3
$\begingroup$

Prove that the profits of the firm weakly decreases with input prices. More formally, suppose that the firm has a production function f, so that its profit function is

π(p, w) = max(x≥0) $pf(x) − w · x$,

where p denotes the output price and w denotes the input price vector. Then show that if $w$ and $w'$ are two input price vectors such that $w'_j$ = $w_j$ for all $j \neq i$ and $w'_i$ > $w_i$ , then $π(p, w') ≤ π(p, w)$

I know that as input prices decrease, the toal costs for the firm decrease as well. Due to the decrease in TC, the firm then produces more output at the same cost. However, due to the increase in supply, the price of the output falls to a new equilibrium, causing profits to fall as well unless the increase in q is proportionate to the decrease in price that would prevent profits from dropping.

However, I am unsure on how to start the proof using $\pi (p, w)$. Since it is $w$ that is changing, would I hold price of output fixed at $p$?

Then I would assume that $x'$ is profit maximizing at:

$$pf(x) − w'· x ≤ pf(x') − w'· x'$$

and $x$ is profit maximizing at:

$$pf(x) − w · x \geq pf(x') − w· x'$$

Then I would multiply the second equation by -1, getting

$$-pf(x) + w · x \leq -pf(x') + w· x'$$

I would then add that to the first equation getting:

$$ (− w'· x) + (w \cdot x) ≤ (− w'· x') + (w \cdot x')$$

which simplifies to:

$$ (w - w') (x - x') \leq 0 $$

Since all components of $w'− w$ are 0 except the ith

$$ (w_i - w_i') (x_i - x_i') \leq 0 $$

We are left with:

$$(x_i - x_i') \leq 0 $$ $$ x_i \leq x_i' $$

Leaving us with the fact that factor input demands at $x' \geq x$ therefore we assume that the demand for output is greater, causing prices to fall.

However, I have no idea how to prove that $π(p, w') ≤ π(p, w)$ if $w'_j$ = $w_j$ for all $j \neq i$ and $w'_i$ > $w_i$.

I understand that the idea is based on the fact that $w'_j$ = $w_j$ and $w'_i$ > $w_i$, $w' > w $, that means that the $w' \cdot x$ in the profit function is greater than $w \cdot x$. Since the cost is greater in $\pi (p, w')$, that means that $π(p, w') ≤ π(p, w)$. However, I am confused as to how I can approach this proposition using the given profit function.

$\endgroup$
  • $\begingroup$ If you don't want this question closed - at the least show how you tried to solve it. $\endgroup$ – user161005 Oct 28 at 10:18
  • 1
    $\begingroup$ Hi, Welcome to Economics SE! we have a policy regarding homework questions where we require that you show some work before an answer is provided. More on this topic here:economics.meta.stackexchange.com/questions/1465/…. I am closing this q because currently it very clearly violates the policy but if you edit it to make it in line with our policy it will be reopened. $\endgroup$ – 1muflon1 Oct 28 at 10:52
  • $\begingroup$ @1muflon1 I have edited the question to show how far I got into it! I am still confused however on the latter part of the question $\endgroup$ – DH00325 Oct 28 at 19:23
  • $\begingroup$ @DH00325 I reopened your Q. $\endgroup$ – 1muflon1 Oct 28 at 19:23
  • $\begingroup$ Hint: differentiate $\pi (\mathbf{x,w})$ w.r.t $w_j$ and use the FOC for profit maximization. $\endgroup$ – Dayne Oct 29 at 1:07
4
$\begingroup$

From FOC, we know that:

\begin{align} \nabla_x\pi(\mathbf{x},\mathbf{w})=p\nabla f(\mathbf{x})-\mathbf{w}=\mathbf{0} \tag{1} \end{align}

This will be true at equilibrium, i.e. for any given $\mathbf{w}$, the input vector $\mathbf{x}$ will adjust so that the above holds.

Now consider $d\pi(\mathbf{x},\mathbf{w})/d w_i$ (and using $(1)$):

\begin{align} \frac{d\pi(\mathbf{x},\mathbf{w})}{d w_i} &=\nabla_x\pi\cdot\nabla_{w_i} \mathbf{x} \, + \nabla_w\pi\cdot\nabla_{w_i} \mathbf{w} \\ &=0 +\nabla_w\pi\cdot\nabla_{w_i} \mathbf{w} \tag{*}\\ &=\nabla_w(pf(\mathbf{x})-\mathbf{w}\cdot \mathbf{x)} \cdot \nabla_{w_i} \mathbf{w} \\ &= \Big(p\nabla_wf(\mathbf{x})-\nabla_w(\mathbf{w}\cdot \mathbf{x)}\Big)\cdot \nabla_{w_i} \mathbf{w} \\ &= \Big(p \frac{\partial\mathbf{x}}{\partial\mathbf{w}}\nabla f(\mathbf{x})-\frac{\partial\mathbf{x}}{\partial\mathbf{w}}\mathbf{w} - \mathbf{x}\Big)\cdot \nabla_{w_i} \mathbf{w}\\ &= \mathbf{J}(p\nabla f(\mathbf{x})-\mathbf{w})\cdot \nabla_{w_i} \mathbf{w} -\mathbf{x} \cdot \nabla_{w_i} \mathbf{w} \\ &= \mathbf{J} \nabla_x\pi(\mathbf{x},\mathbf{w})\cdot \nabla_{w_i} \mathbf{w}-\mathbf{x} \cdot \nabla_{w_i} \mathbf{w} \tag{**}\\ &= -\mathbf{x} \cdot \nabla_{w_i} \mathbf{w} \\ &= -x_i \end{align}

where, in steps $(*), (**)$, FOC is used and $\mathbf{J}$ is the Jacobian matrix.

Since, $x_i \geq0$, we have:

$$\frac{d\pi(\mathbf{x},\mathbf{w})}{d w_i} \leq0$$

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This is correct, but perhaps you would want to consider adding a line to show clearly why $\nabla_w(pf(\mathbf{x})-\mathbf{w}\cdot \mathbf{x)} = -\mathbf{x}$ $\endgroup$ – Alecos Papadopoulos Oct 29 at 13:10
  • $\begingroup$ Ok. Thanks for the advice. Will edit accordingly. $\endgroup$ – Dayne Oct 29 at 13:44
2
$\begingroup$

(Without using differentiation) When $w \leq w'$ it follows that $pf(x) − w · x \geq pf(x) − w' · x$ and so $\pi(p,w) \geq \pi(p,w')$.

EDIT 1. The last inequality (first left as an exercise) can be justified as follow: $w \leq w'$ implies that $$pf(x) − w · x \geq pf(x) − w' · x$$ for any $x \geq 0$ and admissible. The inequality is in particular true for $x=x^*(p,w')$ and so $$ pf(x^*(p,w')) − w · x^*(p,w') \geq pf(x^*(p,w')) − w' · x^*(p,w').$$ However, $x^*(p,w')$ does not maximize profits for input prices equal to $w$ and so $$ pf(x^*(p,w)) − w · x^*(p,w) \geq pf(x^*(p,w')) − w' · x^*(p,w')$$ or equivalently $\pi(p,w) \geq \pi(p,w')$.

EDIT 2. If the output price $p$ is endogenous and adjusts to aggregate output supply and demand, as it seems to be the case in your question, the issue has been treated by:
Heiner, R. A. (1982): “Theory of the Firm in “Short-Run” Industry Equilibrium,” American Economic Review, 72, 555-62.
Braulke, M. (1984): “The Firm in Short-Run Industry Equilibrium: Comment,” American Economic Review, 74, 750-753.

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Note that $\pi(p,w)={\color{red}{\max_{x\ge0}}pf(x)-w\cdot x}=pf(x^*(p,w))-w\cdot x^*(p,w)$, where $x^*(p,w)$ is the maximizer given $p$ and $w$. So it not necessary that $x^*(p,w)=x^*(p,w')$ when $w\le w'$. $\endgroup$ – Herr K. Oct 30 at 20:54
  • $\begingroup$ @Herr K. Yes of course, this is not necessary (and I do not assume it). I do not understand your comment very well... $\endgroup$ – Bertrand Nov 1 at 22:04
  • $\begingroup$ Perhaps you can try to provide a more explicit explanation of the first sentence in your answer. Specifically, how did you derive the last two inequalities from knowing $w\le w'$? $\endgroup$ – Herr K. Nov 1 at 22:18
  • 2
    $\begingroup$ Thanks for the update. But I think there is still some inconsistencies. By definition, $x^*(p,w')$ maximizes profit when output price is $p$ and input prices are $w'$. If you agree with this interpretation of notation, then how is it possible to have $$ pf(x^*(p,w')) − w · x^*(p,w') \color{red}{\overset{??}{\geq}} pf(x^*(p,w')) − w' · x^*(p,w').$$ $\endgroup$ – Herr K. Nov 1 at 23:14
  • $\begingroup$ @Herr K. This is due to $w \leq w' \implies w · x^*(p,w') \leq w' · x^*(p,w')$, I think this point was clear in the edit. $\endgroup$ – Bertrand Nov 2 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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