# For what demand function is a monopoly most harmful?

Consider a firm with zero marginal cost. If it gives the product for free, then all the demand is satisfied and the social welfare increases by the maximum possible amount; call this increase $W$.

But because the firm is a monopoly, it reduces the demand and increases the price in order to optimize its revenue. Now the social welfare increases by a smaller amount, say, $V$.

Define the relative loss of welfare (deadweight loss) as: $W/V$. This ratio depends on the shape of the demand function. So my question is: is this ratio bounded, or can it be arbitrarily large? In particular:

• If $W/V$ is bounded, then for what demand function is it maximized?
• If $W/V$ is unbounded, then for what family of demand functions can it become arbitrarily large?

Here is what I tried so far. Let $u(x)$ be the consumers' marginal utility function (which is also the inverse demand function). Assume that it is finite, smooth, monotonically decreasing, and scaled to the domain $x\in[0,1]$. Let $U(x)$ be its anti-derivative. Then: • $W = U(1)-U(0)$, the total area under $u$.
• $V = U(x_m)-U(0)$, where $x_m$ is the amount produced by the monopoly. This is the area under $u$ except the "deadweight loss" part.
• $x_m = \arg \max (x \cdot u(x))$ = the quantity which maximizes the producer's revenue (the marked rectangle).
• $x_m$ can usually be calculated using the first-order condition: $u(x_m) = -x_m u'(x_m)$.

To get some feeling of how $W/V$ behaves, I tried some function families.

Let $u(x)=(1-x)^{t-1}$, where $t>1$ is a parameter. Then:

• $U(x)=-(1-x)^{t}/t$.
• The first-order condition gives: $x_m=1/t$.
• $W=U(1)-U(0) = 1/t$
• $V=U(x_m)-U(0)=(1-(\frac{t-1}{t})^{t})/t$
• $W/V=1/[1-(\frac{t-1}{t})^{t}]$

When $t\to\infty$, $W/V \to 1/(1-1/e)\approx 1.58$, so for this family, $W/V$ is bounded.

But what happens with other families? Here is another example:

Let $u(x)=e^{-t x}$, where $t>0$ is a parameter. Then:

• $U(x)=-e^{-t x}/t$.
• The first-order condition gives: $x_m=1/t$.
• $W=U(1)-U(0) = (1-e^{-t})/t$
• $V=U(x_m)-U(0)=(1-e^{-1})/t$
• $W/V=(1-e^{-t})/(1-e^{-1})$

When $t\to\infty$, again $W/V \to 1/(1-1/e)\approx 1.58$, so here again $W/V$ is bounded.

And a third example, which I had to solve numerically:

Let $u(x)=\ln(a-x)$, where $a>2$ is a parameter. Then:

• $U(x)=-(a-x)log(a-x)-x$.
• The first-order condition gives: $x_m=(a-x_m)\ln(a-x_m)$. Using this desmos graph, I found out that $x_m \approx 0.55(a-1)$. Of course this solution is only valid when $0.55(a-1)\leq 1$; otherwise we get $x_m=1$ and there is no deadweight loss.
• Using the same graph, I found out that $W/V$ is decreasing with $a$, so its supremum value is when $a=2$, and it is approximately 1.3.

Is there another family of finite functions for which $W/V$ can grow infinitely?

• Zero marginal cost does not imply zero production cost. Who bears the burden of this cost if the product is given away for free, and in what sense does social welfare is maximized then? – Alecos Papadopoulos Apr 16 '15 at 18:15
• "Let u(x) be the consumers' utility function (which is also the inverse demand function)." $$.$$Isn´t it the consumers $\texttt{marginal}$ utility function ? – callculus Apr 16 '15 at 18:18
• Without having read most of it, harmful depends on the concept of social welfare, and how we weight those two. If we only look at household surplus, a smaller price-elasticity allows the firms to reap more of the surpluses. Consequently, the demand function D(p) = x, is "worst", if we focus consumer surplus. – FooBar Apr 16 '15 at 18:38
• @AlecosPapadopoulos By $W$ I meant increase in social welfare due only to the trade (maybe I should have called it $\Delta W$). In this sense, the production costs are irrelevant. – Erel Segal-Halevi Apr 16 '15 at 19:28
• @calculus You are right, I corrected this, thanks! – Erel Segal-Halevi Apr 16 '15 at 19:36

$P=\begin{cases} \frac{1}{Q} & \text{if } Q>1 \\ 2-Q & \text{if } Q\leq 1 \\ \end{cases}$.
The monopolist prices at $P=1$, but the consumers' surplus if $P=0$ is infinite, because the area under the demand curve contains $\int_1^\infty \frac{1}{Q}dQ=\infty$.