1 À ρ2 σY 2π

ð7:60Þ

By (7.14), the integral in braces with respect to x is equal to 1 and the outer
integral with respect to y is also equal to 1 to yield the right-hand side of the above
to be equal to 1.

Q.E.D.

Theorem 7.3.2 If RVs X and Y are jointly normal

ðX; YÞ $ À μY ; σ 2X ; σY2 ; Á ð7:61Þ
N μX; ρ

The marginal distributions of X and Y are normal with means and variances
μX, μY, σX2 and σ2Y, respectively, as follows:

2

X $ À σ 2X Á f X ðxÞ ¼ p1 ffiffiffiffiffi eÀ12 xÀμX ð7:62Þ
N μX; σX 2π σX

7.3 Two Jointly Gaussian RVs 199

2

Y $ À ; σ 2 Á f Y ðxÞ ¼ p1 ffiffiffiffiffi eÀ21 yÀμY ð7:63Þ
N μY Y σY 2π σY

This theorem states that, if two RVs X and Y are jointly normal, X and Y are
individually normal. However, the converse is not necessarily true. In other words,
two normally distributed RVs X and Y are not necessarily jointly normal.

Proof By (4.60), the marginal pdf of X is given by

Z þ1

f YðyÞ ¼ f XYðx; yÞdx
À1
& 2 2 '
Z þ1
¼ p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi À2ð1À1 ρ2 xÀμX À 2ρðxÀμXÞðyÀμY Þ þ yÀμY
Þ σX σXσY σY dx
e
À1 2πσXσY 1 À ρ2

Through the process of (7.58) through (7.60), the bivariate normal pdf in the
above integral can be expressed by (7.58). Substituting (7.58) into the above
equation, we obtain the following equation:

& '2 2 !

Z1 p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi À12 pxÀffiAffiffiffiffiffiffi p1 ffiffiffiffiffieÀ12 yÀμY
pffiffiffiffiffi σY dx
f YðyÞ ¼ e σX 1Àρ2
¼
À1 2πσX 1 À ρ2 σY 2π '2
&
2
yÀμY Z 1 À21 pxÀffiAffiffiffiffiffiffi
p1 ffiffiffiffiffieÀ21 σY p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffi e σX 1Àρ2 dx ð7:64Þ

σY 2π À1 2πσX 1 À ρ2

The integrand pinffiffiffiffiffitffihffiffieffiffiffiffiffilast integral is in the form of a normal pdf with
μ ¼ A and σ ¼ σX 1 À ρ2. Therefore, as proven by (7.14), the integral is equal
to 1 and (7.64) becomes

2

f Y ðyÞ ¼ p1 ffiffiffiffiffi eÀ12 yÀμY
σY 2π σY

This proves (7.63). Eq. (7.62) is proven similarly.

Theorem 7.3.3 If RVs X and Y are jointly normal

ðX; YÞ $ À μY ; σX2 ; σ2Y ; Á ð7:65Þ
N μX; ρ

ρ is the correlation coefficient of X and Y, and the covariance of X and Y is given by

qffiffiffiffiffiffiffiffiffiffi

cXY ¼ CovðX; YÞ ¼ ρ σ2XσY2 ¼ ρσXσY ð7:66Þ

200 7 Gaussian Distributions

ρ ¼ cXY ð7:67Þ
σXσY

Proof By (5.33), we have the following equation: ð7:68Þ
cXY ¼ CovðX; YÞ ¼ EðXYÞ À μXμY

Z þ1Z þ1

EðXYÞ ¼ xyf XYðx; yÞdxdy

À1 À1

& 2 2 '

Z þ1Z þ1 p1 ffiffiffiffiffiffiffiffiffiffiffiffiffi À2ð1À1 ρ2 xÀμX À 2ρðxÀμXÞðyÀμY Þ þ yÀμY
¼ xy Þ σX σXσY
e σY dxdy
À1 À1 2πσXσY 1 À ρ2

ð7:69Þ

Through the process of (7.58) through (7.60), the last integral can be expressed

as the following where A is given by (7.59)

Z þ1 <8Z þ1 n o2 9 2
EðXYÞ ¼ À1 : À1 px ffiffiffiffiffiffiffiffiffiffiffiffiffieÀ21 σX pxÀffiAffiffiffiffiffi =
pffiffiffiffiffi dx;σ py ffiffiffiffiffi eÀ12 yÀμY dy
2πσ 1 À ρ2 1Àρ2 2π σY

X Y

ð7:70Þ

Through the process of (7.21) through (7.24), the integral in braces in the above
integral is the expected value of X given Y fixed at y as follows:

Z þ1 n o2
pffiffiffiffiffi
px ffiffiffiffiffiffiffiffiffiffiffiffiffi eÀ21 pxÀffiAffiffiffiffiffi dx ¼ A

σX 1Àρ2

À1 2πσX 1 À ρ2

where A is given by (7.59) ð7:71Þ
A ¼ ρσσXY y þ μX À ρσσXY μY

Substituting (7.71) into (7.70), we obtain

Z þ1 & ' yÀμY 2
ρσσXY y σY σY
EðXYÞ ¼ À1 þ μX À ρσσXY μY py ffiffiffiffiffieÀ21 dy
2π
2 2
Z þ1 yÀμY & ' Z þ1 yÀμY
σX py2ffiffiffiffiffieÀ12 σY μX ρσσXY μY py ffiffiffiffiffieÀ12 σY
¼ ρ σY À1 dy þ À À1 dy
σY 2π σY 2π

ð7:72Þ

7.4 Vector Gaussian RV 201

By (7.32), the first and the second integral in the above equation are, respec-
tively, given by the following:

Z þ1 2

À1 py2ffiffiffiffiffi yÀμY EÀY2Á
σY 2π σY
e dyÀ21 ¼ ¼ σY2 þ μY2 ð7:73Þ
ð7:74Þ
Z þ1 2

À1 py ffiffiffiffiffi yÀμY
σY 2π σY
e dyÀ21 ¼ EðYÞ ¼ μY

Substituting (7.73) and (7.74) into (7.72), we have

EðXYÞ ¼ ρσσXY Àσ2Y þ μY2 Á þ μY ¼ ρσ X σ Y þ μX μY ð7:75Þ
μX À ρσσXY μY

Substituting (7.75) into (7.68), we have

cXY ¼ CovðX; YÞ ¼ EðXYÞ À μXμY ¼ ρσXσY þ μX μY À μXμY ¼ ρσXσY

The above proof also proves that the coefficient ρ is the correlation coefficient

ρ ¼ cXY
σXσY

Hence, both (7.66) and (7.67) are proven.

Q.E.D.

7.4 Vector Gaussian RV

In the general discussion of RVs in earlier chapters, the RVs are not a function of
time. However, in this section, we will consider the RVs of the specific time points
of a stochastic process. The vector RVs discussed in this section follow the
definitions given in Sect. 6.3 and, therefore, have the time points as the arguments.

The vector RV X(t1, t2, . . . , tn) is said to be jointly normal if its n-dimensional
joint pdf is given by the following equation:

f Xðt1, t2, ..., tnÞðx1, x2, . . . :, xn; t1, t2, . . . , tnÞ ¼ pðffiffi2ffiffiffiπffiffiÞffiffin1ffiffidffiffiffieffiffitffiffiffiffiXffiffiffiXffiffi eÀ12ðxÀμXÞTÀXX1 ðxÀμXÞ
ð7:76Þ

where the vector of specific values x and the vector RV X are as defined by (6.32a)
and (6.35), respectively, and the mean vector μX and the autocovariance matrix XX

202 7 Gaussian Distributions

are given by (6.164a) and (6.168), respectively. By (2.33) and (2.41), we have the
following equations for the terms occurring in (7.76):

detXX≜ X ðÀ1ÞtðjÞC1j1 C2j2 : : :Cnjn ð7:77Þ

j

ÀXX1 ¼ 1 adjXX ð7:78Þ
detXX

where adj XX is defined by (2.40) and t( j) is the total number of inversions of the
permutation j defined by (2.34).

Consider the dimensions of the matrices in the following matrix multiplication

term in the exponent in (7.76):

ðx À μXÞTXÀX1 ðx À μXÞ ð7:79Þ

The following are the dimensions of the three matrices in the above multiplication:

ðx À μXÞT 1  n XÀX1 n  n ðx À μXÞ n  1

The multiplication of ðn  nÞ matrix XÀX1 and ðn  1Þ matrix ðx À μXÞ yields
a n  1 column vector. Therefore, multiplication of ð1  nÞ row vector ðx À μXÞT
and ðn  1Þ column vector fÀXX1 ðx À μXÞg yields a scalar. The determinant detXX
is a scalar. Therefore, the multivariate joint normal pdf given by (7.76) is a scalar
function.

By evaluating the matrix multiplication as discussed above, we may rewrite the

multivariate joint pdf given by (7.76) as follows with the exponent in an expanded form:

f Xðt1, t2, ..., tnÞðx1, x2, . . . :, xn; t1, t2, . . . , tnÞ

Pn Pn
¼ pffiðffi2ffiffiffiπffiffiÞffiffin1ffiffidffiffieffiffiffitffiffiffiffiXffiffiffiXffiffi eÀ12Âdet1XX i¼1 ðxi Àμi ÞAij ðxj Àμj Þ
ð7:80Þ
j¼1

where detXX and cofactor Aij are gYiÀvetn10 ; by t(0mÀ7Á.7a7re) and (2.39), respectively.
Two vector RVs Xðt1; ::; tnÞ and ::; jton;int10tl;y::G; tam0 uÁssidaenfi,nifedtheb(ym(þ6.3n3))-

dimensional concatenated vector RV ZXY t1; ::;

consisting of m and n RVs have the following (m þ n)-dimensional multivariate

normal pdf

f ZXYðt1, ::, tn, t10 , ::, tm0 Þðx1, ::, xn, y1, ::, ym; t1, ::, tn, t10 , ::, tm0 Þ ð7:81Þ
¼ pðffiffi2ffiffiffiπffiffiÞffiffinffiffidffiffi1ffieffiffitffiffiffiffiZffiffiffiXffiffiYffiffiZffiffiXffiffiYffi eÀ21ðzXYÀμZXY ÞTÀZX1 YZXY ðzXYÀμZXY Þ

where zXY is the vector of specific values of the vector RV ZXY defined by (6.35) and
μZXY and ZXYZXY are defined by (6.165) and (6.191), respectively.

7.5 Characteristic Function of a Gaussian RV 203

Example 7.4.1 "# ð7:82Þ
Consider the case of n ¼ 3. μ1

" X1 # μX ¼ EðXÞ ¼ μμ23
X ¼ X2

X3

XX ¼ n À μXÞðX À o ¼ EÀXXTÁ À μXμXT
E ðX μXÞT

where

" X1 # X2 2 X1X2 3
XXT ¼ X2 ½X1 X1X1 X2X2 X1X3
X2X3 5
X3Š ¼ 4 X2X1

X3 X3X1 X3X2 X3X3
2 3
"# μ1μ1 μ1μ2 μ1μ3
μ1 μ2μ2 μ2μ3 5
μXμXT ¼ μμ23 ½ μ1 μ2 μ3 Š ¼ 4 μ2μ1 μ3μ2 μ3μ3
μ3μ1
82 39
< ðX1 À μ1ÞðX1 À μ1Þ ðX1 À μ1ÞðX2 À μ2Þ ðX1 À μ1ÞðX3 À μ3Þ =
¼ E:4 5
XX 2 ðX2 À μ2ÞðX1 À μ1Þ ðX2 À μ2ÞðX2 À μ2Þ ðX2 À μ2ÞðX3 À μ3Þ ;
ðX3 À μ3Þð3X1 À μ1Þ ðX3 À μ3ÞðX3 À μ3Þ ðX3 À μ3ÞðX1 À μ1Þ

c11 c12 c13
¼ 4 c21 c22 c23 5

c31 c32 c33

ð7:83Þ

7.5 Characteristic Function of a Gaussian RV

This section derives the characteristic function of a Gaussian RV first for a scalar
RV and then for a vector RV.

7.5.1 Characteristic Function of a Scalar Gaussian RV

We first consider a scalar RV X. The characteristic function of the normal distri-
bution is, by definition, given by

204 7 Gaussian Distributions

EÈejωX É Z þ1 ejωx p1ffiffiffiffiffi e ð ÞÀ21xÀμ2 Z þ1 p1ffiffiffiffiffi e ð ÞjωxÀ21xÀμ2
σ 2π σ σ 2π σ
ΨXðωÞ ¼ ¼ À1 dx ¼ À1 dx

ð7:84Þ

To evaluate the above integral, complete the square in the exponent in the
integral as follows:

¼¼¼¼¼¼¼jωxÀÀÀÀÀÀÀÀ12222222111111σσσσσσ21x222222 ÂnnnÀnÀxxxÂÂÂx2xxxÀ2ÀðσÀjÀÀÀÀωÀμσjjσ 2ÀÀÀωω2jjjÀ2ωωω2σjþ¼ωσ2σσσ2σþ222μxj2ωÞþþþÀμ!þxÁ2μμμ2ÃÀμþÁÁÁ2μÃÃÃÁx2À222xj1ωσþÀþþþ122μÀωμωωÀÀxÀj22j222ωωÁσσσ12Àσ2σ44¼ω2þ2ÀÀ22þÀþμσj22x2ω2μμjj1σþωωμÁÁ222μμÈμþσσÀx222Á2μÀÀÀÀ2jωo2σμμÀ222jωþþþσμ2μÁμ2þ22oðþμþÁμxμ2Þþo2oμ2É

Substituting the last expression in the exponent in (7.84), we obtain the follow-
ing equation:

Z þ1 p1ffiffiffiffiffieÀ12 xÀðjωσ2þμÞ!2 þ jωμ À 21ω2σ2
¼
ΨXðωÞ σ dx

À1 σ 2π xÀðjωσ2þμÞ!2

ω2 σ 2 Z þ1 p1ffiffiffiffiffieÀ21 σ
2
¼ ejωμÀ dx ð7:85Þ
À1 σ 2π

By (7.15) through (7.19), we have shown that the last integral, which is the
Gaussian integral, is equal to 1. Therefore, we have the following characteristic
function of the normal scalar RV X:

ψXðωÞ ¼ ejωμÀω22σ2 ð7:86Þ

7.5.2 Characteristic Function of a Gaussian Vector RV

In this section, the time points as the arguments of the RV X(t1, .., tn) are not shown
for simplicity. By the definition given by (6.62), the characteristic function of a
Gaussian vector RV X is as follows:

7.5 Characteristic Function of a Gaussian RV 205

ΨXðω1, ω2, . . . , ωnÞ≜EfejωTXg ¼ Efejðω1X1þω2X2þÁÁÁþωnXnÞg ð7:87Þ
Z þ1 Z þ1 ð7:88Þ
¼ . . . ejωTxpðffiffi2ffiffiffiπffiffiÞffiffin1ffiffidffiffiffieffiffitffiffiffiffiffiXffiffiXffiffi eÀ21ðxÀμXÞTXÀX1 ðxÀμXÞdx1dx2 . . . dxn
À1 À1

To evaluate the above integral, let us first evaluate the following simpler integral
and then return to complete the above integral:

Z þ1 Z þ1 ð7:89Þ
ΨXðω1; ω2; . . . ; ωnÞ ¼ . . . ejωTxeÀ12xTxdx1dx2 . . . dxn

À1 À1

Consider the second exponent without -½ in the above integral, xTx, in which
 is assumed to be an (n  n) symmetric matrix. Since À1 ¼ , by
pre-multiplying and post-multiplying  by À1, we have the following expression:

xTx ¼ xT ÀℙℙÀ1 ÁÀℙℙÀ1 Á ¼ xTℙÀℙÀ1ℙÁℙÀ1x
x

In the above equation, we choose ℙ to diagonalize  so that ÀℙÀ1 Á ¼ ,


yielding the following expression:

xTx ¼ xTℙÀ1x ð7:90Þ

where

 ¼ ℙÀ1, det ℙ ¼6 0 ð7:91Þ

 is a diagonal matrix with the eigenvalues of  as the diagonal elements as
follows:

23 ð7:92Þ
λ1 Á Á Á 0

 ¼ 4 ⋮ ⋱ ⋮ 5 , λi ¼ eigenvalue, i ¼ 1, . . . , n:
0 Á Á Á λn

From (7.91), we have

 ¼ ℙÀ1 ð7:93Þ

Taking the transposition of both sides of the above and using the matrix identity
(2.20d), we have

T ¼ ÀℙℙÀ1ÁT ¼ ÀℙÀ1ÁTTℙT ¼ ÀℙÀ1ÁTℙT ð7:94Þ

Since  is symmetric, T ¼ , and, so, by equating (7.93) with (7.94), we obtain
the following equation:

ℙÀ1 ¼ ÀℙÀ1ÁTℙT ð7:95Þ

or

206 7 Gaussian Distributions

 ¼ ℙÀ1ÀℙÀ1ÁTℙTℙ ð7:96Þ

From the above, we derive the following relationship that holds true with the
diagonalizing matrix ℙ:

ℙTℙ ¼ 

By left multiplying both sides, we obtain ð7:97Þ
ℙÀ1 ¼ ℙT

The above result shows that the matrix ℙ diagonalizing  is an orthogonal matrix
if  is a symmetric matrix. In (7.90), let

23
z1
6466666 7777577
z ¼ z2 ¼ À1x ð7:98Þ
:

zi
:

zn

Then, we have

zT ¼ ÀℙÀ1xÁT ¼ xTÀℙÀ1ÁT ð7:99Þ

Substituting (7.97) into the above equation, we have ð7:100Þ
zT ¼ ½ z1 z2 : zi : zn Š ¼ xTÀℙTÁT ¼ xTℙ

Substituting (7.98) and (7.100) into (7.90), we obtain the following equation:

Xn ð7:101Þ
xTx ¼ zTz ¼ λizi2

i¼1

Following the method described in Sect. 2.2.6, we can find the diagonalizing
matrix ℙ as follows. First, find n eigenvalues of  by solving the following
equation:

det ð À λÞ ¼ 0 ð7:102Þ

Then, obtain the eigenvectors corresponding to the eigenvalues from the fol-
lowing n linearly independent relationships:

bi ¼ λibi, i ¼ 1, . . . :, n ð7:103Þ

where the components of the eigenvectors are denoted as follows:

7.5 Characteristic Function of a Gaussian RV 207

23
b1i
6666664 7777577,
bi ¼ b2i i ¼ 1, : : :, n ð7:104Þ
:

bki
:

bni

For the symmetric matrix , its diagonalizing matrix ℙ is orthonormal. The
eigenvectors are orthogonal to each other and their norms are unity. Therefore, the
inner products, h., .i, of the eigenvectors are given by

Xn ð7:105Þ
bi; bj ¼ bkibkj ¼ δij ð7:106Þ
ð7:107Þ
k¼1

The diagonalizing matrix ℙ is given by

23
b11 : b1i : b1n
64666 77775
ℙ ¼ ½b1 b2 : : : bnŠ ¼ : : : : :
: :
bk1 : bki : bkn
: : :

bn1 : bni : bnn
2 3
b11 : bk1 : bn1
66466 57777
ℙT ¼ ℙÀ1 ¼ : : : : :
: :
b1i : bki : bni
: : :

b1n : bkn : bnn

Substituting (7.107) into (7.98), the transformation of x by ℙT to z is given by

23 2 32 3
z1 b11 : bj1 : bn1 x1
46666 77757 66646 7777566664 77577
z ¼ : ¼ ℙTx ¼ : : : : : :
: :
zi b1i : bji : bni xj
: : : : :

zn b1n : bjn : bnn xn
2 Xn 3

¼ 2 b11x1 þ :: þ bj1xj þ :: þ bn1xn 3 ¼ 66664666666666 bj1xj 57777777777777 ð7:108Þ
66664 b1ix1 þ :: þ : þ :: þ bnixn 57777
b1nx1 þ :: þ þ :: þ bnnxn j¼1 :
bjixj
: :
Xn
bjnxj
bjixj

j¼1 :

:
Xn

bjnxj

j¼1

Now, consider the first exponent in (7.89)