用Gamma分布构造Dirichlet分布

令是相互独立的随机变量，每个变量的伽玛分布参数为表示，与 $X_1,\dots,X_{k+1}$ $\alpha_i,i=1,2,\dots,k+1$ $Y_i=\frac{X_i}{X_1+\cdots+X_{k+1}},i=1,\dots,k$ $\text{Dirichlet}(\alpha_1,\alpha_2,\dots,\alpha_k;\alpha_{k+1})$

的联合PDF。然后找到关节 pdf文件，我找不到jacobian即 $(X_1,\dots,X_{k+1})=\frac{e^{-\sum_{i=1}^{k+1}x_i}x_1^{\alpha_1-1}\dots x_{k+1}^{\alpha_{k+1}-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)\dots \Gamma(\alpha_{k+1})}$ $(Y_1,\dots,Y_{k+1})$ $J(\frac{x_1,\dots,x_{k+1}}{y_1,\dots,y_{k+1}})$

— ha
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请参阅本文档的第13-14页。

@Procrastinator非常感谢您，对于我的问题，文档是最好的答案。

— Argha 2012年

@Procrastinator-也许您应该将此作为答案，因为OP对此很满意，并添加了几句话，这样您就不会遇到“我们希望提供多句子答案”的警告了吗？

— jbowman 2012年

该文档现在是404

— 无法解决。–麻烦

Wayback机器进行救援：pdf

— mobeets'Apr 19'17 at

雅各比学派（Jacobians）-变量函数变化的绝对决定因素-看起来很强大并且可能很复杂。然而，它们是变量多元变化的计算中必不可少的部分。似乎没有什么可做的，只是写下一个 $k+1$ x $k+1$ 的导数矩阵并进行计算。

有更好的方法。 它在“解决方案”部分的末尾显示。因为这篇文章的目的是向统计学家介绍对于许多人来说可能是一种新方法，所以它的大部分致力于解释解决方案背后的机制。这是微分形式的代数。（差异形式是一个在多个维度上集成的事物。）其中包含一个详细而有效的示例，以帮助使其更加熟悉。

背景

一个多世纪以前，数学家发展了微分代数理论，以处理多维几何中出现的“高阶导数”。行列式是由此类代数操纵的基本对象的特例，这些代数通常是交替的多线性形式。 这样做的好处在于计算变得如此简单。

这就是您需要了解的所有内容。

甲差动是如下形式的表达“ $dx_i$ ”。它是“ $d$ ”与任何变量名的串联。
一种形式是微分的线性组合，例如 $dx_1+dx_2$ 或什至 $x_2 dx_1 - \exp(x_2) dx_2$ 。即，系数是变量的函数。
可以使用记为的楔积乘以“乘积”形式。该乘积是反交换的（也称为交替的）：对于任何两种形式的和， $\wedge$ $\omega$ $\eta$

$ω \land η = - η \land ω .$ $\omega \wedge \eta = -\eta \wedge \omega.$
此乘法是线性且关联的：换句话说，它以熟悉的方式工作。一个直接的后果是，，这意味着任何一个形式的平方始终为零。这使得乘法非常容易！ $\omega \wedge \omega = -\omega \wedge \omega$
用于操纵所述积的出现在概率计算的目的，就像表达式可以理解为。 $dx_1 dx_2 \cdots dx_{k+1}$ $|dx_1\wedge dx_2 \wedge \cdots \wedge dx_{k+1}|$
当是一个函数时，则其微分由微分给出： $y = g(x_1, \ldots, x_n)$

$d y = d g (x_{1}, \dots, x_{n}) = \frac{\partial g}{\partial x_{1}} (x_{1}, \dots, x_{n}) d x_{1} + \dots + \frac{\partial g}{\partial x_{1}} (x_{1}, \dots, x_{n}) d x_{n} .$ $dy = dg(x_1, \ldots, x_n) = \frac{\partial g}{\partial x_1}(x_1, \ldots, x_n) dx_1 + \cdots + \frac{\partial g}{\partial x_1}(x_1, \ldots, x_n) dx_n.$

与雅可比行列式的联系是：变换的雅可比行列式直到符号为止，只是的系数 $(y_1, \ldots, y_n) = F(x_1, \ldots, x_n) = (f_1(x_1, \ldots, x_n), \ldots, f_n(x_1, \ldots, x_n))$ 显示在计算 $dx_1\wedge \dots \wedge dx_n$

d y_{1} \land \dots \land d y_{n} = d f_{1} (x_{1}, \dots, x_{n}) \land \dots \land d f_{n} (x_{1}, \dots, x_{n})

$dy_1 \wedge \cdots \wedge dy_n = df_1(x_1,\ldots, x_n)\wedge \cdots \wedge df_n(x_1, \ldots, x_n)$

在将每个为规则（5）中的线性组合之后。 $df_i$ $dx_j$

例

雅可比定律的简单定义很吸引人。尚未确信这值得吗？考虑将二维积分从笛卡尔坐标为极坐标的众所周知的问题，其中。以下是上述规则的完全机械应用，其中“ $(x, y)$ $(r,\theta)$ $(x,y) = (r\cos(\theta), r\sin(\theta))$ $(*)$ “用于简化将明显凭借规则（3），这意味着消失表达式。 $dr\wedge dr = d\theta\wedge d\theta = 0$

\begin{aligned} d x d y & = | d x \land d y | = | d (r \cos (θ)) \land d (r \sin (θ)) | \\ = | (\cos (θ) d r - r \sin (θ) d θ) \land (\sin (θ) d r + r \cos (θ) d θ | \\ = | (*) d r \land d r + (*) d θ \land d θ - r \sin (θ) d θ \land \sin (θ) d r + \cos (θ) d r \land r \cos (θ) d θ | \\ = | 0 + 0 + r \sin^{2} (θ) d r \land d θ + r \cos^{2} (θ) d r \land d θ | \\ = | r (\sin^{2} (θ) + \cos^{2} (θ)) d r \land d θ) | \\ = r d r d θ \end{aligned} .

$\eqalign{ dx dy &= |dx\wedge dy| = |d(r\cos(\theta)) \wedge d(r\sin(\theta))| \\ &= |(\cos(\theta)dr - r\sin(\theta)d\theta) \wedge (\sin(\theta)dr + r\cos(\theta)d\theta| \\ &= |(*)dr\wedge dr + (*) d\theta\wedge d\theta - r\sin(\theta)d\theta\wedge \sin(\theta)dr + \cos(\theta)dr \wedge r\cos(\theta) d\theta| \\ &= |0 + 0 + r\sin^2(\theta) dr\wedge d\theta + r\cos^2(\theta) dr\wedge d\theta| \\ &= |r(\sin^2(\theta) + \cos^2(\theta)) dr\wedge d\theta)| \\ &= r\ dr d\theta }.$

这样做的重点是可以轻松执行此类计算，而不会弄乱矩阵，行列式或其他此类多指标对象。您只是将事情乘以，记住楔是反交换的。它比高中代数中的授课要容易。

初赛

让我们看看这个微分代数的作用。 在这个问题中，的联合分布的PDF 是单独的PDF的产物（因为被假设为独立的）。为了处理变量的变化，我们必须明确要积分的微分元素。这些构成了术语 $(X_1, X_2, \ldots, X_{k+1})$ $X_i$ $Y_i$ $dx_1 dx_2 \cdots dx_{k+1}$ . Including the PDF gives the probability element

\begin{aligned} f_{X} (x, α) d x_{1} \dots d x_{k + 1} & \propto (x_{1}^{α_{1} - 1} \exp (- x_{1})) \dots (x_{k + 1}^{α_{k + 1} - 1} \exp (- x_{k + 1})) d x_{1} \dots d x_{k + 1} \\ = x_{1}^{α_{1} - 1} \dots x_{k + 1}^{α_{k + 1} - 1} \exp (- (x_{1} + \dots + x_{k + 1})) d x_{1} \dots d x_{k + 1} . \end{aligned}

$\eqalign{ f_\mathbf{X}(\mathbf{x},\mathbf{\alpha})dx_1 \cdots dx_{k+1} &\propto \left(x_1^{\alpha_1-1}\exp\left(-x_1\right)\right)\cdots \left(x_{k+1}^{\alpha_{k+1}-1}\exp\left(-x_{k+1}\right) \right)dx_1 \cdots dx_{k+1} \\ &= x_1^{\alpha_1-1}\cdots x_{k+1}^{\alpha_{k+1}-1}\exp\left(-\left(x_1+\cdots+x_{k+1}\right)\right)dx_1 \cdots dx_{k+1}. }$

(The normalizing constant has been ignored; it will be recovered at the end.)

Staring at the definitions of the $Y_i$ a few seconds ought to reveal the utility of introducing the new variable

Z = X_{1} + X_{2} + \dots + X_{k + 1},

$Z = X_1 + X_2 + \cdots + X_{k+1},$

giving the relationships

X_{i} = Y_{i} Z .

$X_i = Y_i Z.$

This suggests making the change of variables $x_i \to y_i z$ in the probability element. The intention is to retain the first $k$ variables $y_1, \ldots, y_k$ along with $z$ and then integrate out $z$ . To do so, we have to re-express all the $dx_i$ in terms of the new variables. This is the heart of the problem. It's where the differential algebra takes place. To begin with,

d x_{i} = d (y_{i} z) = y_{i} d z + z d y_{i} .

$dx_i = d(y_i z) = y_i dz + z dy_i.$

Note that since $Y_1+Y_2+\cdots+Y_{k+1}=1$ , then

0 = d (1) = d (y_{1} + y_{2} + \dots + y_{k + 1}) = d y_{1} + d y_{2} + \dots + d y_{k + 1} .

$0 = d(1) = d(y_1 + y_2 + \cdots + y_{k+1}) = dy_1 + dy_2 + \cdots + dy_{k+1}.$

Consider the one-form

ω = d x_{1} + \dots + d x_{k} = z (d y_{1} + \dots + d y_{k}) + (y_{1} + \dots + y_{k}) d z .

$\omega = dx_1 + \cdots + dx_k = z(dy_1 + \cdots + dy_k) + (y_1+\cdots + y_k) dz.$

It appears in the differential of the last variable:

\begin{aligned} d x_{k + 1} & = z d y_{k + 1} + y_{k + 1} d z \\ = - z (d y_{1} + \dots + d y_{k}) + (1 - y_{1} - \dots y_{k}) d z \\ = d z - ω . \end{aligned}

$\eqalign{ dx_{k+1} &= z dy_{k+1} + y_{k+1}dz \\ &= -z(dy_1 + \cdots + dy_k) + (1-y_1-\cdots y_k)dz \\ &= dz - \omega. }$

The value of this lies in the observation that

d x_{1} \land \dots \land d x_{k} \land ω = 0

$dx_1 \wedge \cdots \wedge dx_k \wedge \omega = 0$

because, when you expand this product, there is one term containing $dx_1 \wedge dx_1 = 0$ as a factor, another containing $dx_2 \wedge dx_2 = 0$ , and so on: they all disappear. Consequently,

\begin{aligned} d x_{1} \land \dots \land d x_{k} \land d x_{k + 1} & = d x_{1} \land \dots \land d x_{k} \land z - d x_{1} \land \dots \land d x_{k} \land ω \\ = d x_{1} \land \dots \land d x_{k} \land z . \end{aligned}

$\eqalign{ dx_1 \wedge \cdots \wedge dx_k \wedge dx_{k+1} &= dx_1 \wedge \cdots \wedge dx_k \wedge z - dx_1 \wedge \cdots \wedge dx_k \wedge \omega \\ &= dx_1 \wedge \cdots \wedge dx_k \wedge z. }$

Whence (because all products $dz\wedge dz$ disappear),

\begin{aligned} d x_{1} \land \dots \land d x_{k + 1} & = (z d y_{1} + y_{1} d z) \land \dots \land (z d y_{k} + y_{k} d z) \land d z \\ = z^{k} d y_{1} \land \dots \land d y_{k} \land d z . \end{aligned}

$\eqalign{ dx_1 \wedge \cdots \wedge dx_{k+1} &= (z dy_1 + y_1 dz) \wedge \cdots \wedge (z dy_k + y_k dz) \wedge dz \\ &= z^k dy_1 \wedge \cdots \wedge dy_k \wedge dz. }$

The Jacobian is simply $|z^k| = z^k$ , the coefficient of the differential product on the right hand side.

Solution

The transformation $(x_1, \ldots, x_k, x_{k+1})\to (y_1, \ldots, y_k, z)$ is one-to-one: its inverse is given by $x_i = y_i z$ for $1\le i\le k$ and $x_{k+1} = z(1-y_1-\cdots-y_k)$ . Therefore we don't have to fuss any more about the new probability element; it simply is

\begin{aligned} (z y_{1})^{α_{1} - 1} \dots (z y_{k})^{α_{k} - 1} {(z (1 - y_{1} - \dots - y_{k}))}^{α_{k + 1} - 1} \exp (- z) | z^{k} d y_{1} \land \dots \land d y_{k} \land d z | \\ = (z^{α_{1} + \dots + α_{k + 1} - 1} \exp (- z) d z) (y_{1}^{α_{1} - 1} \dots y_{k}^{α_{k} - 1} {(1 - y_{1} - \dots - y_{k})}^{α_{k + 1} - 1} d y_{1} \dots d y_{k}) . \end{aligned}

$\eqalign{ &(z y_1)^{\alpha_1-1}\cdots (z y_k)^{\alpha_k-1}\left(z(1-y_1-\cdots-y_k)\right)^{\alpha_{k+1}-1}\exp\left(-z\right)|z^k dy_1 \wedge \cdots \wedge dy_k \wedge dz| \\ &= \left(z^{\alpha_1+\cdots+\alpha_{k+1}-1}\exp\left(-z\right) dz\right)\left( y_1^{\alpha_1-1} \cdots y_k^{\alpha_k-1}\left(1-y_1-\cdots-y_k\right)^{\alpha_{k+1}-1}dy_1 \cdots dy_k\right). }$

That is manifestly a product of a Gamma $(\alpha_1+\cdots+\alpha_{k+1})$ distribution (for $Z$ ) and a Dirichlet $(\mathbf\alpha)$ distribution (for $(Y_1,\ldots, Y_k)$ ). In fact, since the original normalizing constant must have been a product of $\Gamma(\alpha_i)$ , we deduce immediately that the new normalizing constant must be divided by $\Gamma(\alpha_1+\cdots+\alpha_{k+1})$ , enabling the PDF to be written

f_{Y} (y, α) = \frac{Γ (α_{1} + \dots + α_{k + 1})}{Γ (α_{1}) \dots Γ (α_{k + 1})} (y_{1}^{α_{1} - 1} \dots y_{k}^{α_{k} - 1} {(1 - y_{1} - \dots - y_{k})}^{α_{k + 1} - 1}) .

$f_\mathbf{Y}(\mathbf{y},\mathbf{\alpha}) = \frac{\Gamma(\alpha_1+\cdots+\alpha_{k+1})}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_{k+1})}\left( y_1^{\alpha_1-1} \cdots y_k^{\alpha_k-1}\left(1-y_1-\cdots-y_k\right)^{\alpha_{k+1}-1}\right).$

— whuber
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