Our techniques of proof require the number of independent tools to be not smaller than the number of goals. That observations implies that a simple redistribution of the numeraire good among all households would not work.

That redistribution has to satisfy the constraintPH

h=1½_{h} = 0; and, therefore,
the number of independent tools is only H ¡ 1, while the number of goals
-the utility levels of all households - is H.

On the other hand, if the planner can tax not only households but also

…rms, more tools become available. In fact, we show that a planner can Pareto improve upon the market outcome if her intervention is as described below:

The planner imposes taxes the use of inputs and the production of output
in a proportion ¿^{c} for each good c = 1; :::; C; G.

De…ne ¿ = (¿^{c})^{C;G}_{c=1}: Observe that the …rm f’s problem becomes:

max(^{y}^{f}^{;y}f^{g})PC;G

c=1(1¡ ¿^{c}) p^{c}y_{f}^{c} s:t: tf

¡yf; y_{f}^{g}¢

= 0:

Therefore to describe equilibria with planner intervention we have to change the equilibrium system as follows:

1. The tax collection has to balance:

XF f =1

XC;G c=1

¿^{c}p^{c}y_{f}^{c} = 0:

2. The First Order Conditions for Firm f becomes
(1¡ ¿) ¤ (p; p^{g}) + ®_{f}Dt_{f}¡

y_{f}; y^{g}_{f}¢

= 0 tf

¡yf; y_{f}^{g}¢

= 0 ;
where, for given x; y 2 R^{n}; x¤y ´ (x^{i}yi)^{n}_{i=1}:

Remark 8 Here we choose to tax each …rm in the same proportion with respect to each good. Another possibility would be to impose taxes (and sub-sidies) which depend on the type of goods and identity of …rms. That would allow substituting for the requirement C ¸ H of the main Theorem of the Section a much weaker requirement of the form F C ¸ H.

We could also let the government have her own demand µ^{g} for the public
good: that would increase by one the number of tools at her disposal.

Finally, we could let the government impose taxes on households; it turns out that this does not have any signi…cant e¤ect.

Another tool which may be e¤ective would be a subsidy/tax on the price of the public good for the households. Observe that taxing only one good does not create any kink in the budget set, but it only rotates the budget plane.

Remark 9 If we use production functions (y^{g}_{f} = tf(yf)) instead of
transfor-mation functions, there are two di¤erent ways to show the above mentioned
generic result:

1. Impose a condition of the following form

ylim^{c}_{f}!0Dy^{c}_{f}tf(yf) =¡1:

2. Write down the problem as

maxyf p^{g}y^{g}_{f}¡ pyf s:t: y^{g}_{f} = tf(yf) °_{f}
yf ¸ 0 ¯_{f}

Then, it is enough to show that generically it cannot be the case that
yf = 0 and ¯_{f} = 0:

In this case, we can partition …rms in the group of ”strictly” active and

”strictly” inactive ones, and proceed taxing and subsidizing the active ones only.

The function F1 de…ning the ”equilibrium with planner intervention” in this case is the left hand side of the following system:

(1¡ ¿) ¤ (p; p^{g}) + ®fDtf

½_{h} 6= 0 i¤ h 2 H^{0}[ f1g :

= C + 1. Therefore, to be consistent with condition (24), we must have

C + 1¸ H + 1; or, C ¸ H: (14)

Remark 10 In fact, as shown in the proof of the main Theorem of this
section, see Case 18 and Remark 12, we have to impose C ¸ H^{+}: Moreover,
since H^{+} is an endogenous variable which at most can be equal to H, we are
going to require C ¸ H.

Denote a H^{+}¡dimensional subvector of ¿ by ¿^{¤} 2 R^{H}^{+}.
Observe that » is an equilibrium i¤ eF (»; (½; ¿ ) = 0; ¼) = 0:

Observe that the Walras’ law holds in the above case. Summing up consumers’ budget constraints, and observing that PH

h=1sh = 1, we get

We now show an important preliminary result.

Lemma 17 There exists an open and dense subset ¦^{0} µ ¦ such that for
every ¼^{0} 2 ¦^{0} and for every »^{0} such that eF (»^{0}; (¿; ½) = 0; ¼^{0}; u) = 0;

D_{(»;¿ ;½)}F (»e ^{0}; 0; ¼^{0}) has full row rank (equal to dim ¥ + 1):

Proof. D(»;½_{1})F (»e ^{0}; 0; ¼^{0}; u) is
D»F1 ¤

0 PF

f=1y_{f}^{g} > 0

¸

;

and the desired result follows. Observe that in the above matrix, D»F_{2} = 0
because it is computed at ¿ = 0:

As usual, from the above result, for every economy ¼; and every »^{0} 2
F_{¼}^{¡1}(0) ; we can then de…ne

bgc : V ! R; bgc : (¿^{¤})7! Gc(h (¿^{¤}) ; ¿^{¤}; ¼)
such that h (0) = (»^{0}; ¿^{¤} = 0) :

Theorem 18 Assume that C ¸ H. For an open and dense subset Sc^{¤} of
the set of the economies, at any equilibrium »^{0}; the function bgc is locally
onto around 0: That is, there exists a choice of taxes on the …rm’s inputs and
outputs which Pareto improves or impairs upon the equilibrium »^{0}: Moreover,
in the subset of the economies for which there are H^{+} contributors, it is
enough to assume that C ¸ H^{+}:

Proof. For simplicity of notation, we take h = 1 as the household whose associated columns are used to clean up the columns of the matrix under analysis; h = 2 as a non-contributor, and h = 3 as a contributor.

The proof of the Theorem goes through the usual steps. Properness follows in the same way as in the other cases. Denseness is proved in detail

below. We have

(i.e. the variables with ^{¤}) are H dimensional vectors of the same type as ¿^{¤}:
Observe that [bf^{0}1 = 0 for some f^{0}] implies that ( bf^{0}2 = 0; dm1 = 0; dm2 = 0,
and therefore) [bf 1 = 0 for each f ].

Remark 11 Moreover, observe that we cannot perturb equations (f1) with-out using the perturbation of the transformation function: bf 1 is going to be used to perturb equations (M:3) : For that reason, we have to consider the two cases bf 1 6= 0 and bf1 = 0. We …rst consider the cases in which bf 16= 0.

As pointed out in Remark 12, the cases under which bf1= 0 reduces to only one case (Case 19 below).

Assuming that bf 16= 0; we summarize the relevant cases in the following table:

Case

1 d11 6= 0; c^{12} 6= 0 c^{21}6= 0 d^{31}6= 0; d^{32}6= 0

2 d316= 0; d^{32}= 0

3 d_{31}= 0; d_{32}= 0

4 c_{21}= 0 d_{31}6= 0; d326= 0

5 d316= 0; d32= 0

6 d31= 0; d32= 0

7 d11 6= 0; c^{12} = 0 d21 6= 0 d^{31}6= 0; d^{32}6= 0

8 d316= 0; d^{32}= 0

9 d_{31}= 0; d_{32}= 0

10 d_{21} = 0 d_{31}6= 0; d326= 0

11 d316= 0; d^{32}= 0

12 d31= 0; d32= 0

13 c11 = 0; c12= 0 d21 6= 0 d^{31}6= 0; d^{32}6= 0

14 d316= 0; d^{32}= 0

15 d_{31}= 0; d_{32}= 0

16 d21 = 0 d316= 0; d^{32}6= 0

17 d316= 0; d^{32}= 0

18 d_{31}= 0; d_{32}= 0

After that we should consider the same case when df1 = 0: As pointed out in Remark 12 , those other Cases reduce to only one (Case 19).

Since the strategy of proof is similar, we analyze the most interesting cases.

Case 1: bf 16= 0; f or h 2 H d^{h1}6= 0; f or h 2 H^{+} d2h 6= 0:^{22}
(f:1) ¢ (b_{f 1})

(f:2) b^{C}_{f 1}
(1:1) ¢ (d_{11})
(1:2) ¢ (d12)
(1:3) d^{C}_{11}
(2:1) ¢ (d_{21})
(2:3) d^{C}_{21}
(3:1) ¢ (d31)
(3:3) d^{C}_{31}
(M:1) d^{n}_{h1}

(M:2) d_{h2}; h2 H^{+}
(M:3) b^{nC}_{f 1}

(M:4) duh as long as d12 6= 0 and d^{h1}6= 0

(15)

We now show in more detail the perturbation of equation (M:4) :
(M:4) Ã d^{uh} Ã (1:2) Ã ¢ (d^{12})

Ã (h:1) Ã ¢ (dh1) : (16)
Observe that we could perturb equation (M:4) also using dm2 - as long
as d126= 0 and dfi 6= 0:^{23}

Case 18. d11= 0; d_{12} = 0; d_{21} = 0; d_{31}= 0; d_{32} = 0:

Lost unkns. # Lost eqns. # # Eqns. we can erase Eqns. we erase #

d11 C C (M:3) H

d_{12} 1 (1:3) 1 (M:4) 1

d21 C (2:3) 1 C¡ 1 …rst (C ¡ 1) in (2:1) C ¡ 1

d31 C C …rst (C ¡ 1) in (3:1) C ¡ 1

d_{32} 1 (3:3)

3C¡¡¡¡ 1 in total 2C + H^{¡¡¡}

(17)

22If c12 = 0; we have to erase (1:2) : To perturb (f:2), we need to use a d^{c}_{f1} with
c6= 1; :::; H. We take d^{C}f 1to perturb (f:2), because d^{C}_{f1} does not appear in (M:3) :

23See Remark 12 for the reason for which we didn’t choose cm2as the perturbing variable.

Then, proceed as illustrated in the following table:

(f:1) ¢ (b_{f1})
(f:2) b^{C}_{f1}
(1:1) d_{m1}; d_{u1}
(1:2) dm2

(1:3) cancelled
(2:1) d_{u2}
(2:3) cancelled
(3:1) du3

(3:3) cancelled
(M:1) d^{nC;G}_{f 1}
(M:2) d^{nC;G}_{f 1}
(M:3) cancelled
(M:4) cancelled

: (18)

Remark 12 We should now analyze the other remaining 18 Cases in which bf1= 0 for each f: But it is enough to observe what follows.

1. If bf 1= 0; then we have the following situation:

Lost unkns. # Lost eqns. # # Eqns. we can erase Eqns. we erase #

bf1 C + 1 (f:1) C + 1 (H:7) H

bf2 1 (f:2) 1

d_{m1} C¡ 1 C¡ 1

dm2 1 1

¡¡¡C in total ^{¡¡¡}H

:

In all …rst 17 cases, we never used dm2 and we used bf1 and bf2 to perturb equations (f:1), (f:2) and (M:3) which we erased.

2. The case in which all the perturbing variables are equal to zero is analyzed below.

Proof. Case 19. bf 1 = 0 for each f; dh1 = 0 for h 2 H; and d^{2h} = 0 for
h2 H^{+}:

>From (f:1) ; bf 2 = 0; dm1 = 0; and dm2 = 0: Therefore, the system

Using the …rst order conditions of households’ problems, we can rewrite the …rst part of the system as follows:

(1:1C) 1 _{¡c13} _{cu1¸1}

>From (3:1C) ; d33= du3¸3:

>From (2:1C), d23 = du2¸2:

>From (1:2) ; du2¸_{2} =¡du3¸_{3}:

Therefore, d13 = d33 =¡d^{23}; or more generally,

d13= d_{h}+03 = dh^{+}1 =¡d^{h}^{0}^{0}^{3} =¡d^{h}^{0}^{3} for h^{+}; h^{+}^{0} 2 H^{+} and h^{0}; h^{0}^{0} 2 H^{0}:
Substituting in (M:1) ; (M:2) and (M:3), and using some obvious notation,
we get
and also d = 0; a contradiction.

To get d13= 0, it must be that either

The …rst condition can be proved to hold true and the second condition
is clearly true, generically in the space of economies and as long as H^{0} 6= ;:

Observe that it cannot be that H^{0} =H:

The case in which H^{0} =; is analyzed below:

If H^{0} = ;, then equations (2:1) and all terms related to household 2 in
the system disappear, and we get the following system.

(1:1) C ¡d13

Using the …rst order conditions of households’ problems, we can rewrite the

…rst part of this system as:

(1:1C) 1 _{¡d13} _{du1¸1}

>From (3:2) ; d13 = d33:

>From (3:1C) ; d33= du3¸3:

>From (1:2) ; du3 = 0:

Therefore, d13 = d33 = du1 = du2 .

Finally with (M:3) ; and recalling that p^{g}y^{g} 6= 0; d^{m3}= 0 and also d = 0;

we get a contradiction.

### 6 Appendix. Di¤erential Analysis on the Equi-librium Manifold

The starting point of the analysis is a function whose zeros describe equilibria:

F : ¥^{n}^{1} £ £ £ U ! R^{n}^{1}; F : (»; µ; u)7! F (»; µ; u) ;

where ¥^{n}^{1} is an open subset ofR^{n}^{1}, the set of endogenous variables, £ is the
set of the exogenous variables, and U is the utility function space.

Then some new variables ¿ 2 T ´ R^{m} are added. T is the set of the
planner’s tools. The function

F_{1} : ¥^{n}^{1} £ T £ £ £ U ! R^{n}^{1}; F_{1} : (»; ¿ ; µ; u)7! F1(»; ¿ ; µ; u)
describes the equilibrium ”with planner intervention”, and the function

F2 : ¥^{n}^{1} £ T ! R^{p}; F2 : (»; ¿) 7! F^{2}(»; ¿ )
describes the constraints on the planner intervention. De…ne

F : ¥e ^{n}^{1} £ T £ £ £ U ! R^{n}^{1}^{+p}^{´n}; eF : (»; ¿ ; µ; u)7! (F^{1}(»; ¿ ; µ; u) ; F2(»; ¿ )) :

The set T of tools can be written as T = T1 £ T^{2} = R^{m}^{¡p} £ R^{p}; with
(¿1; ¿2) 2 T; and where ¿^{1} can be interpreted as the vector of independent
tools and ¿2 as the vector of dependent tools.

Step 1. There exists ¿1 such that for each (µ; u) 2 £ £ U
f» 2 ¥^{n}^{1} : F (»; µ; u) = 0g

=n

»2 ¥^{n}^{1} :9! ¿2 such that eF (»; ¿_{1}; ¿_{2}; µ; u) = 0o (23)
That is, we have ”An equilibrium without planner intervention is an
equilib-rium with planner intervention when the planner decides not to intervene”.

In fact, we want to study the e¤ects of changes in ¿1 around ¿1: Finally, G describes the goals of the planner:

G : ¥^{n}^{1} £ T £ £ £ U ! R^{k}; G : (»; ¿ ; µ; u)7! G (»; ¿; µ; u) :

Step 2. For every u 2 U; there exists an open and full measure
sub-set £u of £ such that for every µ^{0} 2 £u and for every »^{0} 2 ¥ such that
F (»e ^{0}; ¿1; ¿2; µ; u) = 0;

rank D_{(»;¿}_{2}_{)}F (»e ^{0}; ¿_{1}; ¿_{2}; µ; u) is full.

Usually, the above result follows from the fact that D(»;¿2)F (»e ^{0}; ¿_{1}; ¿_{2}; µ; u) is
D»F1 D¿2F1

D»F2 D¿2F2

¸

;

and, from a Regularity Lemma, D»F_{1} has full row rank in an open and full
measure subset of £u, the fact that D»F2 = 0; and that D¿2F2 has full row
rank.

De…ning

bg : T1 ! R^{k}; bg : ¿1 7! G (» (¿^{1}) ; ¿2(¿1) ; ¿1) ;

we want to show that in an open and dense set of economies dbg^{¿}1 is onto
and, therefore, bg is locally onto around ¿1: As explained in Citanna, Kajii
and Villanacci (1998), that condition is implied by the following one.

There exists an open and dense subset S^{¤} µ £ £ U such that for every
(µ^{0}; u^{0})2 £ £ U and for every »^{0} 2 ¥^{n}^{1} such that eF (»^{0}; ¿_{1}; ¿_{2}; µ; u) = 0

rank h

D_{(»;¿ )}F (»e ^{0}; ¿_{1}; ¿_{2}; µ; u)i

(n1+p+k)£(n1+m) = n_{1}+ p + k:

The above condition implies that it must be m ¸ p + k;

i.e.,

(number of tools ¸ (number of constraints on planner intervention) + (number of goals)

(24)

or

m¡ p ¸ k;

i.e., (number of independent tools)¸ (number of goals).

The above statement is equivalent to showing that for (µ; u) 2 S^{¤} the
following system has no solutions (»; c) 2 ¥^{n}^{1} £ R^{n}^{1}^{+p+k} :

or, using condition (23), that the following system has no solutions:

8> is closed, it is su¢cient to show that the following function is proper:

pr : F^{¡1}(0)! £ £ U; pr : (»; µ; u) 7! (µ; u) :
Step 4. Density of S^{¤}:

We apply the Parametric Transversality Theorem to the function de…ned by the left hand side of system (25). That amounts to show that the following matrix has full row rank:

» c ; ®_{u}

where ®u is an element of an Euclidean space which is a …nite dimensional local parametrization of the utility function space.

Again from a Regularity result, D»F has full row rank in an open and dense subset of £ £ U: Moreover, it is crucial to have B (»; c; µ; ua) = 0: If that is the case, to get the result in Step 4, it is enough to show that the following matrix has full row rank

A (»; ¿ ; µ; u) =

" h

D_{(»;¿ )}³
F ; Ge ´

(»; ¿ ; µ; u)iT

N (®u)

c 0

# :

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