Welfare implications of naive and sophisticated saving

Abstract

Agents with declining discount rates who are unable to commit to their future decisions can either be sophisticated—meaning that they anticipate their future behavior and take it into account in their consumption choices—or naive. Previous studies have shown that sophistication may lead to higher or lower consumption rates, but have not resolved the implications for welfare, arguably the most important question from an economic point of view. Since neither the sophisticated nor the naive solutions are Pareto-efficient, their own welfare ranking is not obvious. This paper shows that the ‘better saver’ amongst the two (lower consumption rate) is always better-off, from all temporal perspectives.

Proofs

Derivation of equation (14)

Let $c_{\gamma }(K_0,t)$ be the consumption plan resulting from applying the CRC $\gamma$ and starting with a stock $K_0$. Let $\epsilon$ be a small time interval, and separate the contribution to present discounted utility into

$$\begin{aligned} W(\gamma ,K)&\approx u(c_0)\epsilon +{\displaystyle \int \nolimits _{\epsilon }^{\infty }} u(c_{\gamma }(K_0,\tau ))D(\tau )d\tau \\&\approx u(c_0)\epsilon +{\displaystyle \int \nolimits _{\epsilon }^{\infty }} u(c_{\gamma }(K_0-c_0\epsilon ,\tau -\epsilon ))D(\tau -\epsilon )d\tau \\&\quad -\epsilon {\displaystyle \int \nolimits _{\epsilon }^{\infty }}u(c_{\gamma } (K_0-c_0\epsilon ,\tau -\epsilon ))d(\tau ))d\tau \\&\approx u(c_0)\epsilon + W(\gamma ,K-c_0\epsilon ) + O(\epsilon ) + O(\epsilon ^2) \end{aligned}$$

where $O(\epsilon )$ is independent of $c_0$. Expanding gives

$$\begin{aligned} W(\gamma ,K) = u(c_0)\epsilon + W(\gamma ,K) - \frac{\partial W}{\partial K} c_o\epsilon + O(\epsilon ) + O(\epsilon ^2) \end{aligned}$$

which upon re-arrangement results in

$$\begin{aligned} \frac{\partial W}{\partial K} c_o\epsilon = u(c_0)\epsilon + O(\epsilon ) + O(\epsilon ^2) \end{aligned}$$

Differentiating w.r.t. $c_0$ and noting that $u'(c)=c^{-\theta }$ and that $c_0=\gamma K$ gives

$$\begin{aligned} \frac{\partial W}{\partial K} \epsilon= & {} u'(c_0)\epsilon + O(\epsilon ^2) \\= & {} \gamma ^{-\theta } K^{-\theta } \epsilon + O(\epsilon ^2) \end{aligned}$$

Taking the limit as $\epsilon \rightarrow 0$ derives Eq. (14).

The following Lemma will be useful in the proofs that follow. It does not appear in the main text.

Lemma 2

Let f(t) be a non-negative function integrable on $[0,\infty )$ and g(t) be a non-negative strictly decreasing function on $[0,\infty )$. Then

$$\begin{aligned} \frac{\int _0^{\infty } f(t)g(t)tdt}{\int _0^{\infty }f(t)g(t)dt} < \frac{\int _0^{\infty } f(t)tdt}{\int _0^{\infty }f(t)dt}. \end{aligned}$$

(18)

Proof

Define the following probability densities on $[0,\infty )$:

$$\begin{aligned} \psi _f(t) = \frac{f(t)}{\int _0^{\infty }f(t)dt}, \psi _{fg}(t) = \frac{f(t)g(t)}{\int _0^{\infty }f(t)g(t)dt}. \end{aligned}$$

(19)

Since g(t) is decreasing, the density $\psi _f$ strictly Likelihood Ratio Dominates$\psi _{fg}$:

$$\begin{aligned} \frac{\psi _f(t')}{\psi _f(t)} > \frac{\psi _{fg}(t')}{\psi _{fg}(t)} , \forall t<t'. \end{aligned}$$

(20)

A well known result (Hickman 2009) implies that $\psi _f$ also strictly First Order Stochastic Dominates$\psi _{fg}$. In particular, it follows that ${\mathbb {E}}_{\psi _{fg}}(t) < {\mathbb {E}}_{\psi _{f}}(t)$, which is the desired result. $\square$

Proof of Proposition 1

Noting that $c(t)=\gamma K(t)$ and that $\frac{dK}{dt}=\alpha K-c(t) = (\alpha -\gamma )K(t)$ and integrating w.r.t to time gives $K(t)=K(0)e^{(\alpha -\gamma )t}$. Substituting into the definition of $W(K,\gamma )$, it is straightforward to calculate that

$$\begin{aligned} W(K,\gamma ) = K^{1-\theta } \frac{\gamma ^{1-\theta }}{1-\theta } \int _0^\infty e^{(\alpha -\gamma ) (1-\theta ) \tau } D(\tau ) d\tau . \end{aligned}$$

(21)

Differentiating w.r.t. to K gives

$$\begin{aligned} \frac{\partial W}{\partial K} = K^{-\theta } {\gamma ^{1-\theta }} \int _0^\infty e^{(\alpha -\gamma ) (1-\theta ) \tau } D(\tau ) d\tau . \end{aligned}$$

(22)

Proposition 1 therefore states that there is a unique solution $\gamma \ge \alpha$ to the equation

$$\begin{aligned} K^{-\theta } {\gamma ^{-\theta }} = K^{-\theta } {\gamma ^{1-\theta }} \int _0^\infty e^{(\alpha -\gamma ) (1-\theta ) \tau } D(\tau ) d\tau . \end{aligned}$$

(23)

or

$$\begin{aligned} 1=F(\gamma ) \equiv \gamma {\displaystyle \int \nolimits _{0}^{\infty }} D(\tau ) e^{(\alpha -\gamma )(1-\theta )\tau } d\tau \end{aligned}$$

(24)

Existence: Note that $F(\gamma )$ is continuous. Suppose $\theta >1$. When $\gamma \rightarrow 0$, $F(\gamma ) \rightarrow 0$ (recall that $\alpha \le \delta _0$, so the integral converges when $\gamma =0$). When $\gamma \rightarrow \alpha - \delta _0/(1-\theta )$ the integral diverges. Thus, because of the continuity of $F(\gamma )$ there is an intermediate value of $\gamma$ for which $F(\gamma )=1$. Suppose now that $\theta <1$. Again, when $\gamma \rightarrow 0$, $F(\gamma ) \rightarrow 0$. When $\gamma \rightarrow \infty$, note that $F(\gamma ) \ge \gamma \int _0^\infty e^{-d(0)t} e^{(\alpha -\gamma )(1-\theta )t} dt = \frac{\gamma }{d(0)+(\gamma -\alpha )(1-\theta )} \rightarrow \frac{1}{1-\theta } > 1$. Again, there is an intermediary value of $\gamma$ for which $F(\gamma )=1$.

Uniqueness: Consider now the derivative of $F(\gamma )$:

$$\begin{aligned} F'(\gamma ) = \int _{0}^{\infty }D(\tau ) e^{(\alpha -\gamma )(1-\theta )\tau } d\tau - \gamma (1-\theta ) \int _{0}^{\infty } D(\tau ) e^{(\alpha -\gamma )(1-\theta )\tau } \tau d\tau . \end{aligned}$$

(25)

Uniqueness will follow from showing that $F'(\gamma ) \ge 0$. When $\theta \ge 1$ this is obvious. When $\theta <1$, we need to show that

$$\begin{aligned} \frac{ \int _{0}^{\infty } D(\tau ) e^{(\alpha -\gamma )(1-\theta )\tau } \tau d\tau }{\int _{0}^{\infty }D(\tau ) e^{(\alpha -\gamma )(1-\theta )\tau } d\tau } \le \frac{1}{\gamma (1-\theta )}. \end{aligned}$$

(26)

Decomposing the discount factor as $D(t) = v(t) e^{-\delta _0 t}$ where v(t) is decreasing, we can rewrite the LHS of (26) as $\frac{ \int _{0}^{\infty } v(\tau ) e^{-x \tau } \tau d\tau }{\int _{0}^{\infty } v(\tau ) e^{-x \tau } d\tau }$ where $x=\delta _0+(\gamma -\alpha )(1-\theta )$. Applying Lemma 2 with $f(t)=e^{-xt}$, $g(t)=v(t)$, we get

$$\begin{aligned} \frac{ \int _{0}^{\infty } v(\tau ) e^{-x \tau } \tau d\tau }{\int _{0}^{\infty } v(\tau ) e^{-x \tau } d\tau } < \frac{\int _0^{\infty } t e^{-xt}dt}{\int _0^{\infty } e^{-xt}dt }= \frac{1}{x} \equiv \frac{1}{\delta _0+(\gamma -\alpha )(1-\theta )} \le \frac{1}{\gamma (1-\theta )} \end{aligned}$$

(27)

where the last inequality follows from the assumption that $\alpha \le \delta _0$. $\square$

Proof of Proposition 2

The proof uses

Holder’s Inequality: Let $0 \le \nu ,\mu \le 1$, $\nu +\mu =1$. Let f(x) and g(x) be two non-negative integrable on $[0,\infty )$. Then:

$$\begin{aligned} {\displaystyle \int \nolimits _{0}^{\infty }} f(x)^{\mu } g(x)^{\nu } dx \le \left( {\displaystyle \int \nolimits _{0}^{\infty }} f(x)dx\right) ^{\mu } \left( {\displaystyle \int \nolimits _{0}^{\infty }} g(x)dx\right) ^{\nu } \end{aligned}$$

(28)

with equality occurring only if f(x) and g(x) are proportional.

To begin the proof, suppose first that $\theta <1$. Rewrite (24) as

$$\begin{aligned} 1= \gamma _S {\displaystyle \int \nolimits _{0}^{\infty }} \left( D(\tau )^{\frac{1}{\theta }} e^{(\alpha )(\frac{1}{\theta }-1)\tau } \right) ^{\theta } \left( e^{-\gamma _S\tau } \right) ^{1-\theta } d\tau \end{aligned}$$

(29)

and apply Holder’s inequality to obtain (note that the inequality is strict since the two integrands cannot be proportional to each other if the discount factor is strictly hyperbolic, as we assume throughout)

$$\begin{aligned} 1 < \gamma _S \left( \displaystyle \int \nolimits _{0}^{\infty } D(\tau )^{\frac{1}{\theta }}e^{(\alpha )(\frac{1}{\theta }-1)\tau } d\tau \right) ^{\theta } \left( \displaystyle \int \nolimits _{0}^{\infty } e^{-\gamma _S\tau }d\tau \right) ^{1-\theta }, \end{aligned}$$

(30)

or

$$\begin{aligned} 1 < \gamma ^{S} \left( \gamma ^{N}\right) ^{-\theta } \left( \frac{1}{\gamma ^{S}}\right) ^{1-\theta } = \left( \frac{\gamma ^{S}}{\gamma ^{N}}\right) ^{\theta }, \end{aligned}$$

(31)

which completes the proof.

Now suppose that $\theta >1$. Rewrite (10) as (the $\gamma$ exponentials cancel out)

$$\begin{aligned} \frac{1}{\gamma _N} = {\displaystyle \int \nolimits _{0}^{\infty }}(D(\tau )e^{(\alpha -\gamma _S)(1-\theta )\tau }) ^{\frac{1}{\theta }} (e^{\gamma _S\tau })^{(1-\frac{1}{\theta })} d\tau . \end{aligned}$$

(32)

Apply Holder’s inequality to to obtain

$$\begin{aligned} \frac{1}{\gamma _N}< & {} \left( {\displaystyle \int \nolimits _{0}^{\infty }}D(\tau )e^{(\alpha -\gamma _S)(1-\theta )\tau }d\tau \right) ^{\frac{1}{\theta }} \left( {\displaystyle \int \nolimits _{0}^{\infty }}e^{-\gamma _S\tau }d\tau \right) ^{(1-\frac{1}{\theta })} \nonumber \\= & {} (\gamma _S)^{-\frac{1}{\theta }} (\gamma _S)^{\frac{1}{\theta }-1}, \end{aligned}$$

(33)

from which it follows that $\gamma _N \ge \gamma _S$. Note that when $\theta =1$, it can be directly seen in (10) and (24) that $\gamma _N = \gamma _S = \frac{1}{{\displaystyle \int \nolimits _{0}^{\infty }}D(\tau )d\tau }.$$\square$

The following two Lemmas will be useful in the proofs that follow. It does not appear in the main text.

Lemma 3

Let H(t) be a (strictly) Hyperbolic discount factor (as defined in (3)). Then

$$\begin{aligned} \int _0^{\infty } H(t)t dt > \left( \int _0^{\infty }H(t)dt \right) ^2. \end{aligned}$$

(34)

Proof

$$\begin{aligned} \left( \int _0^{\infty }H(t)dt \right) ^2 = \int _0^{\infty }\int _0^{\infty } dtds H(t)H(s) < \int _0^{\infty }\int _0^{\infty } dtds H(t+s). \end{aligned}$$

(35)

Changing the integration variables to $x=t+s, y=t-s$, for which the Jacobian is $-1/2$, we see that the (RHS) equals $\frac{1}{2} \int _0^{\infty }dx H(x) \int _{-x}^{x} dy = \int _0^{\infty }H(x)xdx$. $\square$

Lemma 4

Let H(t) be a hyperbolic discount factor (as defined in (3)) and $x > 0$.

$$\begin{aligned} \frac{1}{\int _0^\infty H(t) e^{-xt}dt } < \frac{1}{\int _0^\infty H(t)dt } + x. \end{aligned}$$

(36)

Proof

Again, note that if H(t) is a Hyperbolic discount factor, then so is $H(t)e^{-xt}$ for any $x\ge 0$. It follows from Lemma 3 that

$$\begin{aligned} \int _0^{\infty }H(t)e^{-xt}dt > \left( \int _0^{\infty }H(t)dt\right) ^2, \end{aligned}$$

(37)

which ensures the derivate of the RHS is always larger than that of the LHS. The result follows from the equality of the LHS and RHS at $x=0$. $\square$

Proof of Proposition 3

(a) Suppose first that $\theta <1$. As $\gamma \rightarrow 0$, $G(\gamma ) \rightarrow 0$. When $\gamma \rightarrow \infty$ the only contribution to the integral in (17) comes from near $t=0$, so we can take $D(t) \approx 1$, and

$$\begin{aligned} G(\gamma ) \approx \frac{\gamma ^{1-\theta }}{1-\theta } \int _0^\infty e^{(\alpha -\gamma )(1-\theta )t} dt = \frac{\gamma ^{1-\theta }}{1-\theta } \frac{1}{(\gamma -\alpha )(1-\theta )} \approx \gamma ^{-\theta } \rightarrow 0. \end{aligned}$$

(38)

Thus, $G(0)=G(\infty )=0$, and since G is non-negative, we deduce the existence of an interior maximum in $(0,\infty )$. Suppose now that $\theta >1$. In this case, $G(\gamma ) \rightarrow -\infty$ when either $\gamma \rightarrow 0$ or $\gamma \rightarrow \infty$ so there is again an interior maximum.

(b) Differentiating (17), we see that every interior maximum $\gamma _M$ satisfies the F.O.C.

$$\begin{aligned} \frac{1}{\gamma _{M}} = \frac{\int _0^{\infty } D(t) e^{(\alpha -\gamma _{M})(1-\theta )t} t dt}{\int _0^{\infty } D(t) e^{(\alpha -\gamma _{M})(1-\theta )t} dt } \equiv F_1(\gamma _{M}). \end{aligned}$$

(39)

Recall that $\gamma _S$ satisfies (24)

$$\begin{aligned} \frac{1}{\gamma _S}={\displaystyle \int \nolimits _{0}^{\infty }} D(\tau ) e^{(\alpha -\gamma _S)(1-\theta )\tau } d\tau \equiv F_2(\gamma ^{S}). \end{aligned}$$

(40)

Note that if H(t) is a Hyperbolic discount factor, then so is $H(t)e^{-xt}$ for any $x\ge -\delta _0$. Thus, Lemma 3 can be applied to $\phi (\tau )=D(\tau ) \gamma e^{(\alpha -\gamma )(1-\theta )\tau }$, and it follows that $F_1(\gamma ) > F_2(\gamma )$ for all $\gamma$. Since $\frac{1}{\gamma }$ intersects $F_1 (\gamma )$ at $\gamma _{M}$ and $F_2 (\gamma )$ at $\gamma ^{S}$ it follows that $\gamma _{M} < \gamma _S$.

(c) Again, let $\gamma _M$ be an interior maximum. When $\theta >1$, the claim follows immediately from Proposition 2 and part (b) of this proposition, proved above: $\gamma _N>\gamma _S>\gamma _{M}$. Assume then that $\theta <1$. Using (39) and Lemmas 3 and 4, applied to $x=(\alpha -\gamma _{M}) (1-\theta )$, we have:

$$\begin{aligned} \gamma _{M}= & {} \frac{\int _0^{\infty } D(t) e^{(\alpha -\gamma _{M})(1-\theta )t} dt }{\int _0^{\infty } D(t) e^{(\alpha -\gamma _{M})(1-\theta )t} t dt} \end{aligned}$$

(41)

$$\begin{aligned}< & {} \frac{1}{\int _0^{\infty } D(t) e^{(\alpha -\gamma _{M})(1-\theta )t} dt } < \frac{1}{\int _0^{\infty } D(t)e^{\alpha (1-\theta )t}dt} + \gamma _{M}(1-\theta ). \end{aligned}$$

(42)

Also, from ‘hyperbolicity’, $D(t)>D(\theta t)^{\frac{1}{\theta }}$ so that

$$\begin{aligned} \gamma _{M}< & {} \frac{1}{\int _0^{\infty } D(\theta t)^{\frac{1}{\theta }}e^{\alpha (1-\theta )t}dt} + \gamma _{M}(1-\theta ) \end{aligned}$$

(43)

$$\begin{aligned}= & {} \frac{\theta }{\int _0^{\infty } D(\theta t)^{\frac{1}{\theta }}e^{\alpha (\frac{1}{\theta }-1)\theta t}d(\theta t)} + \gamma _{M}(1-\theta ) = \theta \gamma _N + (1-\theta ) \gamma _{M}. \end{aligned}$$

(44)

It follows that $\gamma _{M}<\gamma _N$. This concludes the proof. $\square$