The case for subsidizing harm: constrained and costly Pigouvian taxation with multiple externalities

Abstract

Many activities are subsidized despite generating negative externalities. Examples include needle exchanges and energy production subsidies. We explain this phenomenon by developing a model in which the policymaker faces constraints or costs. We highlight three examples. First, it may be optimal to subsidize a harmful activity if the policymaker cannot set the first-best tax on an externally harmful substitute. Second, it may be optimal to subsidize a harmful production process if the activity mix at lower levels of output uses more harmful activities than the activity mix at higher levels of output. Third, it may be optimal to subsidize a harmful activity if there is a large administrative cost associated with taxing a harmful substitute. We also show how the functional form of the cost of administering a Pigouvian tax affects the optimal tax. When administrative cost is a function of only tax rates, the policymaker should tax each activity. However, an increase in the tax presents a trade-off: lower externality but higher administrative cost. A subsidy may be optimal for some externally harmful activities. When administrative cost is a function of only activity levels, it may not be optimal to tax every activity. If it is optimal to tax each activity, the policymaker should set the tax equal to the externality plus the marginal administrative cost. If it is not optimal to tax every activity, the complementarity between activities comes into play, and it may be optimal to subsidize externally harmful activities.

Appendix

1.1 A Net benefit and utility maximization

Let $u(x,x_{n+1}):\mathbb {R}^{n+1} \rightarrow \mathbb {R}$ be a strictly increasing, strictly concave representative agent’s utility function over $n+1$ activities. Let $p(x,x_{n+1}):\mathbb {R}^{n+1} \rightarrow \mathbb {R}$ be a strictly increasing, weakly convex function, such that $p(x,x_{n+1}) = 0$ defines a production possibility frontier. Then

• (i) There exists a function $f(x):\mathbb {R}^{n} \rightarrow \mathbb {R}$ such that $p(x,x_{n+1})=0$ if and only if $x_{n+1} = f(x)$;

• (ii) $u(x,f(x)) = b(x)$ defines a strictly concave net benefit function; and

• (iii) Assuming activity $n+1$ generates no external harm and is untaxed, the agent’s best response to a vector of activity taxes is the same under either the constrained utility maximization problem or the unconstrained net benefit maximization problem.

Proof

• (i) Under the implicit mapping theorem, there exists a function f such that $p(x,x_{n+1})=0$ if and only if $x_{n+1} = f(x)$ and $f_i = -p_i/p_{n+1}$.

• (ii) First, the weak convexity of p implies the weak concavity of f.

  $$\begin{aligned} p(y,y_{n+1}) - p(x,x_{n+1})&\ge \sum _{i=1}^{n+1} p_i(x,x_{n+1})(y_i - x_i) \end{aligned}$$

  (73)

  Thus, $\forall (x,x_{n+1}),(y,y_{n+1})$ such that $p(x,x_{n+1}) = p(y,y_{n+1}) = 0$, we have

  $$\begin{aligned} 0&\ge \sum _{i=1}^{n+1} p_i(x,x_{n+1})(y_i - x_i) \implies \end{aligned}$$

  (74)
  $$\begin{aligned} -p_{n+1}(x,x_{n+1})(y_{n+1} - x_{n+1})&\ge \sum _{i=1}^{n} p_i(x,x_{n+1})(y_i - x_i) \implies \end{aligned}$$

  (75)
  $$\begin{aligned} (y_{n+1} - x_{n+1})&\le \sum _{i=1}^{n} (-p_i(x,x_{n+1}))/(p_{n+1}(x,x_{n+1}))(y_i - x_i) \implies \end{aligned}$$

  (76)
  $$\begin{aligned} f(y) - f(x)&\le \sum _{i=1}^{n} f_i(x) (y_i - x_i) \end{aligned}$$

  (77)

  Second, the weak concavity of f and strict concavity of u imply the strict concavity of b.

  $$\begin{aligned} \forall \theta \in [0,1] : b \left( \theta x + [1-\theta ] y \right)&= u(\theta x + [1-\theta ] y,f(\theta x + [1-\theta ] y)) \end{aligned}$$

  (78)
  $$\begin{aligned}&\ge u(\theta x + [1-\theta ] y,\theta f(x) + [1-\theta ] f(y)) \end{aligned}$$

  (79)
  $$\begin{aligned}&= u(\theta x + [1-\theta ] y,\theta x_{n+1} + [1-\theta ] y_{n+1}) \end{aligned}$$

  (80)
  $$\begin{aligned}&> \theta u(x, x_{n+1}) + [1-\theta ] u(y, y_{n+1}) \end{aligned}$$

  (81)
  $$\begin{aligned}&= \theta b(x) + [1-\theta ] b(y) \end{aligned}$$

  (82)

• (iii) With a vector of taxes t, the agent’s utility maximization problem is a Lagrangian with Lagrange multiplier $\mu$,

  $$\begin{aligned} \mathcal {L} = u(x,x_{n+1}) - t^\top x -\mu p(x,x_{n+1}) \end{aligned}$$

  (83)

  with first-order conditions

  $$\begin{aligned} u_i -t_i - \mu p_i = 0\text { for all }i \in \{1, \hdots , n\}\hbox { and } u_{n+1} - \mu p_{n+1} = 0 \end{aligned}$$

  (84)

  which may be combined to yield

  $$\begin{aligned} u_i -t_i - u_{n+1} \frac{p_i}{p_{n+1}} = 0 \end{aligned}$$

  (85)

  With a vector of taxes t, the agent’s net benefit maximization problem is $\max _x u(x,f(x)) - t^\top x$ with first-order conditions

  $$\begin{aligned} u_i +u_{n+1}f_i -t_i = 0\hbox { for all}\ i \in \{1, \hdots , n\} \end{aligned}$$

  (86)

  From (i), $f_i = -p_i/p_{n+1}$. Thus,

  $$\begin{aligned} u_i -t_i - u_{n+1} \frac{p_i}{p_{n+1}} = 0 \end{aligned}$$

  (87)

  Since the two problems have the same first-order conditions, they imply the same agent best response function. $\square$

1.2 B Private net benefit and output

A rigorous investigation of the optimal output tax requires a formal relationship between private net benefit and output. We assume that private net benefit is the benefit of output less the cost of activities. Thus, when output is explicitly incorporated into the model, we require two additional functions. Strictly increasing, strictly concave v maps output to private benefit. Strictly increasing, strictly convex g maps the activities to private cost. Thus, $b(x) = v(q(x)) - g(x)$ and the private market’s problem is

$$\begin{aligned} \max _x v(q(x)) - g(x) -\tau q(x) \end{aligned}$$

(88)

with first-order condition

$$\begin{aligned} v'(q(x(\tau ))q'(x(\tau )) - g'(x(\tau )) -\tau q'(x(\tau )) = 0 \end{aligned}$$

(89)

Under the implicit mapping theorem, $x(\tau )$ exists and is continuously differentiable everywhere that the derivative of first-order condition with respect to x has a nonzero determinant. The derivative of first-order condition with respect to x is

$$\begin{aligned}&v'(q(x))q''(x) + (q'(x))^\top v''(q(x)) q'(x) - g''(x) - \tau q''(x) \end{aligned}$$

(90)

$$\begin{aligned}&\quad =[v'(q(x))-\tau ]q''(x) + v''(q(x)) (q'(x))^\top q'(x) - g''(x) \end{aligned}$$

(91)

$v'(q(x))-\tau > 0$ under the first-order condition. $q''(x)$ is negative semidefinite by assumption. $v''(q(x)) < 0$ by assumption. $(q'(x))^\top q'(x)$ is positive semidefinite because for any vector v, $(v^\top q'(x)^\top )(q'(x)v) \ge 0$. And $g''(x)$ is positive definite by assumption. Thus,

$$\begin{aligned} =&\underbrace{[v'(q(x))-\tau ]}_+ \underbrace{q''(x)}_\text {NSD} + \underbrace{v''(q(x))}_{-} \underbrace{(q'(x))^\top q'(x)}_\text {PSD} - \underbrace{g''(x)}_\text {PD} \end{aligned}$$

(92)

Because a positive (semi)definite matrix multiplied by a negative scalar is a negative (semi)definite matrix and since the sum of a negative definite matrix and a negative semidefinite matrix is a negative definite matrix, the sum here is negative definite. Since a negative definite matrix has a nonzero determinant, $x(\tau )$ exists and is continuously differentiable. We have shown that $b''(x) - \tau q''(x)$ is negative definite. Thus, its inverse, A, is also negative definite.

The additional structure that v and g provide to the problem makes it clearer under what circumstances the level of an activity increases with the output tax. In our solar and coal power example, we have

$$\begin{aligned} x'(\tau ) =&\text {A}q'(x(\tau ))^\top \end{aligned}$$

(93)

$$\begin{aligned} =&\frac{1}{\det (\text {A}^{-1})}\left( \begin{array}{cc} (v'-\tau ) q_{cc} + v''q_cq_c - g_{cc} &{} -(v'-\tau ) q_{sc} - v''q_sq_c + g_{sc} \\ -(v'-\tau ) q_{cs} - v''q_cq_s + g_{cs} &{} (v'-\tau ) q_{ss} + v''q_sq_s - g_{ss} \end{array} \right) \left( \begin{array}{c} q_s \\ q_c \end{array} \right) \end{aligned}$$

(94)

$$\begin{aligned} =&\frac{1}{\det (\text {A}^{-1})}\left( \begin{array}{c} q_s(v'-\tau ) q_{cc} + q_sv''q_cq_c - q_sg_{cc} -q_c(v'-\tau ) q_{sc} - q_cv''q_sq_c + q_cg_{sc} \\ -q_s(v'-\tau ) q_{cs} - q_sv''q_cq_s + q_sg_{cs} +q_c(v'-\tau ) q_{ss} + q_cv''q_sq_s - q_cg_{ss} \end{array} \right) \end{aligned}$$

(95)

$$\begin{aligned} \tfrac{\partial {x_c}}{\partial {\tau }} =&\frac{1}{\det (\text {A}^{-1})}\left( -q_s(v'-\tau ) q_{cs} + q_sg_{cs} +q_c(v'-\tau ) q_{ss} - q_cg_{ss} \right) \end{aligned}$$

(96)

Since $\frac{1}{\det (\text {A}^{-1})}>0$, $\tfrac{\partial {x_c}}{\partial {\tau }}>0$ requires

$$\begin{aligned} \underbrace{-q_s(v'-\tau ) }_\text {negative}q_{cs} + \underbrace{q_s}_\text {positive}g_{cs} + \underbrace{q_c(v'-\tau ) q_{ss}}_\text {negative} +\underbrace{ - q_cg_{ss}}_\text {negative} > 0 \end{aligned}$$

We know that $(v'-\tau )>0$, $q_i >0$, $q_{ss}<0$, and $g_{ss}>0$. Thus, for $\tfrac{\partial {x_c}}{\partial {\tau }}>0$ we require either $q_{sc}<0$ or $g_{sc}>0$. Having $q_s$ be large, $q_c$ be small, $q_{ss}$ be small, and $g_{ss}$ be small will also help push $\tfrac{\partial {x_c}}{\partial {\tau }}$ above zero.