134
H. Matsuzoe and A. Takatsu
b
1
0 = 1,
b
2
0 = a(q − 1) + 1,
b
3
0 = {2a(q − 1) + 1}{a(q − 1) + 1},
b
2
1 = a − 1,
b
3
1 = (a − 1){(4a + 1)(q − 1) + 3},
b
3
2 = (a − 1)(2a − 1).
Corollary 1 For a ∈ R \ {0} and n ∈ N, then b
n
0 (1, a) = 1.
Proof It follows from Lemma 1 that
b
n+1
0 (1, a) = {na(1 − 1) + 1}b
n
0 (1, a) = b
n
0 (1, a) = · · · = b
1
0 (1, a) = 1.
Corollary 2 Let q ∈ R and n ∈ N. For 1 ≤ j < n, then b
n
j (q, 1) = 0.
Proof This holds for 1 = j < n = 2 by Remark 3. For n ≥ 2, if b
n
j (q, 1) = 0 holds
for
1 ≤ j ≤ n − 1,
then
Lemma
1
implies
that b
n+1
n (q, 1) = (na − 1)b
n
n−1 (q, 1) = 0. For 2 ≤ j ≤ n − 1, we have that
b
n+1
j (q, 1) = {(n + j)(q − 1) + 1}b
n
j (q, 1) + ( j − 1)b
n
j−1 (q, 1) = 0
by the assumption b
n
k (q, 1) = 0 for 1 ≤ k ≤ n − 1. For j = 1, we have that
b
n+1
1 (q, 1) = {(n + 1)(q − 1) + 1}b
n
1 (q, 1) + (1 − 1)b
n
0 (q, 1) = 0.
6.2 Escort Expectations
The ordinary expectation of a random variable is the integral of the random variable
with respect to its law. An introduction to escort expectations admits us to replace
the law by any other measures. The literature is very large and we just refer to the
paper by Naudts [4].
For a probability space ((, p) and a probability measure r on , the expectation of
a random variable on with respect to r is called the escort expectation of the random
variable with respect to r. We emphasize that the expectation does not depend on the
probability measure p. Moreover, it is possible to replace the probability measure r
with an arbitrary measure on .
Definition 1 For a measure ν on a measurable space , the escort expectation of a
function f ∈ L
1
(ν) with respect to ν is defined by
E ν [ f ] :=
f (ω)dν(ω).
(6.3)
In this section, we fix a manifold S consisting of positive probability densities on
a measure space ((, m). Take T ∈ (0, ∞] such that
H. Matsuzoe and A. Takatsu
b
1
0 = 1,
b
2
0 = a(q − 1) + 1,
b
3
0 = {2a(q − 1) + 1}{a(q − 1) + 1},
b
2
1 = a − 1,
b
3
1 = (a − 1){(4a + 1)(q − 1) + 3},
b
3
2 = (a − 1)(2a − 1).
Corollary 1 For a ∈ R \ {0} and n ∈ N, then b
n
0 (1, a) = 1.
Proof It follows from Lemma 1 that
b
n+1
0 (1, a) = {na(1 − 1) + 1}b
n
0 (1, a) = b
n
0 (1, a) = · · · = b
1
0 (1, a) = 1.
Corollary 2 Let q ∈ R and n ∈ N. For 1 ≤ j < n, then b
n
j (q, 1) = 0.
Proof This holds for 1 = j < n = 2 by Remark 3. For n ≥ 2, if b
n
j (q, 1) = 0 holds
for
1 ≤ j ≤ n − 1,
then
Lemma
1
implies
that b
n+1
n (q, 1) = (na − 1)b
n
n−1 (q, 1) = 0. For 2 ≤ j ≤ n − 1, we have that
b
n+1
j (q, 1) = {(n + j)(q − 1) + 1}b
n
j (q, 1) + ( j − 1)b
n
j−1 (q, 1) = 0
by the assumption b
n
k (q, 1) = 0 for 1 ≤ k ≤ n − 1. For j = 1, we have that
b
n+1
1 (q, 1) = {(n + 1)(q − 1) + 1}b
n
1 (q, 1) + (1 − 1)b
n
0 (q, 1) = 0.
6.2 Escort Expectations
The ordinary expectation of a random variable is the integral of the random variable
with respect to its law. An introduction to escort expectations admits us to replace
the law by any other measures. The literature is very large and we just refer to the
paper by Naudts [4].
For a probability space ((, p) and a probability measure r on , the expectation of
a random variable on with respect to r is called the escort expectation of the random
variable with respect to r. We emphasize that the expectation does not depend on the
probability measure p. Moreover, it is possible to replace the probability measure r
with an arbitrary measure on .
Definition 1 For a measure ν on a measurable space , the escort expectation of a
function f ∈ L
1
(ν) with respect to ν is defined by
E ν [ f ] :=
f (ω)dν(ω).
(6.3)
In this section, we fix a manifold S consisting of positive probability densities on
a measure space ((, m). Take T ∈ (0, ∞] such that
