Proof of Pearson-Aitken selection formula I am trying to understand the proof of the Pearson-Aitken selection formula, widely used in statistical genetics. A proof that the formula is general is given by Aitken (1936). However, I failed to understand the derivation of equation (4) in this paper, given below: 
Let $\phi(\boldsymbol{x},\boldsymbol{y})$ denote the probability density function (pdf) a standard zero-mean unit-variance multivariate Normal distribution, with two components $\boldsymbol{x}$ and $\boldsymbol{y}$. The moments generating function (mgf) is given by: 
\begin{align} 
G(\boldsymbol{\alpha}, \boldsymbol{\beta}) &= \int \phi(\boldsymbol{x},\boldsymbol{y}) \exp(\boldsymbol{\alpha}^T\boldsymbol{x}+\boldsymbol{\beta}^T\boldsymbol{y}) d\boldsymbol{x} d\boldsymbol{y} \\\\
& = 1 + \frac{1}{2}(\boldsymbol{\alpha}, \boldsymbol{\beta})^T\boldsymbol{V}(\boldsymbol{\alpha}, \boldsymbol{\beta}) + \cdots
\end{align}
Denoting the variance and covariance matrices of $\boldsymbol{x}$ and $\boldsymbol{y}$ by $\boldsymbol{V}_x$, $\boldsymbol{V}_y$, and $\boldsymbol{V}_{xy}$,  Aitken (1936) then writes in equation (4): 
\begin{equation}
G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = F(\boldsymbol{\alpha}, \boldsymbol{\beta}) \int \phi(\boldsymbol{x}) \exp((\boldsymbol{\alpha} + \boldsymbol{V}_x^{-1} \boldsymbol{V}_{xy}\boldsymbol{\beta})^T\boldsymbol{x}) d \boldsymbol{x}
\end{equation}
where he calls $F(\boldsymbol{\alpha}, \boldsymbol{\beta})$ the "partial generating function" which does not contain $\boldsymbol{\alpha}$ in its terms of first and second degree. 
Can anyone give me a more detailed explanation of this derivation?  
Many thanks in advance. 
 A: Actually, I found out that the above can be proved using basic multivariate calculus. 
Let $\phi(\boldsymbol{x})$ and $\phi(\boldsymbol{y|x})$ denote the marginal and conditional distribution, respectively. We can write: 
\begin{equation}
  G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \int \phi(\boldsymbol{x})\phi(\boldsymbol{y}|\boldsymbol{x}) \exp (\boldsymbol{\alpha}^T\boldsymbol{x})\exp(\boldsymbol{\beta}^T\boldsymbol{y})d\boldsymbol{x} d\boldsymbol{y}
\end{equation}
Noting that the conditional mean of $\boldsymbol{y}$ given $\boldsymbol{x}$ is given by 
$ \boldsymbol{V}_{yx} \boldsymbol{V}_x^{-1} \boldsymbol{x}$,
 we let $\boldsymbol{z} = \boldsymbol{y} - \boldsymbol{V}_{yx}\boldsymbol{V}_x^{-1}\boldsymbol{x}$, and observe that 
\begin{equation}
\left |\frac{d\boldsymbol{z}}{d\boldsymbol{y}}\right | = |\boldsymbol{I}| = 1
\end{equation}
Thus, 
\begin{equation}
G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \int \phi(\boldsymbol{x})\phi_{\boldsymbol{V}_{y|x}}(\boldsymbol{z}) \exp (\boldsymbol{\alpha}^T\boldsymbol{x})\exp(\boldsymbol{\beta}^T(\boldsymbol{z} + \boldsymbol{V}_{yx}\boldsymbol{V}_x^{-1}\boldsymbol{x}))\left |\frac{d\boldsymbol{y}}{d\boldsymbol{z}}\right | d\boldsymbol{z} d\boldsymbol{x}
\end{equation}
where $\phi_{\boldsymbol{V}_{y|x}}(\boldsymbol{z})$ denotes a zero-centred multivariate Normal with variance $\boldsymbol{V}_{y|x} = \boldsymbol{V}_y - \boldsymbol{V}_{yx}\boldsymbol{V}_x^{-1}\boldsymbol{V}_{xy}$. Thus writing $F(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \int \phi_{\boldsymbol{V}_{y|x}}(\boldsymbol{z}) \exp(\boldsymbol{\beta}^T\boldsymbol{z})d\boldsymbol{z}$, we obtain the required. 
