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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.

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

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?

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' 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.

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

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.

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

Consider a firm with zero marginal cost. If it gives the product for free, then all the demand is satisfied and the social welfare is maximized at, say, $W$.

But because the firm is a monopoly, it reduces the demand and increases the price in order to optimize its revenue. This reduces the welfare to, 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' 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.

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

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' 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.

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