# Modifying Hotelling's lemma: Is this valid?

Hotelling's lemma is stated as:

$$\frac{\partial \pi}{\partial p}=y$$

knowing however that on the more basic level, output $y$ is determined by the input(s) $x(p,w)$,let the profit function be defined as:

$$\pi=py(x(p,w))-wx(p,w)$$

taking the derivative with respect to $p$

$$\frac{\partial\pi}{\partial p}=y(x(p,w))+p\frac{\partial y(x(p,w))}{\partial x(p,w)}\frac{\partial x(p,w)}{\partial p}-w \frac{\partial x(p,w)}{\partial p}$$

Wouldn't this be a more accurate definition of hotelling's lemma?

I'm speculating considering some of the critsisim I read on hotelling's lemma in applied work. Namely: Duality, Optimization, and Microeconomic Theory: Pitfalls for the Applied Researcher

• The wikipedia page for Hotelling's Lemma is abysmal, but the one for the Envelope theorem is good, you should read that one. – Giskard Apr 27 '18 at 19:52

You are basically on the right track but might have been carried away by not writing down function arguments. The profit function $\pi$ has arguments $p$ and $w$; don't drop them.

In Hotelling's lemma, $y$ is the output. On the other hand, you write $y$ for the production function. Let's make things clear by writing neutrally $f$ for the production function. The profit-maximizing input choice is $x(p,w)$, so the quantity produced is $f\big(x(p,w)\big)$, the correct meaning of $y$ in Hotelling's lemma.

Hotelling's lemma then becomes $$\frac{\partial \pi(p,w)}{\partial p}=f\big(x(p,w)\big).$$ The profit function is given by $$\pi(p,w)=pf\big(x(p,w)\big)-wx(p,w).$$

Taking the derivative with respect to $p$ gives us $$\frac{\partial \pi(p,w)}{\partial p}=f\big(x(p,w)\big)+pf'\big(x(p,w)\big)\frac{\partial x(p,w)}{\partial p}-w\frac{\partial x(p,w)}{\partial p}.$$ This is pretty much what you got there above.

Now $x(p,w)$ is by definition a maximizer of $pf(x)-wx$. We have the first order condition $pf'(x^*)=w$. Therefore, $$pf'\big(x(p,w)\big)-w=0,$$ which implies $$pf'\big(x(p,w)\big)\frac{\partial x(p,w)}{\partial p}-w\frac{\partial x(p,w)}{\partial p}=0$$ This reduces our expression for ${\partial \pi(p,w)}/{\partial p}$ to Hotelling's lemma.

This tell's us also what the important message of Hotelling's lemma is: Indirect effects don't matter. A marginal change in prices has a direct effect on profits since the output gets multiplied by a different number. There is also an indirect effect that comes from firms adjusting how much to produce. Hotelling's lemma just tells us that the latter effect is zero for a profit-maximizing firm, which need not do much adjusting. At a profit maximum, the marginal effect of adjusting production on profits is zero.

• "$pf'\big(x(p,w)\big)\frac{\partial x(p,w)}{\partial p}-w\frac{\partial x(p,w)}{\partial p}=0$ This reduces our expression for ${\partial \pi(p,w)}/{\partial p}$ to Hotelling's lemma." This is astounding. I'm simulating this result and am amazed with it. – EconJohn Apr 29 '18 at 2:31

With subjects like Hotelling's lemma precise notation is crucial.

The function $\Pi(a,b,c,d)$ is defined as $$\Pi(a,b,c,d) \triangleq a \cdot c - b \cdot d.$$ Now if we want to make this the profit function $$\Pi (p,w,y(p,w),x(p,w)) = p \cdot y(p,w) - w \cdot x(p,w),$$ we should use the input where $y(p,w)$ and $x(p,w)$ are solutions of the constrained optimization problem \begin{align*} \max_{y,x} \ & p \cdot y - w \cdot x \\ s.t. \ & f(x) = y \end{align*}

There is difference between $$\frac{\text{d}\Pi (p,w,y(p,w),x(p,w))}{\text{d}p}$$ and $$\frac{\partial\Pi (p,w,y(p,w),x(p,w))}{\partial p}.$$

You seem to be doing $$\frac{\text{d}\Pi (p,w,y(p,w),x(p,w))}{\text{d}p} = y(p,w) + p\frac{\partial y(p,w)}{\partial p} -w \frac{\partial x(p,w)}{\partial p}$$ whih is not trivially the same as $$\frac{\partial \Pi (p,w,y(p,w),x(p,w))}{\partial p} = y(p,w).$$

A trivial example to illustrate the difference:

Let the function $A(x,y) \triangleq x \cdot y$ measure the area of a rectangle with sides the length of $x$ and $y$. Then the partial derivate w.r.t. $x$ $$\frac{\partial A(x,y)}{\partial x} = y$$ measures how much the size of the rectangle would increase if I were to increase size x. Now imagine that I am interested in squares specifically, so I look at rectangles where if one size has length $x$, the length $y$ of the other size is $y(x) = x$. Then the function becomes $$A(x,y(x)) = x \cdot y(x) = x^2.$$ If I now consider how much the size of the square would increase if I were to increase only one side, the answer is still $$\frac{\partial A(x,y(x))}{\partial x} = y(x) = x.$$ (This is the definition of partial derivatives.) But if I want to consider how much the area of the square would change if I were to increase the parameter determining the length of both sides I should take the full derivative $$\frac{\text{d} A(x,y(x))}{\text{d} x} = \frac{\text{d} x^2}{\text{d} x} = 2x.$$ Notice that in this case $$\frac{\text{d} A(x,y(x))}{\text{d} x} \neq \frac{\partial A(x,y(x))}{\partial x}.$$ Hotelling's lemma tells you that some types of constrained optimization problems are magic because there this equation holds.