A. The determinant of a matrix with special form
To prove the lemma described in eqs. (3–5), we first establish a general equality in linear algebra:
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\det \left[ \begin{matrix}{\bf A} & {\bf B} \\ {\bf C} & {\bf
D}\end{matrix} \right] = \det {\bf A} \cdot \det \left[ {\bf D} -
{\bf C} {\bf A}^{- 1} {\bf B} \right] = \det {\bf D} \cdot \det
\left[ {\bf A} - {\bf B} {\bf D}^{- 1} {\bf C} \right] , \tag{30}
\end{align*}\end{document}
where A and D are invertible square matrices, while B and C are rectangular matrices. The equality (30) follows from the decompositions
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\left[ \begin{matrix}{\bf A} & {\bf B} \\ {\bf C} & {\bf
D}\end{matrix} \right] = \left[ \begin{matrix}{\bf A} & {\bf 0} \\
{\bf C} & {\bf I}\end{matrix} \right] \left[ \begin{matrix}{\bf I}
& {\bf A}^{- 1}{\bf B} \\ {\bf 0} & {\bf D} - {\bf CA}^{- 1} {\bf
B}\end{matrix} \right] = \left[ \begin{matrix}{\bf I} & {\bf B} \\
{\bf 0} & {\bf D}\end{matrix} \right] \left[ \begin{matrix}{\bf A}
- {\bf BD}^{- 1} {\bf C} & {\bf 0} \\ {\bf D}^{- 1}{\bf C} & {\bf
I}\end{matrix} \right] ,
\end{align*}\end{document}
and the fact that det[XY] = det X · det Y when X and Y are square matrices of the same dimension.
To proceed, we rewrite the matrix M of eq. (3) as the product
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
{\bf M} = {{\bf d}} \cdot \left[ {\bf I} - {\bf bb}^T \right]
\cdot {\bf d} , \tag{31}
\end{align*}\end{document}
where d is a diagonal matrix with elements
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$d_{j , j^{\prime}} = \delta_{j , j^{\prime}} \sqrt{D_j}$$\end{document}
, and b is a vector whose components are
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$b_j = \sqrt{D / D_j}$$\end{document}
. Eq. (31) implies that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\det {\bf M} = \det {\bf d} \cdot \det \left[ {\bf I} - {\bf b}
{\bf b}^T \right] \cdot \det {\bf d}.\tag{32}
\end{align*}\end{document}
Using eq. (30), we observe that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\det \left[ {\bf I} - {\bf b} {\bf b}^T \right] = \left(1 - {\bf
b}^T {\bf b} \right) \tag{33}
\end{align*}\end{document}
by considering the determinant of the matrix
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\left[ \begin{matrix}1 & {\bf b}^T \\ {\bf b} & {\bf
I}\end{matrix} \right].
\end{align*}\end{document}
Since matrix d consists of diagonal elements only, we may write the determinant of M as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\det {\bf M} = \left(\prod_{j = 1}^L \sqrt {D_j} \right)^2 \left[
1 - \sum_{j = 1}^L \left(\sqrt {\frac {D} {D_j}} \right)^2
\right] = F \prod_{j = 1}^L D_j ,
\end{align*}\end{document}
as stated in eqs. (4) and (5).
B. The expectation of a function of cj
To evaluate the asymptotic behavior of Dj and F, it is necessary to calculate expectation values of functions that depend only on the count vector
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$\vec{c}$$\end{document}
. For this purpose, it is convenient to elaborate eq. (6) as follows:
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
P (\vec y) & = \frac {\Gamma (\alpha)} {\Gamma (\alpha + c)}
\prod_{j = 1}^L \frac {\Gamma (c_j + \alpha_j)} {\Gamma
(\alpha_j)} \\ & = \frac {\Gamma (\alpha)} {\prod_{j = 1}^{L}
\Gamma (\alpha_j)} \int_{\Omega_L} \prod_{j = 1}^{L} x_j^{c_j +
\alpha_j - 1} d \vec x . \tag{34}
\end{align*}\end{document}
Focusing now on
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$\vec{c}$$\end{document}
, eq. (34) implies that we may write the expectation value of a function
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$K (\vec{c})$$\end{document}
as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
& E [ K (\vec {c}) ] = \sum_{\vec {y}} P (\vec {y}) K (\vec {c})
\\ & \frac {\Gamma (\alpha)} {\prod_{j = 1}^{L} \Gamma
(\alpha_j)} \sum_{{c_1 , c_2 , \ldots , c_L \ge 0 \atop \Sigma_j
c_j = c}} \frac {c!} {\prod_{j = 1}^L c_j!} \int_{\Omega_L}
\prod_{j = 1}^{L} x_j^{c_j + \alpha_j - 1} K (\vec c) d \vec {x} .
\tag{35}
\end{align*}\end{document}
If K depends on only one of the cj, we may integrate away all the xj′≠j to simplify eq. (35). This amounts to writing
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\sum_{{c_1 , c_2 , \ldots , c_L \ge 0 \atop \Sigma_j c_j = c}} =
\sum_{c_j = 0}^c \sum_{{\{c_{j^{\prime} \ne j} \ge 0 \} \atop
\Sigma_{j^{\prime}} c_{j^{\prime}} = c - c_j}} , \tag{36}
\end{align*}\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\frac {c!} {\prod_{j = 1}^L c_j !} = \frac {c!} {c_j! (c - c_j)
!} \frac {(c - c_j) !} {\prod_{j^{\prime} \ne j} c_{j^{\prime}}
!} , \tag{37}
\end{align*}\end{document}
so that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E [ K (c_j) ] & = \sum_{\vec y} P (\vec y) K (c_j)
\\ & = \frac {\Gamma (\alpha)} {\prod_{j = 1}^{L} \Gamma (\alpha_j)}
\sum_{c_j = 0}^c {c! K (c_j)} {c_j! (c - c_j) !} \int_{\Omega_L}
x_j^{c_j + \alpha_j - 1} (1 - x_j)^{c - c_j} \prod_{j^{\prime} \ne
j} x_{j^{\prime}}^{\alpha_{j^{\prime}} - 1} d \vec {x} \\ & =
\frac {\Gamma (\alpha)} {\prod_{j = 1}^{L} \Gamma (\alpha_j)}
\sum_{c_j = 0}^c \frac {c! K (c_j)} {c_j! (c - c_j) !} \int_0^1
x_j^{c_j + \alpha_j - 1} (1 - x_j)^{c - c_j + \sum_{j^{\prime} \ne
j} \alpha_{j^{\prime}}} d x_j \\ & \quad \times \int_0^{1} \delta
\left((1 - x_j) \left[ \sum_{j^{\prime} \ne j} \tilde
x_{j^{\prime}} - 1 \right] \right) \prod_{j^{\prime} \ne j} \left[
\tilde x_{j^{\prime}}^{\alpha_{j^{\prime}} - 1} d \tilde
x_{j^{\prime}} \right] \\ & = \frac {\Gamma (\alpha)} {\prod_{j =
1}^{L} \Gamma (\alpha_j)} \sum_{c_j = 0}^c \frac {c! K (c_j)}
{c_j! (c - c_j) !} \int_0^1 x_j^{c_j + \alpha_j - 1} (1 - x_j)^{c
- c_j + \alpha - \alpha_{j}} d x_j \\ & \quad \times (1 - x_j)^{-
1} \int_{\Omega_{L - 1}} \prod_{j^{\prime} \ne j} \left[ \tilde
x_{j^{\prime}}^{\alpha_{j^{\prime}} - 1} d \tilde x_{j^{\prime}}
\right] \\ & = \frac {\Gamma (\alpha) / \Gamma (\alpha_j)} {\Gamma
(\alpha - \alpha_j)} \sum_{c_j = 0}^c \frac {c! K (c_j)} {c_j! (c
- c_j) !} \int_0^1 x^{c_j + \alpha_j - 1} (1 - x)^{c + \alpha -
c_j - \alpha_{j} - 1} dx. \tag{38}
\end{align*}\end{document}
C. The asymptotic behavior of D, Dj, and F
As mentioned in the main text, to obtain F to the order α−2, we need to calculate D
j
and D to the order α−4. Here we provide the details needed for this task.
One way to express the trigamma function is
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\psi^{\prime} (x) = \sum_{k = 0}^\infty \frac {1} {(x + k)^2} ,
\end{align*}\end{document}
which in the large x limit yields the asymptotic form
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\psi^{\prime} (x) = \frac {1} {x} + \frac {1} {2 x^2} + \frac
{1} {6 x^3} + {\cal O} (x^{- 5}) . \tag{39}
\end{align*}\end{document}
For notational convenience, the nth derivative of the digamma function will be written as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$\Psi^{(n)} (x) \equiv d^n \Psi (x) / dx^n$$\end{document}
. That is, ψ′(x) and ψ(1)(x) represent the same function dψ(x)/dx. To seek the asymptotic behavior of Dj, we introduce
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$k_j \equiv E[c_j]$$
\end{document}
, and expand the function ψ(1)(α
j
+ cj) around αj + kj prior to taking the average. In other words, we write
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\psi^{(1)} (\alpha_j + c_j) = \psi^{(1)} (\alpha_j + k_j) +
\sum_{\ell = 1}^\infty \frac {(c_j - k_j)^{\ell}} {\ell !}
\psi^{(\ell + 1)} (\alpha_j + k_j) , \tag{40}
\end{align*}\end{document}
which leads to
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ \psi^{(1)} (\alpha_j + c_j) \right] = \psi^{(1)}
(\alpha_j + k_j) + \sum_{\ell = 2}^\infty \frac {E \left[ (c_j -
k_j)^{\ell} \right]} {\ell !} \psi^{(\ell + 1)} (\alpha_j + k_j)
. \tag{41}
\end{align*}\end{document}
The fact that ψ(ℓ+1)(α
j
+ kj) is of order α−ℓ−1 makes eq. (41) an asymptotic expansion, provided that E[(cj − kj)ℓ] becomes independent of α in the large-α limit. We shall show this later by evaluating the leading behavior of E [(cj − kj)ℓ]. From eq. (41), it is evident that we need to compute up to at least the third central moment to obtain Dj to the order α−4.
Introducing the new variable
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}
$$v \equiv c_j - 1$$
\end{document}
, we may now use eq. (38) of Appendix B to calculate kj:
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
k_j & \equiv \sum_{\vec y} P (\vec y) c_j = {\Gamma (\alpha) /
\Gamma (\alpha_j)} {\Gamma (\alpha - \alpha_j)} c \sum_{v = 0}^{c
- 1} C^{c - 1}_v \int_0^1 x^{v + \alpha_j} (1 - x)^{c - 1 - v +
\alpha - \alpha_{j} - 1} dx \\ & = \frac {\Gamma (\alpha) / \Gamma
(\alpha_j)} {\Gamma (\alpha - \alpha_j)} c \int_0^1 x^{\alpha_j}
(1 - x)^{\alpha - \alpha_{j} - 1} dx \\ & = c \frac {\Gamma
(\alpha)} {\Gamma (\alpha_j) \Gamma (\alpha - \alpha_j)} \frac
{\Gamma (\alpha_j + 1) \Gamma (\alpha - \alpha_j)} {\Gamma
(\alpha + 1)} = c \frac {\alpha_j} {\alpha} , \tag{42}
\end{align*}\end{document}
as mentioned in the main text.
To compute the second central moment, we write
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ (c_j - k_j)^2 \right] = E [ c_j (c_j - 1) ] + E [ c_j ] -
k_j^2 = E [ c_j (c_j - 1) ] + k_j - k_j^2 , \tag{43}
\end{align*}\end{document}
and
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E [ c_j (c_j - 1) ] & = \frac {\Gamma (\alpha) / \Gamma
(\alpha_j)} {\Gamma (\alpha - \alpha_j)} c (c - 1) \sum_{v = 0}^{c
- 2} C^{c - 2}_v \int_0^1 x^{v + \alpha_j + 1} (1 - x)^{c - 2 - v
+ \alpha - \alpha_{j} - 1} dx \\ & = \frac {\Gamma (\alpha)}
{\Gamma (\alpha_j) \Gamma (\alpha - \alpha_j)} c (c - 1) \int_0^1
x^{\alpha_j + 1} (1 - x)^{\alpha - \alpha_{j} - 1} d x \\ & =
\frac {\Gamma (\alpha)} {\Gamma (\alpha_j) \Gamma (\alpha -
\alpha_j)} c (c - 1) \frac {\Gamma (\alpha_j + 2) \Gamma (\alpha -
\alpha_j)} {\Gamma (\alpha + 2)} \\ & = c (c - 1) \frac {\alpha_j
(\alpha_j + 1)} {\alpha (\alpha + 1)} . \tag{44}
\end{align*}\end{document}
Therefore,
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E [ (c_j - k_j)^2 ] = \frac {c \alpha_j} {\alpha} \left[ \frac
{(c - 1) (\alpha_j + 1)} {\alpha + 1} + 1 - \frac {c \alpha_j}
{\alpha} \right] = c q_j (1 - q_j) \frac {\alpha + c} {\alpha +
1} . \tag{45}
\end{align*}\end{document}
The calculation of the ℓth central moment of cj employs the same idea. We first express
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
c_j^r & = \sum_{m = 0}^{r - 1} g_{m;r} c_j (c_j - 1) \cdots (c_j -
m) \tag{46} \\ & = \sum_{m = 0}^{r - 1} g_{m;r} \frac {c_j !}
{(c_j - m - 1) !} ,
\end{align*}\end{document}
where the coefficients gm;r depend on r and can be obtained in the following way. By setting cj = 1 in eq. (46), we zero all terms except g0;r on the right hand side, and therefore see that g0;r = 1. With g0;r known, we may then set cj = 2; the only nonzero terms on the right hand side of eq. (46) are then 2g0;r + 2!g1;r. With the left hand side now equal to 2
r
, and g0;r = 1, we easily solve for g1;r = 2r−1 − 1. We may then set
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$c_j = 3 , 4 , \ldots$$\end{document}
etc. to solve respectively for
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$g_{2;r} , g_{3;r} , \ldots$$\end{document}
etc.
Note that it is possible to express
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ c_j (c_j - 1) \cdots (c_j - m) \right] & = {\Gamma
(\alpha) c (c - 1) \cdots (c - m)} {\Gamma (\alpha_j) \Gamma
(\alpha - \alpha_j)} \int_0^1 x^{\alpha_j + m} (1 - x)^{\alpha -
\alpha_{j} - 1} d x \\ & = \frac {\Gamma (\alpha) c (c - 1) \cdots
(c - m)} {\Gamma (\alpha_j) \Gamma (\alpha - \alpha_j)} \frac
{\Gamma (\alpha_j + m + 1) \Gamma (\alpha - \alpha_j)} {\Gamma
(\alpha + m + 1)} \\ & = c (c - 1) \cdots (c - m) \frac {\alpha_j
(\alpha_j + 1) \cdots (\alpha_j + m)} {\alpha (\alpha + 1) \cdots
(\alpha + m)} . \tag{47}
\end{align*}\end{document}
This means that for ℓ ≥ 2 we may write
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ (c_j - k_j)^{\ell} \right] & = (- 1)^{\ell - 1} (\ell -
1) k_j^{\ell} + \sum_{r = 2}^\ell C^{\ell}_r E \left[ c_j^r
\right] (- k_j)^{\ell - r} \\ & = (- 1)^{\ell - 1} (\ell - 1)
k_j^{\ell} + \sum_{r = 2}^{\ell} C^{\ell}_r (- k_j)^{\ell - r} \\
& \times \left\{k_j + \sum_{m = 1}^{r - 1} g_{m;r} \frac {c!} {(c
- m - 1) !} \frac {\alpha_j (\alpha_j + 1) \cdots (\alpha_j + m)}
{\alpha (\alpha + 1) \cdots (\alpha + m)} \right\} . \tag{48}
\end{align*}\end{document}
As an example, we may use this approach to obtain the third central moment
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ (c_j - k_j)^3 \right] = c \ q_j (1 - q_j) (1 - 2 q_j)
\frac {(\alpha + c) (\alpha + 2c)} {(\alpha + 1) (\alpha + 2)} .
\tag{49}
\end{align*}\end{document}
Because c is a fixed positive integer, in the large-α limit we may write the leading term in eq. (48) as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ (c_j - k_j)^{2 \ell} \right] \quad \approx \qquad (2 \ell
- 1) !! \ \frac {c^{\ell} \alpha^{2 \ell - 2} \ (\alpha + c)}
{(\alpha + 1) (\alpha + 2) \cdots (\alpha + 2 \ell - 1)} \
q_j^\ell (1 - q_j)^\ell \tag{50}
\end{align*}\end{document}
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
{\mathop {\longrightarrow} \limits_{\alpha \rightarrow \infty}}
\quad (2 \ell - 1) !! \ c^{\ell} \ q_j^\ell (1 - q_j)^{\ell} ,
\tag{51}
\end{align*}\end{document}
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
E \left[ (c_j - k_j)^{2 \ell + 1} \right] \quad \approx \qquad
\frac {\ell} {3} \ (2 \ell + 1) !! \ \frac {c^{\ell} \ \alpha^{2
\ell - 1} \ (\alpha + c) \ (1 - 2 q_j)} {(\alpha + 1) (\alpha +
2) \cdots (\alpha + 2 \ell)} q_j^\ell (1 - q_j)^\ell \tag{52}
\end{align*}\end{document}
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
{\mathop {\longrightarrow} \limits_{\alpha \rightarrow \infty}}
\quad \frac {\ell} {3} (2 \ell + 1) !! \ c^\ell \ (1 - 2 q_j) \
q_j^\ell (1 - q_j)^\ell. \tag{53}
\end{align*}\end{document}
In the limit of large αj (or kj), we note that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\frac {\psi^{(1 + \ell)} (\alpha_j + k_j)} {\ell !} = \frac {(-
1)^{\ell}} {q_j^{\ell + 1} (\alpha + c)^{\ell + 1}} \left[ 1 +
\frac {\ell + 1} {2 q_j (\alpha + c)} + {\cal O} \left(\frac {1}
{(\alpha_j + k_j)^{2}} \right) \right] . \tag{54}
\end{align*}\end{document}
Therefore, the absence of α dependence and the presence of the factorials in eqs. (51) and (53) means that the expansion in eq. (41) is asymptotic in α. That is, the larger α, the more terms in the expansion one may retain to improve accuracy before the series becomes divergent.
With the aim of obtaining D and the Dj to the order α−4, we now continue the investigation of their large-α behavior. Using eq. (39), we write the asymptotic expression for D as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
D & = \psi^{(1)} (\alpha) - \psi^{(1)} (\alpha + c)
\\ & = \left({1} {\alpha} + \frac {1} {2 \alpha^2} + \frac {1} {6
\alpha^3} + {\cal O} (\alpha^{- 5}) \right) - \left(\frac {1}
{\alpha + c} + \frac {1} {2 (\alpha + c)^2} + \frac {1} {6
(\alpha + c)^3} + {\cal O} (\alpha^{- 5}) \right) \\ & = \frac {c}
{\alpha (\alpha + c)} \left[ 1 + \frac {1} {2} \left(\frac {1}
{\alpha} + \frac {1} {\alpha + c} \right) + \frac {1} {6}
\left(\frac {1} {\alpha^2} + \frac {1} {\alpha (\alpha + c)} +
\frac {1} {(\alpha + c)^2} \right) \right] + {\cal O} (\alpha^{-
5}) . \tag{55}
\end{align*}\end{document}
Next, noting that kj = cqj, we find that
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
D_j & = \psi^{(1)} (\alpha_j) - E [ \psi^{(1)} (\alpha_j + c_j) ]
\\ & = \frac {c} {\alpha_j (\alpha + c)} \left[ 1 + \frac {1}
{2q_j} \left(\frac {1} {\alpha} + \frac {1} {\alpha + c} \right)
+ {1} {6q_j^2} \left(\frac {1} {\alpha^2} + \frac {1} {\alpha
(\alpha + c)} + \frac {1} {(\alpha + c)^2} \right) \right] \\ &
\quad - \frac {E [ (c_j - k_j)^2 ]} {2!} \psi^{(3)} (\alpha_j +
k_j) - \frac {E [ (c_j - k_j)^3 ]} {3!} \psi^{(4)} (\alpha_j +
k_j) + {\cal O} (\alpha^{- 5}) \\ & = \frac {c} {\alpha_j (\alpha
+ c)} \left[ 1 + \frac {1} {2q_j} \left(\frac {1} {\alpha} +
\frac {1} {\alpha + c} \right) + \frac {1} {6q_j^2} \left(\frac
{1} {\alpha^2} + \frac {1} {\alpha (\alpha + c)} + \frac {1}
{(\alpha + c)^2} \right) \right] \\ & \quad - \left[ c q_j (1 -
q_j) \frac {\alpha + c} {\alpha + 1} \right] \left[ \frac {1}
{q_j^3 (\alpha + c)^3} + \frac {3} {2 q_j^4 (\alpha + c)^4}
\right] \\ & \quad - \left[ c q_j (1 - q_j) (1 - 2q_j) \frac
{(\alpha + c) (\alpha + 2c)} {(\alpha + 1) (\alpha + 2)} \right]
\frac {(- 1)} {q_j^4 (\alpha + c)^4} + {\cal O} (\alpha^{- 5}) \\
& = \frac {c} {\alpha_j (\alpha + c)} \left\{1 + \frac {1} {2q_j}
\left(\frac {1} {\alpha} + \frac {1} {\alpha + c} \right) + \frac
{1} {6q_j^2} \left(\frac {1} {\alpha^2} + \frac {1} {\alpha
(\alpha + c)} + \frac {1} {(\alpha + c)^2} \right) \right. \\ &
\left. \quad - \frac {\alpha (1 - q_j)} {q_j (\alpha + c) (\alpha
+ 1)} - \frac {3} {2} \frac {\alpha (1 - q_j)} {q_j^2 (\alpha +
c)^2 (\alpha + 1)} + \frac {\alpha (1 - q_j) (1 - 2q_j) (\alpha +
2c)} {q_j^2 (\alpha + c)^2 (\alpha + 1) (\alpha + 2)} \right\} +
{\cal O} (\alpha^{- 5}) \\ & = \frac {c} {\alpha_j (\alpha + c)}
\left\{1 + \frac {\alpha} {(\alpha + c) (\alpha + 1)} + \frac {2
\alpha (\alpha + 2c)} {(\alpha + c)^2 (\alpha + 1) (\alpha + 2)}
+ \frac {1} {2q_j} \left[ \frac {1} {\alpha} + \frac {1}
{\alpha + c} \right. \right. \\ & \left. \quad - \frac {2 \alpha}
{(\alpha + c) (\alpha + 1)} + \frac {3 \alpha} {(\alpha + c)^2
(\alpha + 1)} - \frac {6 \alpha (\alpha + 2c)} {(\alpha + c)^2
(\alpha + 1) (\alpha + 2)} \right] + \frac {1} {6q_j^2} \left[
\frac {1} {\alpha^2} + \frac {1} {\alpha (\alpha + c)} \right. \\
& \left. \left. \qquad + \frac {1} {(\alpha + c)^2} - \frac {9
\alpha} {(\alpha + c)^2 (\alpha + 1)} + \frac {6 \alpha (\alpha +
2c)} {(\alpha + c)^2 (\alpha + 1) (\alpha + 2)} \right] \right\} +
{\cal O} (\alpha^{- 5}) . \tag{56}
\end{align*}\end{document}
When summing the terms associated with
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inside the second pair of square brackets, we find that the final contribution becomes of order α−5, and thus may be dropped from the analysis. Also, the terms associated with
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}$$q_j^{- 1}$$\end{document}
, when added, can be rearranged as
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\frac {1} {\alpha} + & \frac {1} {\alpha + c} - \frac {2 \alpha}
{(\alpha + c) (\alpha + 1)} + \frac {3 \alpha} {(\alpha + c)^2
(\alpha + 1)} - \frac {6 \alpha (\alpha + 2c)} {(\alpha + c)^2
(\alpha + 1) (\alpha + 2)} \\ & = \frac {c} {\alpha (\alpha + c)}
- \frac {1} {(\alpha + c) (\alpha + 1)} + {\cal O} (\alpha^{- 3})
. \tag{57}
\end{align*}\end{document}
Consequently, we have
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
D_j & = {c} {\alpha_j (\alpha + c)} \left\{1 + \frac {\alpha}
{(\alpha + c) (\alpha + 1)} + \frac {2 \alpha (\alpha + 2c)}
{(\alpha + c)^2 (\alpha + 1) (\alpha + 2)} \right.
\\ & \left. + \frac {1} {2q_j} \left[ \frac {c} {\alpha (\alpha
+ c)} - \frac {1} {(\alpha + c) (\alpha + 1)} \right] \right\} +
{\cal O} (\alpha^{- 5}) . \tag{58}
\end{align*}\end{document}
Eqs. (55) and (58) allow us to express F to the order α−2. We first calculate
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
\frac {D} {D_j} & = \frac {\alpha_j} {\alpha} \left[ 1 + \frac
{1} {2} \left(\frac {1} {\alpha} + {1} {\alpha + c} \right) +
\frac {1} {6} \left(\frac {1} {\alpha^2} + \frac {1} {\alpha
(\alpha + c)} + \frac {1} {(\alpha + c)^2} \right) \right]
\\ & \qquad \times \left\{1 - \frac {\alpha} {(\alpha + c) (\alpha +
1)} - \frac {2 \alpha (\alpha + 2c)} {(\alpha + c)^2 (\alpha + 1)
(\alpha + 2)} \right. \\ & \left. \qquad - \frac {1} {2q_j}
\left[ \frac {c} {\alpha (\alpha + c)} - \frac {1} {(\alpha + c)
(\alpha + 1)} \right] + \frac {\alpha^2} {(\alpha + c)^2 (\alpha
+ 1)^2} \right\} + {\cal O} (\alpha^{- 3}) \\ & = q_j \left\{1 +
\frac {1} {2} \left[ \frac {1} {\alpha} + \frac {1} {\alpha +
c} - \frac {2 \alpha} {(\alpha + c) (\alpha + 1)} \right] \left(1
- \frac {\alpha} {(\alpha + c) (\alpha + 1)} \right) \right. \\ &
\qquad + \frac {1} {6} \left[ \frac {1} {\alpha^2} + \frac {1}
{\alpha (\alpha + c)} + \frac {1} {(\alpha + c)^2} \right] -
\frac {2 \alpha (\alpha + 2c)} {(\alpha + c)^2 (\alpha + 1)
(\alpha + 2)} \\ & \left. \qquad - \frac {1} {2q_j} \left[ \frac
{c} {\alpha (\alpha + c)} - \frac {1} {(\alpha + c) (\alpha +
1)} \right] \right\} + {\cal O} (\alpha^{- 3}) \\ & = q_j \left\{1
+ \frac {1 - q_j^{- 1}} {2} \left[ \frac {c} {\alpha (\alpha +
c)} - \frac {1} {(\alpha + c) (\alpha + 1)} \right] + \frac {3}
{2} \frac {1} {(\alpha + c) (\alpha + 1)} \right. \\ & \left.
\qquad + \frac {1} {6} \left[ \frac {1} {\alpha^2} + \frac {1}
{\alpha (\alpha + c)} + \frac {1} {(\alpha + c)^2} \right] -
\frac {2 \alpha (\alpha + 2c)} {(\alpha + c)^2 (\alpha + 1)
(\alpha + 2)} \right\} + {\cal O} (\alpha^{- 3}) \tag{59}
\end{align*}\end{document}
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
= q_j \left\{1 + \frac {1 - q_j^{- 1}} {2} \left[ \frac {c}
{\alpha (\alpha + c)} - \frac {1} {(\alpha + c) (\alpha + 1)}
\right] \right\} + {\cal O} (\alpha^{- 3}) , \tag{60}
\end{align*}\end{document}
where the final expression comes from the fact that the last three terms in eq. (59) sum to order α−3. We may now compute F to order α−2:
\documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland,xspace}\usepackage{amsmath,amsxtra}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document}\begin{align*}
F & = 1 - D \sum_{j = 1}^L D_j^{- 1} = \left(\frac {c} {2 \alpha
(\alpha + c)} - \frac {1} {2 (\alpha + c) (\alpha + 1)} \right)
\sum_{j = 1}^L (1 - q_j) + {\cal O} (\alpha^{- 3}) \\ & = (L - 1)
\left(\frac {c} {2 \alpha (\alpha + c)} - \frac {1} {2 (\alpha +
c) (\alpha + 1)} \right) + {\cal O} (\alpha^{- 3}) \\ & = \frac
{(L - 1) (c - 1)} {2 \alpha (\alpha + c)} + {\cal O} (\alpha^{-
3}) ,
\end{align*}\end{document}
the result shown in eq. (15) of the main text.
