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


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.

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Fig. 1


  1. See, e.g., Krueger and Kubler (2006), Ludwig and Reiter (2010), Hasanhodzic and Kotlikoff (2013) for social security analyses in settings with aggregate risk where social security improves intergenerational risk sharing and İmrohoroğlu et al. (1995, 1998) and Conesa and Krueger (1999) as examples of studies where social security provides insurance against idiosyncratic productivity risk.

  2. This general equilibrium model can be seen as an extension of the standard Diamond (1965) model with aggregate and idiosyncratic risk. The setup is similar to Huffman (1987) with three important differences: First, we extend his work by taking into account idiosyncratic risk. Second, we do not only consider positive labor income in the first period of life but rather have two periods with positive labor income. Third, we stick to a two-period structure while Huffman (1987) has many periods.

  3. Most real-world pension systems feature some distributional components. Almost all systems have a minimum pension. In fact, our system features strong similarities to the Danish public pension system.

  4. This assumption is also made by Gordon and Varian (1988), Ball and Mankiw (2007), Matsen and Thogersen (2004), Krueger and Kubler (2006), Harenberg and Ludwig (2015), among others.

  5. This terminology is borrowed from statistical data analysis. To measure how both risks increase welfare gains and whether there is more than an additive effect, an econometrician would consider in a (linear) regression as an interaction term the product of risk measures, i.e., the product of variances.

  6. This is done in HL by conditioning higher moments of idiosyncratic productivity risk on the aggregate state of the economy via a countercyclical cross-sectional variance of idiosyncratic productivity risk, cf., e.g., Mankiw (1986), Constantinides and Duffie (1996), Storesletten et al. (2004, 2007), and, more recently, Guvenen et al. (2014).

  7. In our setup, a PAYG pension system would not provide insurance against the risk of longevity even when annuity markets are missing as long as accidental bequests are redistributed, as was shown by Caliendo et al. (2014). We therefore do not model survival risk which would, in any case, lead us on a sidetrack.

  8. In our proof of equilibrium dynamics, we require a homothetic structure. We do not get that with idiosyncratic risk in the first period and a lump-sum pension payment in the second, because the first-period wage poor save less than the first-period wage rich. This could be made homothetic by assuming that pension payments do not redistribute across types but then social security no longer insures against idiosyncratic risk.

  9. For results when households face idiosyncratic risk only in the first period of life, see Harenberg and Ludwig (2015).

  10. In the Supplementary Appendix B.1, we show that the results from Subsection 3.1 would go through in a three-generations model.

  11. The independent, mean-zero shock \(\zeta _t\) is the multiplicative background risk, because it multiplies the market risk \(\tilde{\varrho }_{t+1}\bar{R}\). The independent, mean-zero shock \(\zeta _{t+1} \eta _{i,2,t+1}\) is the additive background risk.

  12. We label the product of any two variables \(x_1\) and \(x_2\) an “interaction term” as it is common in statistical data analysis.

  13. See the product formula of variances derived in Goodman (1960).

  14. In a partial equilibrium model with pension income in the second period—and/or with positive second-period labor income in case \(\lambda >0\)—, the human capital wealth effect inhibits closed-form solutions for the saving rate. Our proof of equilibrium dynamics uses the fact that both the interest rate and the wage rate, on which pension payments are based, are functions of the capital stock in general equilibrium. This enables us to conveniently rewrite the discounted value of second-period labor income (=human capital) so that we can derive closed-form solutions for the saving rate and the equilibrium dynamics.

  15. It is crucial that income and substitution effects of changing interest rates offset each other.

  16. In the Supplementary Appendix B.2, we examine the special case of \(\lambda =0\), which yields concise equations without the need for an additional assumption.

  17. In our proof of Proposition 6 we also characterize a lower bound on the semi-elasticity such that the derivative is negative.

  18. Indeed, from a quantitative perspective, it may be necessary to model both risks to find net welfare gains in general equilibrium, as we do in our quantitative analysis in Harenberg and Ludwig (2015).

  19. Recall from Proposition 4 that the aggregate productivity shocks drop out in general equilibrium. Therefore, \(\sqrt{AR}=\sigma _\varrho \).

  20. We use that, by the familiar envelope condition, \({\mathbb {E}} \left[ \frac{\partial s}{\partial \tau } \left( \frac{\partial u(c_{1,t})}{\partial c_{1,t}} \frac{\partial c_{1,t}}{\partial s} + \beta {\mathbb {E}}_t \left[ \frac{\partial u(c_{2,t+1})}{\partial c_{2,t+1}} \frac{\partial c_{2,t+1}}{\partial s} \right] \right) \right] = 0\) because \(\frac{\partial c_{1,t}}{\partial s}=-1\) and \(\frac{\partial c_{2,t+1}}{\partial s}=R_{t+1}\), hence the term in brackets is just the Euler equation.


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Daniel Harenberg gratefully acknowledges financial support of the Sonderforschungsbereich SFB 884 (German National Research Foundation), and of Swiss Re. Alex Ludwig gratefully acknowledges research support from the Research Center SAFE, funded by the State of Hessen initiative for research LOEWE and financial support by the German National Research Foundation under SPP 1578.

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Correspondence to Daniel Harenberg.

Additional information

This is a companion paper to our quantitative work in Harenberg and Ludwig (2015). The authors wish to thank the editor and two anonymous referees for helpful comments.

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Supplementary material 1 (pdf 99 KB)

Appendix: proofs

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. 1.

    Recursive substitution of Eq. (9) gives

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

    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. 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:

    $$\begin{aligned} c_{1,t}&= \left( 1-s\right) \left( 1-\tau \right) \varUpsilon _t \zeta _t \left( 1-\alpha \right) k_{ms}^\alpha d\left( \zeta ,t-1\right) ^\alpha \equiv c_{1,t}\left( \tau ,k_{ms},s\right) \end{aligned}$$
    $$\begin{aligned} c_{i,2,t+1}&= \left( s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} + \lambda \eta _{i,2,t+1} (1+g) + \tau \left( (1+g)(1+\lambda (1-\eta _{i,2,t+1}))\right. \right. \nonumber \\&\qquad \left. \left. - s \varrho _{t+1} \alpha k_{ms}^{\alpha -1} \right) \right) \cdot \varUpsilon _t (1-\alpha )k_{ms}^\alpha \zeta _t \zeta _{t+1} d(\zeta ,t-1)^{\alpha } d(\zeta ,t)^{\alpha -1}\nonumber \\&\equiv c_{2,t+1}(\tau ,k_{ms},s). \end{aligned}$$
  3. 3.

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

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

    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 conditionFootnote 20

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

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

  4. 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

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

    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

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

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. 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. 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 \(\frac{\partial \left. \epsilon _{s,\tau }\right| _{\tau =0}}{\partial \sigma _\eta ^2}\lesseqgtr 0\):

    1. (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

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

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

    2. (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,

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

      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)— \( \left. \epsilon _{s,\tau }\right| _{\tau =0} > - \frac{3\,a\,\left( \sigma _{\varrho }^2+1\right) ^3-b+2}{2 (1+\beta \varPhi ) (1+b)}. \) \(\square \)

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Harenberg, D., Ludwig, A. Social security in an analytically tractable overlapping generations model with aggregate and idiosyncratic risks. Int Tax Public Finance 22, 579–603 (2015).

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  • Social security
  • Idiosyncratic risk
  • Aggregate risk
  • Welfare
  • Insurance

JEL Classification

  • C68
  • E27
  • E62
  • G12
  • H55