3
$\begingroup$

I am trying to understand how the first order conditions for an interior solution of a maximization problem were derived using the substitution method.

The problem is: $$\max\limits_{x\ge0,y\ge0}P(a-x)+(1-P)(b-y)$$ subject to $$Pf(x)+(1-P)f(y)=c$$ where: $a,b,c>0$, $P\in (0,1)$, $f:[0,+\infty]\to[0,+\infty]$, increasing and strictly concave over its domain.

I can see how this is solved using a lagrangian to find from the first order conditions that $f'(x^*)=f'(y^*)$. Strict concavity of $f$ then implies $x^*=y^*$. But I'm at a loss as to how we can solve it by substituting the constraint in the objective function. Since $f$ is invertible, if $y$ did not appear in the constraint I would find $x$ from the constraint by inverting $f$ and substitute it in the objective function. Doing this here leads to complications that seem unnecessary for this simple problem. There has to be a simpler way that I cannot figure out: what is it? Thanks!

$\endgroup$

2 Answers 2

4
$\begingroup$

Here are two methods. First method: the substitution can be made by inverting $f$. Since $f$ is strictly increasing and continuous, $f^{-1}$ is well-defined. The constraint can therefore be written \begin{equation*} x = f^{-1}\Big(\dfrac{c-(1-P)f(y)}{P}\Big) \end{equation*}

The objective becomes \begin{equation*} \max_{y \geq 0}{P \Big[a-f^{-1}\Big(\dfrac{c-(1-P)f(y)}{P}\Big)\Big]+(1-P) (b-y)} \end{equation*}

The derivative of this expression with respect to $y$ equals \begin{align*} & (1-P) f'(y) (f^{-1})^{'}(\dfrac{c-(1-P)f(y)}{P})-(1-P) \\ = & (1-P) \dfrac{f'(y)}{f^{'} \circ f^{-1}(\dfrac{c-(1-P)f(y)}{P})}-(1-P) \\ & = (1-P) (\dfrac{f'(y)}{f'(x)}-1) \end{align*} And thus $f'(y^{*})=f'(x^{*})$ at the optimum.

Second method: since $f$ is invertible, we can do a change in variables and define $w=f(x)$ and $z=f(y)$. The problem then becomes \begin{equation*} \max_{w \geq 0, z\geq 0}{P(a-f^{-1}(w))+(1-P)(b-f^{-1}(z))} \end{equation*} subject to \begin{equation*} P w +(1-P)z=c \end{equation*} Substituing $w=(c-(1-P)z)/P$ in the problem delivers \begin{equation*} \max_{z\geq 0}{P(a-f^{-1}(\dfrac{c-(1-P)z}{P})+(1-P)(b-f^{-1}(z))} \end{equation*} Differentiating with respect to $z$ yields the same solution.

$\endgroup$
2
  • $\begingroup$ This was of great help. Now I understand, thanks! $\endgroup$
    – jlol
    Apr 11, 2017 at 19:23
  • $\begingroup$ @jlol my pleasure, I'm glad it was helpful. $\endgroup$
    – Oliv
    Apr 11, 2017 at 20:46
4
$\begingroup$

Let $g$ be the inverse function of $f$ defined over range of $f$. Notice that $g$ is increasing and strictly convex. We can rewrite the maximization problem as: \begin{eqnarray*} \max\limits_{u\geq 0, \ v \geq 0} & P(a - g(u)) + (1-P)(b-g(v)) \\ \text{s.t.} & Pu + (1-P)v = c\end{eqnarray*} where $u=f(x)$ and $v = f(y)$. Solving above is equivalent to solving \begin{eqnarray*} \min\limits_{u\geq 0, \ v \geq 0} & Pg(u) + (1-P)g(v) \\ \text{s.t.} & Pu + (1-P)v = c\end{eqnarray*} Now you may substitute $v= \displaystyle\frac{c - Pu}{1-P}$ and rewrite the problem as: \begin{eqnarray*} \min\limits_{0\leq u \leq \frac{c}{P} } & Pg(u) + (1-P)g\left(\frac{c - Pu}{1-P}\right) \end{eqnarray*} Differentiating with respect to $u$, we get the FOC as:

$\displaystyle Pg'(u) - Pg'\left(\frac{c - Pu}{1-P}\right) = 0$

Since $g$ is strictly convex, solution is: $u = \displaystyle\frac{c - Pu}{1-P}$ i.e. $u = v = c$. Therefore, at the optimum $x = y$ holds.

$\endgroup$
1
  • $\begingroup$ This is an excellent answer. Very helpful, thanks! $\endgroup$
    – jlol
    Apr 11, 2017 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.