アスをもてば互いに打ら消し合い（$\hat{\beta}_{1}-\hat{\beta}_{2}$）のバイアスは小さくなる。しかし，逆の場合にはバイアスは大きくなる。以下，(1)の体系において，$\hat{\beta}_{1}$ のバイアスは正，$\hat{\beta}_{2}$ のバイアスは負となり，したがって予測量(4)がよりバイアスを助長する方向にあることを示そう。以下の議論は大標本特性についてである。

まず共分散オペレーター $ S_{xy} $ を

$$ \begin{aligned}Sxy&=\frac{1}{T}\sum_{i=1}^{T}\ (x_{i}-\bar{x})\ (y_{i}-\bar{y})\\\bar{x},\ \bar{y}&: 平均 \\T&\quad:: サンプル数 \end{aligned} $$

と定義する。体系(1)における $ \beta_{1} $ ， $ \beta_{2} $ の最小二乗推定量は，

$$ \hat{\beta}_{1}{=}\frac{S_{p y}S_{11}{-}S_{1y}S_{p1}}{S_{p p}S_{11}{-}S_{p1}S_{p1}} $$

$$ \left\{\begin{aligned}p=&-\frac{\gamma_{1}}{\beta_{1}-\beta_{2}}z_{1}+\frac{\gamma_{2}}{\beta_{1}-\beta_{2}}z_{2}-\frac{e_{1}}{\beta_{1}-\beta_{2}}+\frac{e_{2}}{\beta_{1}-\beta_{2}}\\ y=&-\frac{\beta_{2}\gamma_{1}}{\beta_{1}-\beta_{2}}z_{1}+\frac{\beta_{2}\gamma_{2}}{\beta_{1}-\beta_{2}}z_{2}-\frac{\beta_{2}e_{1}}{\beta_{1}-\beta_{2}}+\frac{\beta_{2}e_{2}}{\beta_{1}-\beta_{2}}\end{aligned}\right. $$

これより。

$$ S_{p p}{=}\frac{1}{(\beta_{1}{-}\beta_{2})^{2}}(\gamma_{1}{}^{2}S_{11}{+}\gamma_{2}{}^{2}S_{22}{-}2\gamma_{1}\gamma_{2}S_{12}{+}\sigma_{1}{}^{2}{+}\sigma_{2}{}^{2}) $$

$$ S_{p y}=\frac{1}{(\beta_{1}-\beta_{2})^{2}}(\beta_{2}\gamma_{1}^{2}S_{11}+\beta_{1}\gamma_{2}^{2}S_{22}-(\beta_{1}+\beta_{2})\gamma_{1}\gamma_{2}S_{12}+\beta_{2}\sigma_{1}^{2}+\beta_{1}\sigma_{2}^{2}) $$

したがって,

$$ \begin{aligned}\hat{\beta}_{1}=&\frac{\beta_{2}\gamma_{1}^{2}S_{11}+\beta_{2}\gamma_{2}^{2}S_{11}S_{22}-(\beta_{1}+\beta_{2})\gamma_{1}\gamma_{2}S_{11}S_{12}}{\gamma_{1}^{2}S_{11}^{2}+\gamma_{2}^{2}S_{11}S_{22}-2\gamma_{1}\gamma_{2}S_{11}S_{12}}\\&+\frac{(\beta_{2}\sigma_{1}^{2}+\beta_{1}\sigma_{2}^{2})S_{11}-(-\beta_{2}\gamma_{1}S_{11}+\beta_{1}\gamma_{2}S_{12})(-\gamma_{1}S_{11}+\gamma_{2}S_{22})}{-(-\gamma_{1}S_{11}+\gamma_{2}S_{12})^{2}+S_{11}(\sigma_{1}^{2}+\sigma_{2}^{2})}\\=&\frac{\beta_{1}\gamma_{2}^{2}(S_{11}S_{22}-S_{12}^{2})+\beta_{2}\sigma_{1}^{2}S_{11}+\beta_{1}\sigma_{2}^{2}S_{11}}{\gamma_{2}^{2}(S_{11}S_{22}-S_{12}^{2})+\sigma_{1}^{2}S_{11}+\sigma_{2}^{2}S_{11}}\end{aligned} $$

したがって

$$ \begin{aligned}&\hat{\beta}_{1}-\beta_{1}=\frac{(\beta_{2}-\beta_{1})\sigma_{1}^{2}S_{11}}{\gamma_{2}^{2}(S_{11}S_{22}-S_{12}^{2})+S_{11}(\sigma_{1}^{2}+\sigma_{2}^{2})}\\ &\hat{\beta}_{1}<0,\ \hat{\beta}_{2}>0,\ S_{11}S_{22}>S_{12}^{2} であるから \\ \end{aligned} $$

結局，

$$ \hat{\beta}_{1}-\beta_{1}>0 $$

同様に,

$$ \hat{\beta}_{1}-\beta_{1}=\frac{(\beta_{2}-\beta_{1})\sigma_{1}^{2}S_{11}}{\gamma_{2}^{2}(S_{11}S_{22}-{S_{12}}^{2})+S_{11}(\sigma_{1}^{2}+{\sigma_{2}}^{2})}<0 $$

となり， $ (\hat{\beta}_{1}-\hat{\beta}_{2}) $ のバイアスは，

$$ \begin{aligned}(\hat{\beta}_{1}-\hat{\beta}_{2})-(\beta_{1}-\beta_{2})=&(\hat{\beta}_{1}-\beta_{1})\\&-(\hat{\beta}_{2}-\beta_{2})>0\end{aligned} $$

$$ \hat{\beta}_{2}=-\frac{S_{py}S_{22}-S_{2y}S_{p2}}{S_{pp}S_{22}-S_{p2}S_{p2}} $$

ただし

$$ \begin{aligned}&S_{11}=S_{z1z1},\ S_{p1}=S_{p z1}\\&S_{12}=S_{z1z2},\ \cdots\\ \end{aligned} $$

と表わすことにする。大標本での議論であるから $ z_{1} $ ， $ z_{2} $ が外生変数であることを考え，

$$ \begin{array}{c}S_{e11}=p l i m S_{e11}=0\\S_{e21}=\underset{\begin{array}{c}\vdots\\\end{array}}{p l i m}S_{e21}=0\end{array} $$

である。また

$$ \begin{aligned}&S_{e1e1}=\sigma_{1}{}^{2}\\&S_{e2e2}=\sigma_{2}{}^{2}\\&S_{e1e2}=0\\ \end{aligned} $$

である。そこで(1)を誘導型にすれば,

となり予測量のバイアスを大きくする方向にあることがわかる。ただし，この場合正確には(4)の分子の性質も同時に調べなくてはならない。

(4)の分子に現われる $ \hat{\gamma}_{1} $ ， $ \hat{\gamma}_{2} $ のバイアスを調べるために，大標本を前提にして以前と同様の演算を行えば，次の結果が得られる。

$$ \begin{aligned}&\hat{\gamma}_{1}=\gamma_{1}+\frac{\sigma_{1}^{2}(\gamma_{2}S_{12}-\gamma_{1}S_{11})-\gamma_{1}\gamma_{2}^{2}S_{11}S_{22}}{\gamma_{2}^{2}(S_{11}S_{22}-S_{1}^{2}2)+(\sigma_{1}^{2}+\sigma_{2}^{2})S_{11}}\\ &\hat{\gamma}_{2}=\gamma_{2}+\frac{\sigma_{2}^{2}(\gamma_{1}S_{12}-\gamma_{2}S_{22})-\gamma_{1}^{2}\gamma_{2}S_{11}S_{22}}{\gamma_{1}^{2}(S_{11}S_{22}-S_{12}^{2})+(\sigma_{1}^{2}+\sigma_{2}^{2})S_{11}}\\ \end{aligned} $$

これらの右辺第2項の分子の符号は確定でき