两个伽马分布之间的Kullback–Leibler散度

选择通过pdf g （x ; b ，c ）= 1参数化伽马分布$\Gamma(b,c)$$g(x;b,c) = \frac{1}{\Gamma(c)}\frac{x^{c-1}}{b^c}e^{-x/b}$ 之间的相对熵$\Gamma(b_q,c_q)$和$\Gamma(b_p,c_p)$是由为[1]中给出

$$\begin{align} KL_{Ga}(b_q,c_q;b_p,c_p) &= (c_q-1)\Psi(c_q) - \log b_q - c_q - \log\Gamma(c_q) + \log\Gamma(c_p)\\ &\qquad+ c_p\log b_p - (c_p-1)(\Psi(c_q) + \log b_q) + \frac{b_qc_q}{b_p} \end{align}$$

我猜$\Psi(x):= \Gamma'(x)/\Gamma(x)$是digamma函数。

这是没有派生的。我找不到任何可以得出这一点的参考。有什么帮助吗？一个好的参考就足够了。困难的部分是将与gamma pdf 集成。$\log x$

[1] WD Penny，法线，伽马，狄利克雷和Wishart密度的KL散度，请访问：www.fil.ion.ucl.ac.uk/~wpenny/publications/densities.ps

KL散度是形式积分的差

$$ \ eqalign {I（a，b，c，d）＆= \ int_0 ^ {\ infty} \ log \ left（\ frac {e ^ {-x / a} x ^ {b-1}} {a ^ b \ Gamma（b）} \ right）\ frac {e ^ {-x / c} x ^ {d-1}} {c ^ d \ Gamma（d）} dx \

＆=-\ frac {1} {a} \ int_0 ^ \ infty \ frac {x ^ de ^ {-x / c}} {c ^ d \ Gamma（d）} \，dx-\ log（a ^ b \ Gamma（b））\ int_0 ^ \ infty \ frac {e ^ {-x / c} x ^ {d-1}} {c ^ d \ Gamma（d）} \，dx \＆\ quad +（b- 1）\ int_0 ^ \ infty \ log（x）\ frac {e ^ {-x / c} x ^ {d-1}} {c ^ d \ Gamma（d）} \，dx \

＆=-\ frac {cd} {a}-\ log（a ^ b \ Gamma（b））+（b-1）\ int_0 ^ \ infty \ log（x）\ frac {e ^ {-x / c } x ^ {d-1}} {c ^ d \ Gamma（d）} \，dx} $$

我们只需要处理通过观察得到的右手积分

$$\eqalign{ \frac{\partial}{\partial d}\Gamma(d) =& \frac{\partial}{\partial d}\int_0^{\infty}e^{-x/c}\frac{x^{d-1}}{c^d}dx\\ =& \frac{\partial}{\partial d} \int_0^\infty e^{-x/c} \frac{(x/c)^{d-1}}{c}\,dx\\ =&\int_0^\infty e^{-x/c}\frac{x^{d-1}}{c^d} \log\frac{x}{c} \,dx\\ =&\int_0^{\infty}\log(x)e^{-x/c}\frac{x^{d-1}}{c^d}dx - \log(c)\Gamma(d). }$$

何处

$$\frac{b-1}{\Gamma(d)}\int_0^{\infty} \log(x)e^{-x/c}(x/c)^{d-1}dx = (b-1)\frac{\Gamma'(d)}{\Gamma(d)} + (b-1)\log(c).$$

插入先前的收益

$$I(a,b,c,d)=\frac{-cd}{a} -\log(a^b\Gamma(b))+(b-1)\frac{\Gamma'(d)}{\Gamma(d)} + (b-1)\log(c).$$

$\Gamma(c,d)$$\Gamma(a,b)$$I(c,d,c,d) - I(a,b,c,d)$

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实施细节

Gamma函数快速增长，因此为避免溢出，请不要计算Gamma并取其对数：而是使用在任何统计计算平台（包括Excel）中都可以找到的log-Gamma函数。

$\Gamma^\prime(d)/\Gamma(d)$$\Gamma,$$\psi,$

R$I$$\psi$

#
# `b` and `d` are Gamma shape parameters and
# `a` and `c` are scale parameters.
# (All, therefore, must be positive.)
#
KL.gamma <- function(a,b,c,d) {
  i <- function(a,b,c,d)
    - c * d / a - b * log(a) - lgamma(b) + (b-1)*(psigamma(d) + log(c))
  i(c,d,c,d) - i(a,b,c,d)
}
print(KL.gamma(1/114186.3, 202, 1/119237.3, 195), digits=12)

The Gamma distribution is in the exponential family because its density can be expressed as:

$$\begin{align} \newcommand{\mbx}{\mathbf{x}} \newcommand{\btheta}{\boldsymbol{\theta}} f(\mbx \mid \btheta) &= \exp\bigl(\eta(\btheta) \cdot T(\mbx) - g(\btheta) + h(\mbx)\bigr) \end{align}$$

Looking at the Gamma density function, its log-normalizer is

$$g(\btheta) = \log(\Gamma(c)) + c\log(b)$$ with natural parameters $$\btheta = \left[\begin{matrix}c-1\\-\frac1 b\end{matrix}\right]$$

All distributions in the exponential family have KL divergence:

$$\begin{align} KL(q; p) &= g(\btheta_p) - g(\btheta_q) - (\btheta_p-\btheta_q) \cdot \nabla g(\btheta_q). \end{align}$$

There's a really nice proof of that in:

Frank Nielsen, École Polytechnique, and Richard Nock, Entropies and cross-entropies of exponential families.