# 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): $$$$G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = F(\boldsymbol{\alpha}, \boldsymbol{\beta}) \int \phi(\boldsymbol{x}) \exp((\boldsymbol{\alpha} + \boldsymbol{V}_x^{-1} \boldsymbol{V}_{xy}\boldsymbol{\beta})^T\boldsymbol{x}) d \boldsymbol{x}$$$$ where he calls $$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?

Let $$\phi(\boldsymbol{x})$$ and $$\phi(\boldsymbol{y|x})$$ denote the marginal and conditional distribution, respectively. We can write: $$$$G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \int \phi(\boldsymbol{x})\phi(\boldsymbol{y}|\boldsymbol{x}) \exp (\boldsymbol{\alpha}^T\boldsymbol{x})\exp(\boldsymbol{\beta}^T\boldsymbol{y})d\boldsymbol{x} d\boldsymbol{y}$$$$ Noting that the conditional mean of $$\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}$$, and observe that $$$$\left |\frac{d\boldsymbol{z}}{d\boldsymbol{y}}\right | = |\boldsymbol{I}| = 1$$$$ Thus, $$$$G(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \int \phi(\boldsymbol{x})\phi_{\boldsymbol{V}_{y|x}}(\boldsymbol{z}) \exp (\boldsymbol{\alpha}^T\boldsymbol{x})\exp(\boldsymbol{\beta}^T(\boldsymbol{z} + \boldsymbol{V}_{yx}\boldsymbol{V}_x^{-1}\boldsymbol{x}))\left |\frac{d\boldsymbol{y}}{d\boldsymbol{z}}\right | d\boldsymbol{z} d\boldsymbol{x}$$$$ where $$\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.