株価とマーケットポートフォリオからベータ値を求める

無リスク金利を\(r_{f}\)とする。
時刻\(k\in\left\{ 1,2,\cdots,n\right\} \)でのリターン\(r_{A,k},r_{M,k}\)は
\[ r_{A,k}=\frac{P_{A,k}}{P_{A,k-1}}-1 \] \[ r_{M,k}=\frac{P_{M,k}}{P_{M,k-1}}-1 \] となり、\(A\)のリターンと\(M\)のリターンを\(R_{A}=\left\{ r_{A,k}\right\} _{k=\left\{ 1,2,\cdots,n\right\} },R_{M}=\left\{ r_{M,k}\right\} _{k=\left\{ 1,2,\cdots,n\right\} }\)とする。
このとき、
\[ S_{k}=r_{A,k}-r_{f}-\left(\beta_{A}\left(r_{M,k}-r_{f}\right)+\alpha_{A}\right) \] として、回帰直線を求めるように、
\[ \sum_{k=1}^{n}S_{k}^{2} \] が最小となる\(a_{A},\beta_{A}\)を求めると、
\begin{align*} 0 & =\frac{\partial}{\partial\alpha_{A}}\sum_{k=1}^{n}S_{k}^{2}\\ & =\frac{\partial}{\partial\alpha_{A}}\sum_{k=1}^{n}\left(r_{A,k}-r_{f}-\beta_{A}\left(r_{M,k}-r_{f}\right)-\alpha_{A}\right)^{2}\\ & =-2\sum_{k=1}^{n}\left(r_{A,k}-r_{f}-\beta_{A}\left(r_{M,k}-r_{f}\right)-\alpha_{A}\right)\\ & =-2\left(nE\left[R_{A}-r_{f}\right]-\beta_{A}nE\left[R_{M}-r_{f}\right]-\alpha_{A}n\right)\\ & =-2n\left(E\left[R_{A}-r_{f}\right]-\beta_{A}E\left[R_{M}-r_{f}\right]-\alpha_{A}\right) \end{align*} \begin{align*} 0 & =\frac{\partial}{\partial\beta_{A}}\sum_{k=1}^{n}S_{k}^{2}\\ & =\frac{\partial}{\partial\beta_{A}}\sum_{k=1}^{n}\left(r_{A,k}-r_{f}-\beta_{A}\left(r_{M,k}-r_{f}\right)-\alpha_{A}\right)^{2}\\ & =-2\sum_{k=1}^{n}\left(r_{M,k}-r_{f}\right)\left(r_{A,k}-r_{f}-\beta_{A}\left(r_{M,k}-r_{f}\right)-\alpha_{A}\right)\\ & =-2\sum_{k=1}^{n}\left(\left(r_{M,k}-r_{f}\right)\left(r_{A,k}-r_{f}\right)-\beta_{A}\left(r_{M,k}-r_{f}\right)^{2}-\alpha_{A}\left(r_{M,k}-r_{f}\right)\right)\\ & =-2\left(nE\left[\left(R_{M}-r_{f}\right)\left(R_{A}-r_{f}\right)\right]-\beta_{A}nE\left[\left(R_{M}-r_{f}\right)^{2}\right]-\alpha_{A}nE\left[R_{M}-r_{f}\right]\right)\\ & =-2n\left(E\left[R_{M}-r_{f}\right]E\left[R_{A}-r_{f}\right]+Cov\left[R_{M}-r_{f},R_{A}-r_{f}\right]\right)-\beta_{A}\left(V\left[R_{M}-r_{f}\right]+E^{2}\left[R_{M}-r_{f}\right]\right)-\alpha_{A}E\left[R_{M}-r_{f}\right]\\ & =-2n\left(E\left[R_{M}-r_{f}\right]E\left[R_{A}-r_{f}\right]+\sigma_{MA}\right)-\beta_{A}\left(\sigma_{M}^{2}+E^{2}\left[R_{M}-r_{f}\right]\right)-\alpha_{A}E\left[R_{M}-r_{f}\right] \end{align*} となるので、
\[ \left(\begin{array}{cc} 1 & E\left[R_{M}-r_{f}\right]\\ E\left[R_{M}-r_{f}\right] & \sigma_{M}+E^{2}\left[R_{M}-r_{f}\right] \end{array}\right)\left(\begin{array}{c} \alpha_{A}\\ \beta_{A} \end{array}\right)=\left(\begin{array}{c} E\left[R_{A}-r_{f}\right]\\ E\left[R_{M}-r_{f}\right]E\left[R_{A}-r_{f}\right]+\sigma_{MA} \end{array}\right) \] これより、\(\alpha_{A},\beta_{A}\)を求めると、
\begin{align*} \left(\begin{array}{c} \alpha_{A}\\ \beta_{A} \end{array}\right) & =\left(\begin{array}{cc} 1 & E\left[R_{M}-r_{f}\right]\\ E\left[R_{M}-r_{f}\right] & \sigma_{M}^{2}+E^{2}\left[R_{M}-r_{f}\right] \end{array}\right)^{-1}\left(\begin{array}{c} E\left[R_{A}-r_{f}\right]\\ E\left[R_{M}-r_{f}\right]E\left[R_{A}-r_{f}\right]+\sigma_{MA} \end{array}\right)\\ & =\frac{1}{\sigma_{M}^{2}}\left(\begin{array}{cc} \sigma_{M}^{2}+E^{2}\left[R_{M}-r_{f}\right] & -E\left[R_{M}-r_{f}\right]\\ -E\left[R_{M}-r_{f}\right] & 1 \end{array}\right)\left(\begin{array}{c} E\left[R_{A}-r_{f}\right]\\ E\left[R_{M}-r_{f}\right]E\left[R_{A}-r_{f}\right]+\sigma_{MA} \end{array}\right)\\ & =\frac{1}{\sigma_{M}^{2}}\left(\begin{array}{c} \sigma_{M}^{2}E\left[R_{A}-r_{f}\right]-E\left[R_{M}-r_{f}\right]\sigma_{MA}\\ \sigma_{MA} \end{array}\right)\\ & =\left(\begin{array}{c} E\left[R_{A}-r_{f}\right]-E\left[R_{M}-r_{f}\right]\frac{\sigma_{MA}}{\sigma_{M}^{2}}\\ \frac{\sigma_{MA}}{\sigma_{M}^{2}} \end{array}\right)\\ & =\left(\begin{array}{c} E\left[R_{A}-r_{f}\right]-E\left[R_{M}-r_{f}\right]\rho_{MA}\frac{\sigma_{A}}{\sigma_{M}}\\ \rho_{MA}\frac{\sigma_{A}}{\sigma_{M}} \end{array}\right) \end{align*} となる。
従って題意は成り立つ。