経済学で出る数学

ワークブックでじっくり攻める：応用問題

CES支出関数．

【問】　次の最小化問題について答えなさい．ただし ${\alpha}, {\beta} > 0, {\rho} < 1$ とする． \begin{align} \min_{x,y} & (px+qy) \\[2ex] s.t. &u=\left\{{{\alpha}}(x)^{\rho}+{{\beta}}(y)^{\rho}\right\}^{\frac{1}{\rho}} \end{align}
支出関数$E(p,q,u)$を求めなさい．

1. ラグランジュ関数を作ると， \[ {\cal L}(x,y,\lambda )= px+qy+{\lambda}[u-[{\alpha}x^{\rho}+{\beta}y^{\rho}]^{\frac{1}{\rho}}] \]


2. 各変数で偏微分してイコールゼロとおくと， \[ \left\{ \begin{align} 0=&p-{\lambda}{\alpha}[{\alpha}x^{\rho}+{\beta}y^{\rho}]^{\frac{1}{\rho}-1}x^{\rho -1} \qquad (1)\\[2ex] 0=&q-{\lambda}{\beta}[{\alpha}x^{\rho}+{\beta}y^{\rho}]^{\frac{1}{\rho}-1}y^{\rho -1} \qquad (2)\\[2ex] 0=&u-[{\alpha}x^{\rho}+{\beta}y^{\rho}]^{\frac{1}{\rho}}\qquad (3) \end{align} \right. \]


3. あとは工夫して解く．$(1)\div(2)$ から， $\Bigl(\dfrac{x}{y}\Bigr)^{\rho -1}=\dfrac{\beta}{\alpha}\cdot\dfrac{p}{q}$． \begin{eqnarray*} x&=&\Bigl(\frac{\beta p}{\alpha q}\Bigr)^{\frac{1}{\rho -1}}y,\\ y&=&\Bigl(\frac{\alpha q}{\beta p}\Bigr)^{\frac{1}{\rho -1}}x, \end{eqnarray*} これを$(3)$に代入すると \begin{eqnarray*} u&=&\Bigl[{\alpha}x^{\rho} +{\beta}\Bigl(\frac{\alpha q}{\beta p}\Bigr)^{\frac{\rho}{\rho -1}}x^{\rho}\Bigr] ^{\frac{1}{\rho}}\\ &=&\Bigl[x^{\rho}p^{\frac{-\rho}{\rho -1}}{\alpha}p^{\frac{\rho}{\rho -1}} +x^{\rho}p^{\frac{-\rho}{\rho -1}}{\beta} \Bigl(\frac{\alpha q}{\beta}\Bigr)^{\frac{\rho}{\rho -1}} \Bigr]^{\frac{1}{\rho}}\\ &=&xp^{\frac{-1}{\rho -1}} \Bigl[{\alpha}p^{\frac{\rho}{\rho -1}} +\Bigl(\frac{\alpha q}{\beta}\Bigr)^{\frac{\rho}{\rho -1}} \Bigr]^{\frac{1}{\rho}}, \end{eqnarray*} より， \[ x=p^{\frac{1}{\rho -1}}\Bigl[{\alpha}p^{\frac{\rho}{\rho -1}} +\Bigl(\frac{\alpha}{\beta}\Bigr)^{\frac{\rho}{\rho -1}} q^{\frac{\rho}{\rho -1}} \Bigr]^{-\frac{1}{\rho}}u, \] が， \begin{eqnarray*} u&=&\Bigl[{\alpha}\Bigl(\frac{\beta p}{\alpha q}\Bigr)^{\frac{\rho}{\rho -1}}y^{\rho} +{\beta}y^{\rho}\Bigr] ^{\frac{1}{\rho}}\\ &=&\Bigl[y^{\rho}q^{\frac{-\rho}{\rho -1}} {\alpha}\Bigl(\frac{\beta p}{\alpha}\Bigr)^{\frac{\rho}{\rho -1}} +y^{\rho}p^{\frac{-\rho}{\rho -1}}{\beta}q^{\frac{\rho}{\rho -1}} \Bigr]^{\frac{1}{\rho}}\\ &=&yq^{\frac{-1}{\rho -1}}\Bigl[\Bigl(\frac{\beta}{\alpha}\Bigr)^{\frac{\rho}{\rho -1}} {\alpha}p^{\frac{\rho}{\rho -1}} +{\beta}q^{\frac{\rho}{\rho -1}} \Bigr]^{\frac{1}{\rho}}, \end{eqnarray*} より， \[ y=q^{\frac{1}{\rho -1}}\Bigl[\Bigl(\frac{\beta}{\alpha}\Bigr)^{\frac{\rho}{\rho -1}} {\alpha}p^{\frac{\rho}{\rho -1}} +{\beta}q^{\frac{\rho}{\rho -1}} \Bigr]^{-\frac{1}{\rho}}u \] が得られる． ここで，$\displaystyle r={\frac{\rho}{\rho -1}}$とおくと， \begin{eqnarray*} \rho &=&\frac{r}{r-1}\\ \rho -1&=&\frac{1}{r-1}\\ \frac{1}{\rho -1}&=&r-1\\ \frac{1}{\rho}&=&\frac{1}{r}-1 \end{eqnarray*} であることより， \begin{eqnarray*} x&=&p^{r-1}\Bigl[{\alpha}p^r +{\beta}\Bigl(\frac{\alpha}{\beta}\Bigr)^rq^r \Bigr]^{\frac{1}{r}-1}u,\\ y&=&q^{r-1}\Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u, \end{eqnarray*} となる．したがって，支出関数は \begin{eqnarray*} E(p,q,u)&=&px+qy\\ &=&p^{r}\Bigl[{\alpha}p^r +{\beta}\Bigl(\frac{\alpha}{\beta}\Bigr)^rq^r \Bigr]^{\frac{1}{r}-1}u +q^{r}\Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u\\ &=&p^r\Bigl(\frac{\alpha}{\beta}\Bigr)^{r(\frac{1}{r}-1)} \Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u +q^r\Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u\\ &=&\Bigl[p^r\Bigl(\frac{\alpha}{\beta}\Bigr)^{r(1-r)}+q^r\Bigr] \Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u\\ &=&\Bigl[\frac{\alpha}{\beta}\Bigl(\frac{\beta}{\alpha}\Bigr)p^r+q^r\Bigr] \Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u\\ &=&\Bigl[{\frac{1}{\beta}\Bigl({\alpha}}\bigl(\frac{\beta}{\alpha}\bigr)^rp^r +{\beta}q^r\Bigr) \Bigr] \Bigl[{\alpha}\Bigl(\frac{\beta}{\alpha}\Bigr)^rp^r+{\beta}q^r \Bigr]^{\frac{1}{r}-1}u\\ &=&{\frac{1}{\beta}\Bigl[\Bigl({\alpha}}\bigl(\frac{\beta}{\alpha}\bigr)^rp^r +{\beta}q^r\Bigr) \Bigr]^{\frac{1}{r}}u\\ &=&\Bigl[{\alpha}^{1-r}p^r+{\beta}^{1-r}q^r\Bigr]^{\frac{1}{r}}u. \end{eqnarray*}