Optimal second best taxation of addictive goods in dynamic general equilibrium: a revenue raising perspective

Proof of Proposition 1

To see that a competitive equilibrium satisfies the IMC and resource constraint, we substitute the factor prices (4.9) and (4.10) into the budget constraint (4.12). Using constant returns to scale, we then have:

(8.1)Rtbbt+F(kt, ht)−τh, tFh(kt, ht)ht=(1+τc, t)ct+(1+τd, t)dt+it+bt+1. (8.1)

Combining the above equation with the government budget constraint (4.22) gives the resource constraint (4.11).

To derive the IMC from the budget constraint, we substitute the household first order conditions (4.13)–(4.15) into the budget constraint (4.12), eliminating the tax rates, so that:

(8.2)λtRtkt+λtRtbbt−λtkt+1−λtbt+1=βt(uc(ct, st, lt)ct+MUd, tdt+ul(ct, st, lt)ht). (8.2)

Next using the first order conditions (4.16) and (4.17), we have:

(8.3)λtRt(kt+bt)−λt+1Rt+1(kt+1+bt+1)=βt(uc(ct, st, lt)ct+MUd, tdt+ul(ct, st, lt)ht). (8.3)

The above equation characterizes a sequence of budget constraints that can be used to recursively eliminate λ_tR_t(k_t+b_t), yielding:

(8.4)λ0(R0k0+R0bb0)−limt→∞λt+1Rt+1(kt+1+bt+1)=∑t=0∞βt(uc(ct, st, lt)ct+MUd, tdt−ul(ct, st, lt)ht). (8.4)

The transversality conditions imply the second term on the left hand side equals zero. Again using the household first order conditions at period zero gives:

(8.5)uc, 0(R0k0+R0bb0)1+τc, 0=∑t=0∞βt(uc(ct, st, lt)ct+MUd, tdt−ul(ct, st, lt)ht), (8.5)

which is the IMC.

We next show that, given allocations which satisfy the IMC and resource constraint, prices and policies exist which, along with the allocations, are a competitive equilibrium. Let {c_t, k_t, h_t, d_t} be a sequence which satisfies the IMC and resource constraint. Then r_t and w_t are defined using equations (4.9) and (4.10). Since τ_{c, 0} is given, we can define λ₀ using equation (4.13). Then λ_t can be defined recursively using equation (4.16). Then Rtb is defined using equation (4.17). Next, we define the government policies:

(8.6)(1+τc, t)=βtuc(ct, st, lt)λt, (8.6)

(8.7)(1−τh, t)=βtul(ct, st, lt)λtFh(kt, ht), (8.7)

(8.8)(1+τd, t)=βtMUd, tλt, (8.8)

Given the above prices and policies, all equations which define a competitive equilibrium are satisfied except the household and government budget constraints. We use b_t to satisfy the household budget constraint:

(8.9)bt=1Rtb(−rtkt−(1−τh, t)wtht+(1+τc, t)ct+(1+τd, t)dt+it+bt+1. (8.9)

We can multiply the above equation by λ_t and recursively eliminate b_t+1 from the above equation. After eliminating prices and policies using the household first order conditions (4.13)–(4.15), b_t is a function of the allocations:

(8.10)bt=(∏i=0t−1(Fk(ki, hi)+1−δ))1+τc, 0τc, 0∑i=t∞βi(uc, ici+MUd, idi−ul, ihi)−kt (8.10)

The above equation is the debt allocation which implies the household budget constraint is satisfied.

Since the budget constraint is satisfied, we simply substitute the resource constraint into the budget constraint to see that the government budget constraint is satisfied. Finally, by substituting the prices and policies into the IMC and reversing the derivation of the IMC, we see that the transversality conditions are satisfied.

Proof of Proposition 2

First, we rewrite equations (5.8) and (5.11), using the σ definitions, so that:

(8.11)∂IMC∂ct=βtuc, t(1−σc+ασsc, t−σhc, t), (8.11)

(8.12)∂IMC∂dt=βtMUd, t(α−ασs,t+σcs,t−σhs,t+βus, t+1s2, t+1MUd, t(ασs, t−ασs, t+1−σcs, t+σcs, t+1+σhs, t−σhs, t+1)). (8.12)

Now since v is homothetic, we know that:

(8.13)vc(ψc, ψs)vs(ψc, ψs)≡vc(c, s)vs(c, s), (8.13)

which implies:

(8.14)vcc(c, s)cvc(c, s)+vcs(c, s)svc(c, s)=vss(c, s)svs(c, s)+vcs(c, s)cvs(c, s), (8.14)

which, using the definition of u(.) in equation (6.1), implies:

(8.15)σsc−σc=σcs−σs. (8.15)

It is also immediate from the definition of u(.) that σ_hc=σ_hs. These facts and equations (5.3), (5.4), (8.11), and (8.12) together imply:

(8.16)MUd, tuc, t=1+μ(1−σs, t+σcs, t−(1−α)σsc, t−σhs, t)1+μ(α−ασs, t+σcs, t−σhs, t+J(αΔσs−Δσcs+Δσhs)). (8.16)

Hence, τ_d,t>τ_c,t if and only if the right hand side is greater than one, or:

(8.17)1−(1−α)σsc, t−σs, t>α−ασs, t+J(αΔσs−Δσcs+Δσhs), (8.17)

which simplifies to the desired result.

Proof of Propositions 3–4

For the CRR case, note that σ_hs=0, σ_s=1–(1–ξ)(1–σ), and σ_sc=(1–ξ)(1–σ), which implies the left hand side of condition (6.3) is zero. The right hand side of (6.3) is also zero since σ_cs and σ_s are constant.

For the steady state case, σ_i,t=σ_{i, t+1} for all i∈{s, sc, cs, hs}, so the result follows immediately from condition (6.3).

An analytical example: the quadratic case

In this section, we consider a linear-quadratic utility function. The linear-quadratic utility, a common specification in the literature (e.g., Becker, Grossman, and Murphy 1994, Gruber and Koszegi 2001), offers several advantages. First, we can analytically obtain a solution for this case. This allows us to study how the dynamics of addictive consumption change optimal taxation. Second, since this specification has no income effects to complicate the dynamics, we can more precisely characterize the relationship between tolerance and addictive taxation.

Assume the subtractive specification (3.6) for effective consumption holds and that the utility and production function are:

(8.18)u(ct, st, lt)=ωct+νst−st22+elt−lt22, e<1, ν>ω1−βγ, (8.18)

(8.19)F(kt, ht)=ktθht1−θ. (8.19)

Proposition 7 requires the above assumptions on e and ν to rule out corner solutions, that is they ensure positive and unique steady state hours and ordinary and addictive consumption.

Using equations (5.3) and (5.4), given the utility function (8.18), reveals that the marginal utility of d_t is constant in the optimal second best allocation. In particular:

(8.20)MUd, t=ω+μ(ω+ν(1−βγ))1+2μ. (8.20)

The marginal utility of d_t divided by MU_{c, t}=ω equals the tax ratio given by equation (4.19). Hence the tax ratio is constant over time. Furthermore, inspection of equations (4.13), (4.16), (5.3), and (5.6) indicates that τ_c is constant over time. Therefore, τ_d and τ_h are also constant over time. Thus the implicit interest tax rate is zero for all t.

Equation (8.20) implies MU_d,t>MU_c,t, and thus τ_d,t>τ_c,t for all t. Hence, we have shown:

Proposition 5Let u(.,.,.) and F(.,.) be given by equations (8.18) and (8.19) and let effective consumption be given by (3.6). Then τ_d,t>τ_c,tfor all t≥1 and the ratio of tax rates1+τd, t1+τc, t.is constant over time.

In the static version of the model without addiction, d_t=s_t has an income elasticity equal to zero whereas the income elasticity of c_t is positive. Further, ordinary consumption and leisure are substitutes, whereas ∂s∂wws=0. Thus, it is optimal to tax d_t at a higher rate because, regardless of k_t or d_t–1, c is more substitutable with leisure.^[23]

It is also clear from equation (8.20) that the second best optimal d_t is the solution to a linear second order difference equation,

(8.21)−βγdt+1+(1+βγ2)dt−γdt−1=1+μ1+2μ(ν(1−βγ)−ω), (8.21)

and that the second best optimal s_t is the solution to a linear first order difference equation. However, before computing the solution to d_t, we must verify that a solution exists for μ. We thus now proceed to prove that a unique, positive solution for μ exists if government spending is not so large as to exhaust the maximum feasible revenue in the economy, and not so small that given initial tax rates are sufficient to pay for all current and future government expenditures.

Proposition 6Let the conditions for Proposition 5 hold. Let g_tbe a stationary sequence with limiting valueg¯.Then there exists an interval [ζ_l,ζ_h], with 0<ζ_l<ζ_h<∞ such that ifG≡∑t=0∞βtgt∈[ζl, ζh],then a unique positive solution for μ exists.

Proof: we derive the solution for μ by starting from the first order conditions for the Ramsey problem (5.3)–(5.6) which now become:

(8.22)ϕtβt=ω(1+μ), (8.22)

(8.23)ϕtβt=ν(1−βγ)(1+μ)−(1+2μ)(st−βγst+1), (8.23)

(8.24)ϕt(1−θ)(ktht)θβt=(e−1)(1+μ)+(1+2μ)ht, (8.24)

(8.25)ϕt(θ(ktht)θ−1+1−δ)=ϕt−1. (8.25)

Using equation (8.22) to eliminate φ_t gives:

(8.26)ω(1+μ)=(ν(1+μ)−μ)(st−βγst+1), (8.26)

(8.27)ω(1+μ)(1−θ)(ktht)θ=(e−1)(1+μ)+(1+2μ)ht, (8.27)

(8.28)β(θ(ktht)θ−1+1−δ)=1. (8.28)

With the subtractive specification equation (8.26) implies:

(8.29)ω(1+μ)=ν(1−βγ)(1+μ)−(1+2μ)(dt−γdt−1−βγ(dt+1−γdt)), (8.29)

which eventually simplifies to (8.21). In the proof for Proposition 7, we are going to show that (8.21) has general solution given by (8.44). In order for Proposition 6 to hold, we now just need to prove that a non-zero and finite solution for μ exists. We proceed to do that in what follows.

Equation (8.28) implies the capital to labor ratio, denoted by A, is constant:

(8.30)A≡(θρ+δ)11−θ. (8.30)

Thus, equation (8.27) implies

(8.31)ht=1+μ1+2μh^  ,  h^≡1−e+ω(1−θ)Aθ, (8.31)

is constant. It then follows that k_t=Ah_t is constant as well and equation (8.44) implies s_t and MU_d,t are constant themselves. Since the elasticity of substitution of consumption over time is infinite, the planner absorbs all changes in g_t by varying c_t. Combining these results with resource constraint (4.11) yields a solution for c_t:

(8.32)ct=1+μ1+2μ(h^(Aθ−δA)−d^(1−γt+1))−γt+1d−1−gt, (8.32)

with

(8.33)d^≡ν(1−βγ)−ω(1−γ)(1−βγ). (8.33)

Now since h₀ enters into the left hand side of the IMC (5.1) and k₀ is given, the solutions for h₀, k₀ and therefore c₀ generally differ from the solutions for t≥1. Therefore, we let x≡1+μ1+2μ and insert the solutions for c_t, d_t, and h_t into the IMC (5.1) for t≥1, so that:

(8.34)R0k0+R0bb01+τc,0=∑t=1∞βt[xω(h^(Aθ−δA)−d^(1−γt+1))−ωγt+1d−1+(1−βγ)(ν−(1−γ)d^x)(d^x(1−γt+1)+γt+1d−1)−(e−1+h^x)h^x]+ωc0+MUd, 0d0−ul,0h0+ωg0−ωG. (8.34)

Next, recall from Proposition 1 that τ_{c, 0} and τ_{h, 0} are given. It follows from equations (4.13) and (4.14) that the planner cannot choose h₀ in this example, and instead takes the solution for h₀ from the competitive equilibrium as given. Further, the terms inside the summation depend on time only through γ^t+1 and β^t, and the equation is quadratic in x. Therefore, after evaluating the summation, we can write (8.34) as:

(8.35)ch(x)≡−ζ1x2+(ζ1−ζ2)x+ζ2−ζ3=(ζ1x+ζ2)(1−x)−ζ3=0, (8.35)

where

(8.36)ζ1≡h^2+(1−γ)2βd^2, (8.36)

(8.37)ζ2≡(1−ββ)(1−γ)γd^d−1, (8.37)

(8.38)ζ3≡1−ββ(ωG−(ω(1+τc,0)−1)(1−δ)k0−R0bb01+τc, 0−ω(1+τc, 0)−θ1+τc, 0k0θh01−θ+ul, 0h0). (8.38)

A solution such that μ>0 is a solution in the range 12<x<1. Now ch(x) attains a maximum at x*=(ζ₁–ζ₂)/(2ζ₁)<1/2 and ch(0)>ch(1). Hence it is immediate that ζ₃<0 is a necessary condition for x<1.

From equation (8.38), ζ₃>0 if and only if:

(8.39)G>ζl≡(ω(1+τc, 0)−1)(1−δ)k0−R0bb0ω(1+τc, 0)+ω(1+τc, 0)−θω(1+τc, 0)k0θh01−θ−ul, 0h0ω. (8.39)

This is the lower bound for government spending, G.

Condition (8.39) implies that ζ₃>0 which in turn implies both roots have modulus less than 1. We are left to show that the roots are real and that one root is greater than one half. Since x*<1/2, the smaller root has modulus less than one half. The larger root is real and greater than one half if and only if:

(8.40)ch(12)=−ζ14+ζ1−ζ22+ζ2−ζ3>0, (8.40)

(8.41)ζ3<ζ1−2ζ24. (8.41)

Using equation (8.38), condition (8.41) holds if and only if:

(8.42)G<ζh≡ζl+β1−β(ζ1+2ζ24ω). (8.42)

Defining ζ_l and ζ_h using equations (8.39) and (8.42) completes the proof.□ Given a unique solution for μ, d_t is the solution to the second order difference equation (8.20) and we can now explicitly characterize that solution.

Proposition 7Let the conditions for Proposition 5 hold. Then the explicit solution for d_tis:

(8.43)dt=ν(1−βγ)−ω(1−γ)(1−βγ)(1+μ1+2μ)(1−γt+1)+d−1γt+1. (8.43)

Proof: We first solve the second order difference equation (8.21):

(8.44)dt+1−(1+βγ2βγ)dt+1βdt−1=−1+μ1+2μ(ν(1−βγ)−ωβγ). (8.44)

It is straightforward to show the general solution of the above difference equation is:

(8.45)dt=Dp+A0γt+A1(βγ)−t, (8.45)

(8.46)Dp≡1+μ1+2μ(ν(1−βγ)−ω(1−βγ)(1−γ)). (8.46)

Following convention, we rule out the explosive, bubble solutions which requires A₁=0. Letting t=–1 implies A₀=γ(d_–1+D_p). Substituting for A₀ and simplifying gives the desired solution.

Having a closed form solution for d_t, given by equation (8.43), allows us to derive some interesting properties of the second best solution, both over time and when compared to the first best solution (μ=0). First, optimal consumption of d_t increases over time, assuming d_–1 is less than the steady state. The planner decreases d_t relative to the first best solution through the tax. However, since 0<μ<∞, equation (8.43) implies d_t in the second best optimum is at least half of the first best level in the steady state. The planner also decreases the growth rate of d_t since:

(8.47)grt=dt−dt−1dt−1=γt(1−γ)(d^x−d−1)d^x(1−γt)+γtd−1, d^≡ν(1−βγ)−ω(1−γ)(1−βγ), x≡1+μ1+2μ, (8.47)

which is decreasing in μ because:

(8.48)∂grt∂μ=−γt(1−γ)d−1dt−12(1+2μ)2<0. (8.48)

Another important advantage of this exercise with linear quadratic preferences is that, since we fully characterize the solution, we can explore how the strength of tolerance affects second best addictive consumption. To this end, since μ(γ) is unique, we can use the implicit function theorem to derive comparative statics using equation (8.20). Our results for other utility functions suggest strong tolerance should moderate the optimal tax ratio, as gains in current tax revenue from taxation of addictive goods are offset by losses in future tax revenues. It turns out that, if d_–1 is sufficiently large, it is indeed true that the optimal tax ratio is inversely related to the degree of tolerance. In particular, we have:

(8.49)d−1>βω(1−βγ)(1−β)⇒∂1+τd1+τc∂γ<0. (8.49)

Condition (8.49) derived explicitly below, is a sufficient condition calculated assuming μ=∞. In practice, for μ small, the optimal tax ratio is decreasing in the degree of tolerance under much less restrictive conditions.

To derive equation (8.49), we rewrite equation (8.20) using the definition of x, which implies:

(8.50)1+τd, t1+τc, t=MUd, tω=x+ν(1−βγ)ω(1−x). (8.50)

Hence:

(8.51)∂∂γ(1+τd, t1+τc, t)=∂x∂γ⋅(1−ν(1−βγ)ω−νβω(1−x))<0  iff,   (8.51)

(8.52)∂x∂γ>βν(1−x)ν(1−βγ)−w. (8.52)

Next, using the implicit function theorem on equation (8.35), we see that:

(8.53)∂x∂γ=(ζ1γx+ζ2γ)(1−x)2ζ1x−ζ1+ζ2. (8.53)

Since x∈(12, 1), the denominator of equation (8.53) is positive. Thus, substituting equation (8.53) into condition (8.52) and cross multiplying results in:

(8.54)[ζ1γ(ν(1−βγ)−ω)+2βνζ1]x+ζ2, γ(ν(1−βγ)−ω)+βνζ2>βνζ1. (8.54)

The coefficient on x in equation (8.54) is positive. Thus, it is sufficient to show:

(8.55)[ζ1γ(ν(1−βγ)−ω)+2βνζ1]12+ζ2, γ(ν(1−βγ)−ω)+βνζ2>βνζ1, (8.55)

(8.56)12ζ1γ(ν(1−βγ)−ω)+ζ2, γ(ν(1−βγ)−ω)+βνζ2>0. (8.56)

Finally substituting in the definitions of ζ using equations (8.36) and (8.37) and the derivatives of ζ:

(8.57)ζ1, γ=−2ω(ν(1−βγ)−ω)(1−βγ)3, ζ2, γ=1−ββ(ν(1−βγ)2−ω(1−βγ)2)d−1, (8.57)

and simplifying yields the desired result.

Table 1 gives parameter values for a numerical example. Table 1 indicates that the optimal tax ratio is decreasing in the degree of tolerance, even though condition (8.49) is violated, since the parameter G, set to 30% of GDP for all t, generates at most a value of only μ=4.74. The planner relies increasingly on labor taxes and less on addictive taxes as the degree of tolerance increases. For γ=0.55, taxation is nearly uniform. Figure 1 shows the time path of the first and second best levels of d for various values of γ. Increasing the level of tolerance severely reduces addictive consumption since the future costs of current consumption are higher. As expected, the difference between first and second best addictive consumption is widest at the steady state.

Additively separable utility

In this section we consider the case in which utility is additively separable. Following a similar procedure as with Proposition 2, we have:

Proposition 8Let assumptions (A1)–(A3) hold. In addition, let u(.) be additively separable in c, s, and l. Then τ_d,t>τ_c,tfor all t≥1 if and only if:

(8.58)ασs, t+1−α−σc, t>J(αΔσs). (8.58)

Proof: If utility is separable, equations (8.11) and (8.12) become:

(8.59)∂IMC∂ct=βtuc, t(1−σc), (8.59)

(8.60)∂IMC∂dt=βtMUd, t(α−ασs, t+J(αΔσs)). (8.60)

Equations (5.3), (5.4), (8.59), and (8.60) together imply:

(8.61)MUd, tuc, t=1+μ(1−σc, t)1+μ(α(1−σs, t)+J(αΔσs)). (8.61)

Hence, d is taxed at a higher rate if and only if the right hand side is greater than one, or:

(8.62)1−σc, t>α(1−σs, t)+J(αΔσs), (8.62)

which simplifies to the desired result.

For the intuition of (8.58), first assume s₂=0, which removes all dynamic effects. Condition (8.58) becomes:

(8.63)σs, ts1, tdtst−s11, tdts1, t−σc, t>0 (8.63)

Specializing further to the multiplicative or subtractive case (s_11,t=0 and s_1,td_t=s_t), gives:

(8.64)σs, t−σc, t>0 (8.64)

This condition is satisfied if and only if s_t=d_t is more complementary with leisure (or income inelastic) than c_t.^[24]

The difference between the left hand side of conditions (8.64) and (8.58) is the tax anticipation effect for the multiplicative case:

(8.65)(1−α)(1−σs, t)>0. (8.65)

Equation (3.3) implies the tax anticipation effect reduces the optimal addictive tax if and only if d_t is addictive.

As in Proposition 2, the dynamic effects in t–1 and t reduce to a single current period term which depends only on the elasticities and homogeneity, and should therefore be straightforward to check in empirical applications.

The addiction stock effect equals 1–α+ασ_s,t₊₁, which offsets current period revenue effects.

For the CRR case, we have:

Proposition 9Let the conditions of Proposition 8 hold, and let u(.)=v¹(c)+v²(s)+v³(l), with v₁and v²(.) CRR. Then τ_d,t>τ_c,tfor all t≥1 if and only if:

(8.66)ασs+1−α>σc. (8.66)

Proof: For CRR preferences, σ_i,t=σ_i,t₊₁ for all i∈{s, c}, so the result follows immediately from condition (8.58).□

Condition (8.66) combines tax anticipation effect with any difference in income elasticities between d_t and c_t. Condition (8.66) requires that d_t and leisure to be sufficiently strong complements relative to c in the model without dynamic effects to overcome the dynamic tax anticipation effect.

The homogeneity of the addiction function affects the strength of the tax anticipation effect. For α=1, as in the subtractive model, the dynamic tax anticipation effect exactly offsets the difference in income elasticities between d and s in the model without dynamic effects. In this case, Proposition 9 says τ_{d, t}>τ_{c, t} if and only if s_t is more complementary with leisure in a static model where s could be taxed. However, in the multiplicative model, higher tolerance implies a lower α and thus makes addictive goods less complementary with leisure, lowering the optimal tax rate, for σ_s,t>1.

Thus, under the conditions outlined above, neglecting the dynamic tax anticipation effect results in overly high tax rates for addictive goods, relative to the optimum, especially for addictive goods that exhibit strong tolerance.

Non-weakly separable utility

For the homothetic and separable cases studied above, weak separability implies that the static income elasticity and the substitutability with leisure are identical. Here we consider a utility function for which the substitutability with leisure does not necessary equal the income elasticity.

To see this in a concise way, let us consider the following class of utility functions:

(8.67)u(st, dt, lt)=q(ct)+v(st, lt). (8.67)

For this specification, we find:

Proposition 10Let assumptions (A1)–(A3) hold. In addition, let u(.) be of the form given by (8.67). Then τ_d,t>τ_c,tfor all t≥1 if and only if:

(8.68)1−α+ασs, t+σhs, t−σc, t>J(αΔσs+Δσhs). (8.68)

Proof: If utility is given by equation (8.67), equations (8.11) and (8.12) become:

(8.69)∂IMC∂ct=βtuc, t(1−σc, t), (8.69)

(8.70)∂IMC∂dt=βtMUd, t(α−ασs, t−σhs, t+J(αΔσs+Δσhs)). (8.70)

Equations (5.3), (5.4), (8.69), and (8.70) together imply:

(8.71)MUd, tuc, t=1+μ(1−σc, t)1+μ(α(1−σs, t)−σhs, t+J(αΔσs+Δσhs)). (8.71)

Hence, d is taxed at a higher rate if and only if the right hand side is greater than one, or:

(8.72)1−σc, t>α(1−σs, t)−σhs, t+J(αΔσs+Δσhs), (8.72)

which simplifies to the desired result.□

The intuition for (8.68) is identical to that of Propositions 2 and 8, with the exception of the additional current period effect on labor tax revenues, reflecting the lack of weak separability.

For constant elasticities, we have:

Proposition 11Let the conditions of Proposition 10 hold, and letv(.)=(sξ(1−l)−(1−ξ))1−σ−11−σ,with1−σ<−11−2ξandξ<12to ensure concavity. Then τ_{d, t}>τ_{c, t}for all t≥1 if and only if:

(8.73)1+(1−σ)(1−ξ(1+α))>σc. (8.73)

Proof: note that σ_s=1–ξ(1–σ), and σ_hs=(1–ξ)(1–σ). Since both terms are constant, we have J(αΔσ_s+Δσ_hs)=0. Substituting in these conditions into equation (8.68) gives (8.73).□

Note that the concavity restrictions imply the left hand side of (8.73) is less than one. Since σ>1 and ξ<12 (both conditions are necessary for concavity) and α≤1, the second term on the left hand side is negative and so σ_c,t≥1 is sufficient for the condition to be violated, and thus τ_{d, t}≤τ_{c, t}.

Proposition 12Let the conditions of Proposition 10 hold. Thenτ¯d>τ¯cif and only if:

(8.74)α(σ¯s−1)+σ¯hs>σ¯c−1. (8.74)

Proof:σ_i,t=σ_i,t₊₁ for all i∈{s, hs} in the steady state, so the result follows immediately from condition (8.68).□

It is possible to construct examples for which d_t is more complementary with leisure and yet the tax anticipation effect implies a lower tax rate for addictive goods. Suppose subtractive model, then in the static model with utility as in Proposition 11, s is more complementary with leisure than c. Yet if (8.73) is violated it is optimal to tax d_t at a lower rate.

Finally, given the subtractive model, the static effects for condition (8.73) are:

(8.75)1−ξ(1−σ)1−γ+(1−ξ)(1−σ)>σc, (8.75)

and condition (8.74) becomes:

(8.76)1−ξ(1−σ)+(1−ξ)(1−σ)>σc. (8.76)

So for σ_c satisfying (8.75), but not (8.76), τ¯d>τ¯c in the model without dynamic effects but τ¯d<τ¯c when the tax anticipation effect is accounted for. Note the range of values satisfying (8.75) but not (8.76) is increasing in γ. Strong tolerance tends to decrease the optimal addictive tax, by strengthening the tax anticipation effect.

Appendix: Tables and Figures

Figure 1

Dynamics of first and second best addictive consumption for various values of γ.