. If 
,
i=1,2, and 
then
(9.2)
Proof: (9.1) Apply theorem 2, noting that 
.
(9.2) Apply theorem 2 noting that 
.
Both for 
.
. If ,
i=1,2, and
then
(10.1)
(10.2)
Proof: (10.1) Write the convolution in its integral form and compare
with equation 9.117
of appendix 9.4.
(10.2) Substitute 
in the result for (10.1).
. If 
,
and 
then
Proof: Apply theorem 10.1 to find the result for the first convolution,
then use the series representation for 
for the second convolution, comparing the result with the definition of 
in appendix 9.4.
Convolving the series term-by-term is ok since the series is uniformly
convergent on [0,1].
Theorem 12 presents preliminary results for the non-overlapping case.
. If the subsets 
defined for 
,
satisfy 
for all 
and if 
for ,
and if 
for 
then
Proof: Assume k=1 and 
to begin. Apply 
with respect to 
to the integral (see appendix 9.6.1
for the definition of the T transform) and evaluate the inner T
transform. Noting 
and that 
for 
,
find
(See appendix 9.7.1
for the justification of the interchange the integral over 
and the T transform.) Now, write the transformed integral above
as the convolution
Use theorem 9.1 9.55
and induction to find (with 
)
Similarly, use theorem 9.2 and induction to find
Substituting the last two expressions into 
yields
The 
transform may now be taken to find (see appendix 9.4,9.7)
Now apply theorem 9.1 in this expression to find for
Refer to appendix 9.8
for the continuation to 
.
Refer to appendix 9.9
for the existence conditions. Now, for k>1 apply the identity operator
k times (with respect to 
respectively) and evaluate only the T transforms initially. Since
for ,
the convolution form of the transformed integral becomes
Now extend the application of theorem 9.2 to the k convolution
products 
for .
Do the substitutions and take the inverse T transforms to find the
result. QED.
Appendix 9.5 contains a derivation of an interesting identity based on an alternate form of the result in theorem 12.
Theorem 13 applies theorem 12 to find non-overlap results needed specifically for the expression of Bayes' estimators for the first two moments of the entropy, mutual information, and various other functions.
. If the subsets
defined for ,
satisfy
for all
and if
for ,
and if
for
then the following hold
(13.1) One logarithm subset sum.
(13.2) Two logarithms of subset sums, different subsets.
(13.3) Squared logarithm of a subset sum.
Proof: The proof is done for theorem 13.1; theorem 13.2 and 13.3
follow in a similar manner. Differentiate both sides of the formula for 
given in theorem 12 with respect to 
using the fact that 
.
(See appendix 9.7.2
for justification of the interchange the integral and derivative.) Doing
this gives the desired result. QED.
Theorems 12 and 13 dealt with non-overlap sums. Theorems 14 and 15
below discuss pair-wise overlap sums. In theorem 14a the non-contained
overlap case is discussed. In theorem 14b the contained overlap case is
discussed. See appendix 9.4
for the definition of the hypergeometric function 
.
. If the subsets 
and 
satisfy 
, 
, 
, 
,
and , ,
then
Proof: To begin, assume that 
,
i=1,2, and that the 
are not integers. Apply 
(
is with respect to ,
see appendix 9.6.1)
to the integral 
.
Evaluating the (non-inverse) T transforms yields the convolution
(see theorem 1)
(See appendix 9.7.1
for justification of the interchange of the integral over
and the T transform.) Apply theorem 9.1 and induction to find (where 
)
Similarly, use theorem 9.2 and induction to find
Substitute the result for theorem 11 into the triple convolution
above, and substitute the last two expressions into the convolution form
of the transformed integral to find
Now, take inverse T transforms and apply theorem 9.1 to find
the desired result. Refer to appendix 9.9
to determine the conditions for the existence of the identity. Refer to
appendix 9.8
for the continuation of the result to 
.
Finally, for values of 
,
refer to appendix 9.10.
QED.
See appendix 9.5
for a derivation of two interesting identities resulting from alternate
forms of this proof. When 
the above result simplifies as in theorem 14b below.
. If the subsets 
satisfy , , 
, 
,
and , ,
then
14b.1
14b.2
Proof: Similar to proof of theorem 14a, but apply theorem 10.1 instead of theorem 11. The second form (theorem 14b.2) of the result is derived by applying Gauss's identity (see appendix 9.5) to the first form of the result above. QED.
Theorems 15a and 15b build upon the results of theorems 14a and 14b
respectively and state results needed to express specific terms of the
various Bayes' estimators. Theorem 15a contains results for the non-contained
overlap case. Theorem 15b contains results for the contained overlap case.
Since we are most directly interested in non-negative integer 
's
and because simplification occurs at those 's,
Theorem 15a is stated only for non-negative integer 's.
. If 
and 
are integers and the conditions for theorem 14a hold then
15a.1
15a.2
15a.3
15a.4
where
and
(a)
(b)
with 
given by
(c) 
is the same as (b) with 
and 
.
(d)
(e)
with 
given by
Proof: The proof is done for theorem 15a.2. The other cases have
similar proofs. Differentiate both sides of the expression for
given in theorem 14a with respect to .
After differentiating, the left-hand side is given by 
.
(The justification of the interchange of the integral and derivative is
given in appendix 9.7.1.)
Write the differentiated right-hand side as
This expands to
.
The derivative of 
is given by 
.
The undifferentiated hypergeometric is evaluated at
and 
using the results in appendix 9.10,
cases 1 and 2. This evaluates to 
defined in (a) above. The derivative of the hypergeometric may be taken
term-by-term (this is justified below). Use the results in appendix 9.10,
cases 1 and 2, equations 9.144
and 9.147,
to evaluate this derivative
and .
Doing this gives the expression 
defined in (b). With these derivatives and evaluations, theorem 15a.2 follows
immediately. Now consider the validity of term-by-term differentiation
of the hypergeometric. There exists a closed neighborhood N containing
the integer
with 
, 
.
The results of appendix 9.10
show that any truncation (in j) of the series for 
(see appendix 9.4)
may be differentiated with respect to x on N. The sequence
of derivatives of the increasing order truncations converges uniformly
on N. (To see this, note that 
is convergent for each i, and 
.
Now, note that 
is a series of terms each monotonic on N with the same monotonicity
in x holding for each term, and that the summation over i
in (b) is finite. These observations and the convergence just established
demonstrate the claim of uniform convergence.) Finally, by theorem 7.17
of [73],
the sequence of derivatives of the increasing order truncations converges
to the derivative of the limit of the series on N, justifying the
term-by-term differentiation of the infinite series. QED.
See appendix 9.5
for some comments regarding alternate forms for the results given above
in theorem 15a. Theorem 15b builds on theorem 14b and states the results
for the case in which there are two subset sums, with the indices of one
subset completely contained in the other. Here, unlike in theorem 15a,
there is no hypergeometric function to consider, so the presentation of
these results is much shorter. Further, unlike theorem 15a, the expressions
given are valid for all 's
in the range specified (not just at nonnegative integers as in theorem
15a) because there are no poles in the expressions being considered at
the integers and therefore no further simplification occurs at these points.
. If the conditions for theorem 14b hold, then
15b.1
15b.2
15b.3
15b.4
15b.5
where
Proof: The proof is done for theorem 15b.2. The proofs of the other
results follow in a similar manner. The result of theorem 14b is
Differentiate both sides of this with respect to .
The left-hand side of the differentiated expression is 
.
(The justification of the interchange of the integral and derivative is
given in appendix 9.7.2.)
The derivative of 
is given by 
.
The derivative of 
is given by 
.
Substituting these expressions for the appropriate derivatives in the overall
derivative of the right-hand side of the equality above for 
gives the claimed result. QED.
