characteristic roots and vectors of
AA
′
.
The inverse square root defined in Section B.7.12 would also provide
a method of transforming
x
to obtain the desired covariance matrix.
18.
The density of the standard normal distribution, denoted
φ
(
x
), is given in (C28).
The function based on
the
i
th derivative of the density given by
H
i
= [(1)
i
d
i
φ
(
x
)/
dx
i
]/
φ
(
x
),
i
= 0,1,2,.
.. is called a
Hermite polynomial
.
By definition,
H
0
= 1.
(a)
Find the next three Hermite polynomials.
(b)
A useful device in this context is the differential equation
d
r
φ
(
x
)/
dx
r
+
xd
r
1
φ
(
x
)/
dx
r
1
+ (
r
1)
d
r
2
φ
(
x
)/
dx
r
2
= 0.
Use this result and the results of part a. to find
H
4
and
H
5
.
The crucial result to be used in the derivations is
d
φ
(
x
)/d
x
=

x
φ
(
x
).
Therefore,
d
2
φ
(
x
)/
dx
2
=
(
x
2
 1)
φ
(
x
)
and
d
3
φ
(
x
)/
dx
3
=
(3
x

x
3
)
φ
(
x
).
The polynomials are
H
1
=
x
,
H
2
=
x
2
 1, and
H
3
=
x
3
 3
x
.
For part (b), we solve for
d
r
φ
(
x
)/
dx
r
=

xd
r
1
φ
(
x
)/
dx
r
1
 (
r
1)
d
r
2
φ
(
x
)/
dx
r
2
Therefore,
d
4
φ
(
x
)/
dx
4
= 
x
(3
x

x
3
)
φ
(
x
)  3(
x
2
 1)
φ
(
x
)
=
(
x
4
 6
x
2
+ 3)
φ
(
x
)
and
d
5
φ
(
x
)/
dx
5
= (
x
5
+ 10
x
3
 15
x
)
φ
(
x
).
Thus,
H
4
=
x
4
 6
x
2
+ 3 and
H
5
=
x
5
 10
x
3
+ 15
x
.
±
19.
Continuation:
orthogonal polynomials
: The Hermite polynomials are orthogonal if
x
has a standard
normal distribution.
That is,
E
[
H
i
H
j
] = 0 if i
≠
j.
Prove this for the
H
1
,
H
2
, and
H
3
which you obtained above.
E
[
H
1
(
x
)
H
2
(
x
)] =
E
[
x
(
x
2
 1)] =
E
[
x
3

x
] = 0
since the normal distribution is symmetric. Then,
E
[
H
1
(
x
)
H
3
(
x
)] =
E
[
x
(
x
3
 3
x
)] =
E
[
x
4
 3
x
2
] = 0.
The fourth moment of the standard normal distribution is 3 times the variance.
Finally,
E
[
H
2
(
x
)
H
3
(
x
)] =
E
[(
x
2
 1)(
x
3
 3
x
)]
=
E
[
x
5
 4
x
3
+ 3
x
]
=
0
because all odd order moments of the normal distribution are zero. (The general result for extending the
preceding is that in a product of Hermite polynomials, if the sum of the subscripts is odd, the product will be a
sum of odd powers of
x
, and if even, a sum of even powers.
This provides a method of determining the higher
moments of the normal distribution if they are needed.
(For example,
E
[
H
1
H
3
] = 0 implies that
E
[
x
4
] =
3
E
[
x
2
].)
20.
If
x
and
y
have means
μ
x
and
μ
y
and variances
and
and covariance
σ
xy
, what is the approximation
of the covariance matrix of the two random variables
f
1
=
x
/
y
and
f
2
=
xy
?
σ
x
2
y
2
The elements of
J
Σ
J
N are (1,1) =
σ
μ
σμ
μ
μ
x
y
y
x
y
xy
x
y
2
2
2
2
43
2
+−
(
1
,
2
)
=
σ

σ
/
x
2
y
2
μ
x
2
μ
y
4
(
2
,
2
)
=
σ μ
+
+ 2
σ
xy
μ
x
μ
y
.
x
2
y
4
σ
y
2
μ
x
2
21.
Factorial Moments
.
For finding the moments of a distribution such as the Poisson, a useful device is the
factorial moment. (The Poisson distribution is given in Example 3.1.)
The density is
f
(
x
)
=
e

λ
λ
x
/
x
!,
x
= 0,1,2,.
..