Social security in an analytically tractable overlapping generations model with aggregate and idiosyncratic risks

Abstract

When markets are incomplete, social security can partially insure against idiosyncratic and aggregate risks. We incorporate both risks into an analytically tractable model with two overlapping generations. We derive the equilibrium dynamics in closed form and show that the joint presence of both risks leads to overproportional risk exposure for households. This implies that the whole benefit from insurance through social security is greater than the sum of the benefits from insurance against each of the two risks in isolation. We measure this through interaction effects which appear even though the two risks are orthogonal by construction. While the interactions unambiguously increase the welfare benefits from insurance, they can increase or decrease the welfare costs from crowding-out of capital formation. The net effect depends on the relative strengths of the opposing forces.

Appendix: proofs

Proof of Proposition 1

Since $\tilde{\beta }=1$, maximizing Eq. (4) amounts to $max {\mathbb {E}}\, c_{i,2,t+1}^{1-\theta }$, where $c_{i,2,t+1}$ is given in (5). Increasing ex-ante utility for a marginal introduction of social security requires the first-order condition w.r.t. $\tau$, evaluated at $\tau =0$, to exceed zero, ${\mathbb {E}} \left. \left[ c_{i,2,t+1}^{-\theta } \frac{ \partial c_{i,2,t+1} }{ \partial \tau } \right] \right| _{\tau =0} > 0.$ We then get ${\mathbb {E}} \left[ \frac{(1+g)\zeta _{t+1} \left( 1+\lambda \left( 1-\eta _{i,2,t+1}\right) \right) - \zeta _t \bar{R} \varrho _{t+1}}{\left( \zeta _t \bar{R} \varrho _{t+1} \!+\! (1+g)\lambda \zeta _{t+1} \eta _{i,2,t+1} \right) ^{\theta } } \right] > 0$, which gives Eq. (7). $\square$

Proof of Proposition 2

Rewrite (8) as $A_{pe} = {\mathbb {E}} \left[ \frac{ a Z_1- b Z_2 - 1 }{1 + b Z_2 } \right] ,$ where $a \equiv (1+\lambda ) \frac{1+g}{\bar{R}}$, $b \equiv \lambda \frac{1+g}{\bar{R}}$, and $Z_1 \equiv \frac{\zeta _{t+1}}{\zeta _t \varrho _{t+1}}$, $Z_2 \equiv \frac{\zeta _{t+1} \eta _{i,2,t+1} }{\zeta _t \varrho _{t+1}}$. Take a second-order Taylor-series approximation around $Z_2=Z_1=1$:

$$\begin{aligned} A_{pe} \approx&\frac{1}{(1+b)^3} \left( a b^2 {\mathbb {E}} [Z_1 Z_2^2]-3 a b^2 {\mathbb {E}} [Z_1 Z_2] \!-\! a b {\mathbb {E}} [Z_1 Z_2]+3 a b^2 {\mathbb {E}} [Z_1]+ \right. \\&\left. 3 a b {\mathbb {E}} [Z_1]+a {\mathbb {E}} [Z_1]-(1+b)^3 \right) \end{aligned}$$

Observe that no interactions are present in term ${\mathbb {E}} Z_1$, and by Assumption 1, there are also no interactions in term ${\mathbb {E}} \left[ Z_1 Z_2\right]$. However,

$$\begin{aligned} {\mathbb {E}} \left[ Z_1 Z_2^2 \right]&={\mathbb {E}} \left[ \frac{\zeta _{t+1}}{\zeta _t \varrho _{t+1}} \frac{\left( \zeta _{t+1}\eta _{2,i,t+1}\right) ^2}{\left( \zeta _t\varrho _{t+1}\right) ^2} \right] \!=\! {\mathbb {E}} \left[ \zeta _{t+1}^3 \right] {\mathbb {E}} \left[ \frac{1}{\zeta _{t}^3} \right] {\mathbb {E}} \left[ \eta _{2,i,t+1}^2 \right] {\mathbb {E}} \left[ \frac{1}{\varrho _{t+1}^3} \right] , \end{aligned}$$

so that an interaction between idiosyncratic and aggregate risks enters only through ${\mathbb {E}} \left[ Z_1 Z_2^2 \right]$. We have ${\mathbb {E}} \left[ \eta _2^2\right] = 1+\sigma _{\eta }^2$, ${\mathbb {E}} \left[ \frac{1}{(\varrho )^3}\right] = \left( 1+\sigma _{\varrho }^2\right) ^6$ and ${\mathbb {E}} \zeta ^3= \left( 1+\sigma _{\zeta }^2\right) ^3$. Therefore ${\mathbb {E}} \left[ Z_1 Z_2^2\right] = \left( 1+\sigma _{\zeta }^2\right) ^9 (1+\sigma _{\varrho }^2)^6 (1+\sigma _\eta ^2)$, from which the cross-derivatives follow. $\square$

Proof of Proposition 3

The proof is by guessing and verifying. As all households are ex-ante identical, we guess that

$$\begin{aligned} a_{2,t+1} = s\left( 1-\tau \right) w_t = s \left( 1-\tau \right) \left( 1-\alpha \right) \varUpsilon _t \zeta _t k_t^\alpha . \end{aligned}$$

If this is correct, then the equilibrium dynamics are given by

$$\begin{aligned} K_{t+1} = a_{2,t+1} = s \left( 1-\tau \right) \left( 1-\alpha \right) \varUpsilon _t \zeta _t k_t^\alpha \end{aligned}$$

As $k_{t+1}=\frac{K_{t+1}}{\varUpsilon _{t+1}(1+\lambda )}$, we get $k_{t+1} = \frac{1}{(1+g)(1+\lambda )} s (1-\tau ) (1-\alpha ) \zeta _t k_t^\alpha$.

To verify (9), notice that our guess for $a_{2,t+1}$ implies that

$$\begin{aligned} c_{1,t}&= (1-s) (1-\tau ) (1-\alpha ) \varUpsilon _t \zeta _t k_t^\alpha \\ c_{i,2,t+1}&= s (1-\tau ) (1-\alpha ) \varUpsilon _t \zeta _t k_t^\alpha \alpha \zeta _{t+1} \varrho _{t+1} k_{t+1}^{\alpha -1} +\\&\quad + (1-\alpha ) \varUpsilon _{t+1} \zeta _{t+1} k_{t+1}^{\alpha } \left( \lambda \eta _{i,2,t+1} + \tau \left( 1+\lambda (1-\eta _{i,2,t+1}) \right) \right) , \end{aligned}$$

where we used the budget constraint. Employing (9), we get

$$\begin{aligned} c_{i,2,t+1}&= \left( \alpha \varrho _{t+1} \left( 1+\lambda \right) + \left( 1-\alpha \right) \left( \lambda \eta _{i,2,t+1} + \tau \left( 1+\lambda \left( 1-\eta _{i,2,t+1}\right) \right) \right) \right) \\&\quad \times \varUpsilon _{t+1} \zeta _{t+1} k_{t+1}^{\alpha }. \end{aligned}$$

Next, notice that the first-order condition of household maximization gives

$$\begin{aligned} 1&= \beta {\mathbb {E}}_t \left[ \frac{c_{1,t}\left( 1+r_{t+1}\right) }{c_{i,2,t+1}}\right] \\&=\beta {\mathbb {E}}_t \left[ \frac{c_{1,t} \alpha \zeta _{t+1} \varrho _{t+1} k_{t+1}^{\alpha -1} }{ \left( \alpha \varrho _{t+1} \left( 1+\lambda \right) + \left( 1-\alpha \right) \left( \lambda \eta _{i,2,t+1} + \tau \left( 1+\lambda \left( 1-\eta _{i,2,t+1}\right) \right) \right) \right) \varUpsilon _{t+1} \zeta _{t+1} k_{t+1}^{\alpha } }\right] \\&= \frac{\beta (1-s) }{s} \varPhi , \end{aligned}$$

where $\varPhi$ is defined in Eq. (11). Equation (10) immediately follows. Since the problem is convex, the solution is unique. The upper bound on $\varPhi$ is $\varPhi = 1$ for $\lambda =0$. For $\lambda > 0$, $\varPhi ={\mathbb {E}}_t \left[ \frac{ 1 }{ 1 + x }\right]$ for $x \equiv \frac{1-\alpha }{\alpha (1+\lambda )\varrho _{t+1}} \left( \lambda \eta _{i,2,t+1} + \tau \left( 1+\lambda (1-\eta _{i,2,t+1}) \right) \right)$. Our assumptions ensure that $x \ge 0$, hence $\varPhi ={\mathbb {E}}_t \left[ \frac{ 1 }{ 1 + x }\right] \le 1$, which implies that $s \le \frac{\beta }{1+\beta }$. $\square$

Proof of Proposition 4

• 1. Recursive substitution of Eq. (9) gives

  $$\begin{aligned} k_{t+1} = \left( \frac{1}{(1+g)(1+\lambda ) } s(\tau ) (1-\tau )(1-\alpha ) \right) ^{\frac{1-\alpha ^{q+1}}{1-\alpha }} \left( \prod _{i=0}^q \zeta _{t-i}^{\alpha ^i} \right) k_{t-q}^{\alpha ^{1+q}} \end{aligned}$$

  (15)

  for any initial capital stock $k_{t-q}$. For $q \rightarrow \infty$, we get

  $$\begin{aligned} k_{t+1}&= \left( \frac{1}{(1+g)(1+\lambda ) } s (1-\tau )(1-\alpha ) \right) ^{\frac{1}{1-\alpha }} \left( \prod _{i=0}^\infty \zeta _{t-i}^{\alpha ^i} \right) \\&=k_{ms} \left( \prod _{i=0}^\infty \zeta _{t-i}^{\alpha ^i} \right) = k_{ms} \, d(\zeta ,t) \end{aligned}$$

  where $k_{ms}$ denotes the capital stock that would obtain in equilibrium if nature would draw $\zeta _t=1$ in all periods $t-q,\ldots ,t$, for $q \rightarrow \infty$ (mean shock equilibrium).

• 2. Rewrite (2b) using the social security budget to make the excess return explicit

  $$\begin{aligned} c_{i,2,t+1}&= \left( s \zeta _t \varrho _{t+1} \bar{R}_{t+1} + \lambda \eta _{i,2,t+1} \frac{\bar{w}_{t+1}}{\bar{w}_t} + \right. \\&\qquad \left. +\,\tau \left( \left( 1+\lambda \left( 1-\eta _{i,2,t+1}\right) \right) \frac{\bar{w}_{t+1}}{\bar{w}_t} - s \zeta _t \varrho _{t+1} \bar{R}_{t+1} \right) \right) \bar{w}_{t} \zeta _{t+1}, \end{aligned}$$

  where $\bar{w}_t =\varUpsilon _t(1-\alpha )k_{t}^\alpha$ and $\bar{R}_{t+1}=\alpha k_{t+1}^{\alpha -1}$. From step 1, we get that in the mean shock equilibrium $\bar{w}_t =\varUpsilon _t(1-\alpha )k_{ms}^\alpha d(\zeta ,t-1)^{\alpha }$, $\bar{R}_{t+1}=\alpha k_{ms}^{\alpha -1} d(\zeta ,t)^{\alpha -1}$ and $\frac{\bar{w}_{t+1}}{\bar{w}_{t}} =(1+g) \left( \frac{d(\zeta ,t)}{d(\zeta ,t-1)}\right) ^{\alpha }$. Noting that $\frac{d(\zeta ,t)}{d(\zeta ,t-1)^\alpha }=\zeta _t$, consumption in $j=1,2$ can be written as:

  $$\begin{aligned} c_{1,t}&= \left( 1-s\right) \left( 1-\tau \right) \varUpsilon _t \zeta _t \left( 1-\alpha \right) k_{ms}^\alpha d\left( \zeta ,t-1\right) ^\alpha \equiv c_{1,t}\left( \tau ,k_{ms},s\right) \end{aligned}$$

  (16)
  $$\begin{aligned} c_{i,2,t+1}&= \left( s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} + \lambda \eta _{i,2,t+1} (1+g) + \tau \left( (1+g)(1+\lambda (1-\eta _{i,2,t+1}))\right. \right. \nonumber \\&\qquad \left. \left. - s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} \right) \right) \cdot \varUpsilon _t (1-\alpha )k_{ms}^\alpha \zeta _t \zeta _{t+1} d(\zeta ,t-1)^{\alpha } d(\zeta ,t)^{\alpha -1}\nonumber \\&\equiv c_{2,t+1}(\tau ,k_{ms},s). \end{aligned}$$

  (17)

• 3. Using (16) and (17) in (4), we can write ex-ante utility as an indirect utility function

  $$\begin{aligned} {\mathbb {E}} u_t&= {\mathbb {E}} \left[ u\left( c_{1,t}\left( \tau ,k_{ms},s\right) \right) + \beta {\mathbb {E}}_t \left[ u\left( c_{2,t+1}\left( \tau ,k_{ms},s\right) \right) \right] \right] \end{aligned}$$

  (18)

  where we use the law of iterated expectations to factor in the conditional expectations operator ${\mathbb {E}}_t$. Maximization of the above with respect to $\tau$ gives rise to the first-order condition²⁰

  $$\begin{aligned}&\underbrace{{\mathbb {E}} \left[ \frac{\partial u(c_{1,t})}{\partial c_{1,t}} \frac{\partial c_{1,t}}{\partial \tau } + \beta {\mathbb {E}}_t \left[ \frac{\partial u(c_{2,t+1})}{\partial c_{2,t+1}} \frac{\partial c_{2,t+1}}{\partial \tau } \right] \right] }_{\equiv A = A_1 + A_2} \nonumber \\&\quad \underbrace{{\mathbb {E}} \left[ \frac{\partial k_{ms}}{\partial \tau } \left( \frac{\partial u(c_{1,t})}{\partial c_{1,t}} \frac{\partial c_{1,t}}{\partial k_{ms}} + \beta {\mathbb {E}}_t \left[ \frac{\partial u(c_{2,t+1})}{\partial c_{2,t+1}} \frac{\partial c_{2,t+1}}{\partial k_{ms}} \right] \right) \right] }_{\equiv B = B_1 + B_2}, \end{aligned}$$

  (19)

  where $A_1$ ($B_1$) and $A_2$ ($B_2$), respectively, capture the effects on the period 1 and period 2, subutility function.

• 4. Using the explicit expressions for consumption in the two periods from (16) and (17) in the above, we get, evaluated at $\tau =0$,

  $$\begin{aligned} A_1&= {\mathbb {E}} \frac{\partial \ln (1-\tau ) }{\partial \tau } = \left. -\frac{1}{1-\tau } \right| _{\tau =0} = - 1, \\ B_1&= \alpha (1+\beta ) \frac{\partial \ln k_{ms}}{\partial \tau } \end{aligned}$$

  and for $A_2$:

  $$\begin{aligned} A_2&= \beta {\mathbb {E}}\left[ \frac{(1+g)\left( 1+\lambda \left( 1-\eta _{i,2,t+1}\right) \right) - s \varrho _{t+1} \alpha k_{ms}^{\alpha -1}}{s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} + \lambda \eta _{i,2,t+1} (1+g)}\right] \\&= \beta {\mathbb {E}}\left[ \frac{\frac{1-\alpha }{\alpha (1+\lambda )}\frac{1+\lambda \left( 1-\eta _{i,2,t+1}\right) }{\varrho _{t+1}}-1 }{1 + \frac{(1-\alpha ) \lambda }{\alpha (1+\lambda )} \frac{\eta _{i,2,t+1}}{\varrho _{t+1}} }\right] . \end{aligned}$$

  Adding term $A_1$, which represents the effects of taxation on income, yields term A in the proposition. Furthermore, we get

  $$\begin{aligned} B_2&\equiv \beta {\mathbb {E}}\left[ \frac{\alpha \varrho _{t+1} s (\alpha -1) k_{ms}^{\alpha -2} \frac{\partial k_{ms}}{\partial \tau } }{s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} + \lambda \eta _{i,2,t+1} (1+g)}\right] \\&=\beta {\mathbb {E}}\left[ -(1-\alpha ) \frac{1}{1+ \frac{(1-\alpha ) \lambda }{\alpha (1+\lambda )} \frac{\eta _{i,2,t+1}}{ \varrho _{t+1}} } \frac{\partial \ln k_{ms}}{\partial \tau }\right] . \end{aligned}$$

  Adding $B_1$ yields

  $$\begin{aligned} B&= \left( \alpha (1+\beta ) - \beta (1-\alpha ) \left. \varPhi \right| _{\tau =0} \right) {\mathbb {E}} \left[ \frac{\partial \ln k_{ms}}{\partial \tau } \right] . \end{aligned}$$

  Turning to $\frac{\partial \ln k_{ms}}{\partial \tau }$, we find that, at $\tau =0$, we have

  $$\begin{aligned} \frac{\partial \ln k_{ms}}{\partial \tau } = \frac{1}{1-\alpha }\left( \frac{\partial \ln s}{\partial \tau } + \frac{\partial \ln (1-\tau )}{\partial \tau } \right) = -\frac{1}{1-\alpha }\left( 1-\left. \epsilon _{s,\tau }\right| _{\tau =0} \right) <0, \end{aligned}$$

  where the sign follows from the fact that the semi-elasticity of the saving rate in $\tau$, $\left. \epsilon _{s,\tau }\right| _{\tau =0}$, is negative. Precisely, it is given by

  $$\begin{aligned} \left. \epsilon _{s,\tau }\right| _{\tau =0}&\equiv \left. \frac{\partial s}{\partial \tau } \frac{1}{s} \right| _{\tau =0} = -\left. \beta (1-s)^2 \frac{1}{s} \varPsi \right| _{\tau =0}= -\frac{\left. \varPsi \right| _{\tau =0}}{(1+\beta \left. \varPhi \right| _{\tau =0})\left. \varPhi \right| _{\tau =0}} < 0, \end{aligned}$$

  (20)

  where $\left. \varPsi \right| _{\tau =0} \equiv \left. -\frac{\partial \varPhi }{\partial \tau }\right| _{\tau =0} = {\mathbb {E}}_t \left[ \frac{\frac{1-\alpha }{\alpha (1+\lambda )} \frac{1+\lambda \left( 1-\eta _{i,2,t+1}\right) }{ \varrho _{t+1}} }{ \left( 1 + \frac{(1-\alpha )\lambda }{\alpha (1+\lambda )} \frac{\eta _{i,2,t+1}}{\varrho _{t+1}} \right) ^{2}} \right] > 0.$ Note that $\varPhi$ is shown in (11) and $\left. \varPhi \right| _{\tau =0} > 0$. Term B in the proposition then follows.

$\square$

Proof of Proposition 5 and Lemma 1

The aggregate resource constraint in the model is $c_{1,t}+c_{2,t}+K_{t+1}=F(K_t,\varUpsilon _t L_t)$, where $c_{2,t}=\int c_{i,2,t} di$. By homogeneity of $F(\cdot ,\cdot )$, maximizing per capita consumption $\bar{c}=\frac{c_{1,t}+c_{2,t}}{2}$ is equivalent to

$$\begin{aligned} \max \left\{ \frac{F(K_t,\varUpsilon _t L_t) }{N_t} - \frac{K_{t+1}}{N_t} \right\} . \end{aligned}$$

(21)

As $N_t=N_{t+1}=2, L_t=1+\lambda$ and recalling that $k_t = \frac{K_t}{\varUpsilon _t L_t}$, we have that $\frac{K_{t+1}}{N_{t+1}} = k_{t+1} \varUpsilon _{t+1} (1+\lambda ) \frac{1}{2}$ and $\frac{L_t}{N_{t}} = (1+\lambda ) \frac{1}{2}$. Maximizing (21) in steady state where $k_{t+1}=k_t=k$ is equivalent to $\max \left\{ f(k) - (1+g) k \right\}$. Using that $f(k))=k^\alpha$, we get the golden rule capital stock $k_{GR} = \left( \frac{\alpha }{1+g}\right) ^{\frac{1}{1-\alpha }}$.

From Eq. (9), we get that the steady-state capital stock in the deterministic $\lambda =0$ economy is $k \!= \! \left( \! \frac{\beta (1-\alpha )}{(1+\beta )(1+g)} \! \right) ^{\!\frac{1}{1-\alpha }}$. Hence, the deterministic $\lambda =0$ economy is dynamically efficient iff $\frac{\beta }{1+\beta } < \frac{\alpha }{1-\alpha }$.

Finally, observe that $B<0$ iff $\alpha (1+\beta ) - \beta (1-\alpha ) \left. \varPhi \right| _{\tau =0} > 0$ which we can write as $\frac{\beta \left. \varPhi \right| _{\tau =0}}{1+\beta } < \frac{\alpha }{1-\alpha }$. Dynamic efficiency of the deterministic $\lambda =0$ economy is a sufficient condition because from $0 < \left. \varPhi \right| _{\tau =0} \le 1$, we have that $\frac{\beta \left. \varPhi \right| _{\tau =0}}{1+\beta } \le \frac{\beta }{1+\beta }< \frac{\alpha }{1-\alpha }$, where the second inequality is condition (14). $\square$

Proof of Proposition 6

• 1. The partial derivative of term A follows immediately from Proposition 2, by setting $\sigma _\zeta ^2=0$ in $\frac{\partial \left. A_{pe} \right| _{\theta =1}}{\partial \sigma _\eta ^2}$.

• 2. The partial derivative of term B is given by

  $$\begin{aligned} \frac{\partial B}{\partial \sigma _\eta ^2}&= \underbrace{\left( \alpha (1+\beta ) - \beta (1-\alpha ) \left. \varPhi \right| _{\tau =0} \right) }_{>0} \frac{1}{1-\alpha } \underbrace{\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2}}_{\lesseqgtr 0} + \beta \underbrace{(1-\left. \epsilon _{s,\tau }\right| _{\tau =0})}_{>0} \underbrace{\frac{\partial \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2}}_{>0} \end{aligned}$$

  where it remains to establish that, indeed, $\frac{\partial \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2}>0$ and $\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2}\lesseqgtr 0$:

  • (a) To evaluate $\frac{\partial \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2}$, let $Z_3=\frac{\eta _{i,2,t+1}}{\varrho _{t+1}}$. Take a second-order Taylor-series approximation of $\left. \varPhi \right| _{\tau =0}$ around $Z_3=1$ to get

    $$\begin{aligned} \left. \varPhi \right| _{\tau =0} \approx \left[ \frac{b^2\,{\mathbb {E}} Z_3^2-\left( 3\,b^2+b\right) \,{\mathbb {E}} Z_3+3\,b^2+3\,b+1}{b^3+3 \,b^2+3\,b+1} \right] , \end{aligned}$$

    where $b \equiv \frac{(1-\alpha )\lambda }{\alpha (1+\lambda )}$. With ${\mathbb {E}} \left[ Z_3\right] = (1+\sigma _{\varrho }^2)$ and ${\mathbb {E}} \left[ Z_3^2\right] = (1+\sigma _\eta ^2)(1+\sigma _{\varrho }^2)^3$ get

    $$\begin{aligned} \frac{\partial \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2} = \frac{b^2\,\left( \sigma _{\varrho }^2+1\right) ^3}{(1+b)^3} > 0 \end{aligned}$$

    (22)

    and $\frac{\partial ^2 \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2 \partial \sigma _\varrho ^2} > 0$.

  • (b) For $\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2}$, we see from Eq. (20) that we need to determine $\frac{\partial \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2}$. Let $Z_4=\frac{1}{\varrho _{t+1}}$ and $a \equiv \frac{(1-\alpha )}{\alpha }$. Take a second-order Taylor-series expansion of $\left. \varPsi \right| _{\tau =0}$ around $Z_3=Z_4=1$ to get

    $$\begin{aligned} \left. \varPsi \right| _{\tau =0} \approx&\frac{1}{(1+b)^4} \left[ \left( 3\,b^2\,a\,{\mathbb {E}} Z_3^2 \!-\!\left( 8\,b^2\!+\!2\,b\right) \,a\,{\mathbb {E}} Z_3\!+\!\left( 6 \,b^2\!+\!4\,b\!+\!1\right) \,a\right) \,{\mathbb {E}} Z_4 \right. \\&+ \left. \left( 2\,b^2-b^3\right) \,{\mathbb {E}} Z_3^2+ \left( 3\,b^3-4\,b^2-b\right) \,{\mathbb {E}} Z_3-3\,b^3 \right] . \end{aligned}$$

    To evaluate this expression under log-normality, recall from above that ${\mathbb {E}} \left[ Z_3^2\right] = (1+\sigma _\eta ^2)(1+\sigma _{\varrho }^2)^3$ and observe that ${\mathbb {E}} \left[ Z_4\right] = 1+\sigma _{\varrho }^2$, ${\mathbb {E}} \left[ Z_3^2 Z_4\right] =(1+\sigma _\eta ^2) (1+\sigma _{\varrho }^2)^6$, and ${\mathbb {E}} \left[ Z_3Z_4\right] = (1+\sigma _{\varrho }^2)^3$. Consequently,

    $$\begin{aligned} \frac{\partial \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2} = \left( \sigma _{\varrho }^2+1\right) ^3 \frac{3\,a\,b^2\,\left( \sigma _{\varrho }^2+1\right) ^3-b^3\,+2\,b^2\,}{\left( b+1\right) ^4} >0 \end{aligned}$$

    (23)

    and $\frac{\partial ^2 \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2 \partial \sigma _\varrho ^2}>0$. The positive sign of $\frac{\partial \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2}$ follows from the fact that

    $$\begin{aligned} 3\,a\,b^2\,\left( \sigma _{\varrho }^2+1\right) ^3-b^3\,+2\,b^2 > 3\,a\,b^2-b^3\,+2\,b^2 >0, \end{aligned}$$

    which results from $3\,a\,b^2-b^3\,+2\,b^2 >0 \Leftrightarrow (3 -\alpha )(1+\lambda ) > (1-\alpha )\lambda ,$ because $\alpha \in (0,1)$ and $\lambda \in (0,1)$. As $\frac{\partial \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2}>0$ and $\frac{\partial \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2}>0$, we have $\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2} \gtreqless 0$. As $\frac{\partial ^2 \left. \varPhi \right| _{\tau =0} }{\partial \sigma _\eta ^2 \partial \sigma _\varrho ^2}>0$ and $\frac{\partial ^2 \left. \varPsi \right| _{\tau =0} }{\partial \sigma _\eta ^2 \partial \sigma _\varrho ^2}>0$, we have that $\frac{\partial ^2 \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2 \partial \sigma _\varrho ^2} \gtreqless 0$.

    We now characterize a lower bound on the semi-elasticity such that $\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2} < 0$. We get the expression $\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2}= \beta ^{-1} s \varPhi ^{-2} \left( 2 \frac{\varPsi }{\varPhi } \frac{\partial \varPhi }{\partial \sigma _\eta ^2} - \frac{\partial \varPsi }{\partial \sigma _\eta ^2} \right)$, which is less than zero if—making use of Eqs. (20), (22), and (23)— $\left. \epsilon _{s,\tau }\right| _{\tau =0} > - \frac{3\,a\,\left( \sigma _{\varrho }^2+1\right) ^3-b+2}{2 (1+\beta \varPhi ) (1+b)}.$ $\square$