A.1 Proof of Proposition 1

Consider problem (1). Let $\nu _{t}^{i}$ and $\mu _{t}^{i}$ be the Lagrange multipliers of the budget constraint and non-negativity bequest constraint, respectively. Maximizing the Lagrangian function of the Kuhn–Tucker program

\begin{equation*} \ln c_{t}^{i}+\beta _{i}\ln \left ( R_{t+1}b_{t+1}^{i}+w_{t+1}\right ) +\nu _{t}^{i}\left ( R_{t}b_{t}^{i}+w_{t}-c_{t}^{i}-b_{t+1}^{i}\right ) +\mu _{t}^{i}b_{t+1}^{i}\, \end{equation*}

we get a system of first-order conditions:

\begin{equation*} \frac {1}{c_{t}^{i}}=\nu _{t}^{i}=\frac {\beta _{i}R_{t+1}}{R_{t+1}b_{t+1}^{i}+w_{t+1}}+\mu _{t}^{i} \end{equation*}

jointly with $\nu _{t}^{i}\geq 0$ , $R_{t}b_{t}^{i}+w_{t}-c_{t}^{i}-b_{t+1}^{i}\geq 0$ , $\nu _{t}^{i}\left ( R_{t}b_{t}^{i}+w_{t}-c_{t}^{i}-b_{t+1}^{i}\right ) =0$ , and $\mu _{t}^{i}\geq 0$ , $b_{t+1}^{i}\geq 0$ , $\mu _{t}^{i}b_{t+1}^{i}=0$ . Since $\nu _{t}^{i}=1/c_{t}^{i}\gt 0$ , a non-negative pair $\left ( c_{t}^{i},b_{t+1}^{i}\right )$ is a solution to (1) if and only if there exists $\mu _{t}^{i}\geq 0$ such that $\mu _{t}^{i}b_{t+1}^{i}=0$ , together with

\begin{equation*} \frac {1}{c_{t}^{i}} =\frac {\beta _{i}R_{t+1}}{R_{t+1}b_{t+1}^{i}+w_{t+1}}+\mu _{t}^{i} \qquad \text {and} \qquad c_{t}^{i}+b_{t+1}^{i} =R_{t}b_{t}^{i}+w_{t} \end{equation*}

The reduced utility function $v\left ( c_{t}^{i}\right ) \equiv \ln c_{t}^{i}+\beta _{i}\ln \left [ R_{t+1}\left ( R_{t}b_{t}^{i}+w_{t}\,-c_{t}^{i}\right ) +w_{t+1}\right ]$ is strictly concave: $v^{\prime \prime }\left ( c_{t}^{i}\right ) =-(c_{t}^{i})^{-2}-\beta _{i}\left ( b_{t+1}^{i}+w_{t+1}/R_{t+1}\right ) ^{-2}\lt 0$ . Then the first-order conditions are necessary and sufficient to utility maximization.

Now, take into account equations (2)–(3) where $k_{t}=\pi _{L}b_{t}^{L}+\pi _{H}b_{t}^{H}$ . If $\mu _{t}^{i}\gt 0$ , then $b_{t+1}^{i}=0$ . If $b_{t+1}^{i}\gt 0$ , then $\mu _{t}^{i}=0$ , so that

\begin{equation*} \frac {\beta _{i}R_{t+1}}{R_{t+1}b_{t+1}^{i}+w_{t+1}}=\frac {1}{c_{t}^{i}}=\frac {1}{R_{t}b_{t}^{i}+w_{t}-b_{t+1}^{i}} \end{equation*}

and the dynamics of bequests are given by

\begin{equation*} b_{t+1}^{i} = \frac {1}{1+\beta _{i}}\left ( \beta _{i}R\left ( k_{t}\right ) \left [ b_{t}^{i}+k_{t}\gamma \left ( k_{t}\right ) \right ] -k_{t+1}\gamma \left ( k_{t+1}\right ) \right ) \end{equation*}

A.2 Proof of Proposition 2

We observe that at the steady state,

\begin{equation*} \left ( 1+\beta _{i}\right ) b^{i}=\max \left \{ 0,\beta _{i}R\left ( k\right ) b^{i}+\left [ \beta _{i}R\left ( k\right ) -1\right ] k\gamma \left ( k\right ) \right \} \end{equation*}

with $i=L, H$ and $k=\pi _{L}b^{L}+\pi _{H}b^{H}$ . In order to have positive bequests for type $i$ , we need (6). In order to have positive bequests for both types, we need

\begin{equation*} \frac {1}{\beta _{L}} \lt R\left ( k\right ) \lt 1+\frac {1}{\beta _{H}} \end{equation*}

Then the steady-state capital stock is a solution to the following equation in $k$ :

\begin{equation*} k=\sum \nolimits _{i}\pi _{i}b^{i}=k\gamma \left ( k\right ) \sum \nolimits _{i}\pi _{i}\max \left \{ 0,\frac {\beta _{i}R\left ( k\right ) -1}{1+\beta _{i}-\beta _{i}R\left ( k\right ) }\right \} \end{equation*}

A.3 Proof of Proposition 3

Let $R\left ( k\right ) =R$ be the steady-state interest rate and $b^{i} \geq 0$ be the steady-state bequests for $i=L,H$ , which are solutions to the following equation:

(A1) \begin{equation} \left ( 1+\beta _{i}\right ) b^{i}=\max \left \{ 0,\beta _{i}Rb^{i}+\left ( \beta _{i}R-1\right ) \gamma k\right \} \end{equation}

If $R\leq 1/\beta _{i}$ , then equation (A1) has a unique solution: $b^{i}=0$ .

If $R\gt 1/\beta _{i}$ , then $b^{i}$ is positive, and we have

(A2) \begin{equation} b^{i}=\gamma k\frac {\beta _{i}R-1}{1+\beta _{i}-\beta _{i}R} \end{equation}

Therefore, if $1/\beta _{i}\lt R\lt 1+1/\beta _{i}$ , equation (A1) has a unique solution given by (A2). If $R\geq 1+1/\beta _{i}$ , equation (A1) has no solutions.

For $1/\beta _{H}\lt R\lt 1+1/\beta _{H}$ , according to (5), we have

\begin{equation*} \frac {b^{i}}{k}=\gamma \max \left \{ 0,\frac {\beta _{i}R-1}{1+\beta _{i}-\beta _{i}R}\right \} \end{equation*}

Observing that $\pi _{L}b^{L}/k+\pi _{H}b^{H}/k=1$ , we have that the steady-state interest rate $R^{*}$ is a solution to the following equation in $R$ :

\begin{equation*} \rho \left ( R\right ) \equiv \pi _{L}\max \left \{ 0,\frac {\beta _{L}R-1}{1+\beta _{L}-\beta _{L}R}\right \} +\pi _{H}\max \left \{ 0,\frac {\beta _{H}R-1}{1+\beta _{H}-\beta _{H}R}\right \} =\frac {1}{\gamma } \end{equation*}

Note that $\rho \left ( R\right )$ is a continuous function which increases with $R$ , and $\rho \left ( 1/\beta _{H}\right ) =0$ and $\lim _{R \rightarrow \left ( 1+1/\beta _{H}\right )} \rho \left ( R\right ) = \infty$ . Then there exists a solution $R^{*}$ to the equation $\rho \left ( R \right ) =1/\gamma$ , and this solution is such that $1/\beta _{H}\lt R^{*}\lt 1+1/\beta _{H}$ . Three cases are possible: (1) $1/\beta _{L}\lt R^{*}\lt 1+1/\beta _{H}$ ; (2) $1/\beta _{H}\lt R^{*}\leq 1/\beta _{L} \lt 1+1/\beta _{H}$ ; and (3) $1/\beta _{H}\lt R^{*}\lt 1+1/\beta _{H} \leq 1/\beta _{L}$ .

Consider first Cases (2) and (3). Let $\pi \equiv \pi _{H}$ . Since in both cases $R \leq 1/\beta _{L}$ , we have

\begin{equation*} \rho \left ( R\right ) =\pi \frac {\beta _{H}R-1}{1+\beta _{H}-\beta _{H}R}\equiv \rho _{2}\left ( R\right ) \end{equation*}

The solution to the equation $\rho _{2}\left ( R\right ) =1/\gamma$ is given by

\begin{equation*} R^{*} = \frac {1}{\beta _{H}}+\frac {1}{1+\gamma \pi } \equiv R_{2}^{*} \end{equation*}

This solution is the steady-state interest rate if and only if $R_{2}^{*} \leq 1/\beta _{L}$ , which is equivalent to $\delta \geq \overline {\delta }$ . Since $f^{\prime }\left ( k \right ) = R$ , the capital stock corresponding to this steady state is $k^{*}_{2} \equiv \left ( \alpha A / R_{2}^{*} \right )^{1/(1-\alpha )}$ . In this case, we also have $b_{2}^{L*}=0$ and $b_{2}^{H*}=k^{*}_{2}/\pi$ . Further,

\begin{equation*} c_{2}^{L*}=\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }=k^{*}_{2}\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha -1}=\gamma k^{*}_{2} R_{2}^{*} \end{equation*}

while

\begin{equation*} c_{2}^{H*}=\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }+\frac {\left ( R_{2}^{*}-1\right ) k^{*}_{2}}{\pi }=\frac {k^{*}_{2}}{\pi }\left ( \gamma \pi R_{2}^{*}+R_{2}^{*}-1\right ) =\frac {k^{*}_{2}}{\pi }\frac {1+\gamma \pi }{\beta _{H}} \end{equation*}

When $\rho \left ( 1/\beta _{L} \right ) \lt 1/\gamma$ , which is equivalent to $\delta \lt \overline {\delta }$ , we are in the conditions of Case (1). We have

\begin{equation*} \rho \left ( R \right ) =\left ( 1-\pi \right ) \frac {\beta _{L}R-1}{1+\beta _{L}-\beta _{L}R}+\pi \frac {\beta _{H}R-1}{1+\beta _{H}-\beta _{H}R}\equiv \rho _{1}\left ( R\right ) \end{equation*}

The steady-state interest rate $R^{*}$ is a solution to the equation $\rho _{1}\left ( R \right )=1/\gamma$ , which can be written as

\begin{equation*} \left ( 1-\pi \right ) \frac {\beta _{L}R-1}{1+\beta _{L}-\beta _{L}R}+\pi \frac {\beta _{H}R-1}{1+\beta _{H}-\beta _{H}R}=\frac {1}{\gamma } \end{equation*}

or, equivalently, as

\begin{eqnarray*} &&\beta _{L}\beta _{H}\left ( 1+\gamma \right ) R^{2} -\left [ \left ( 1+\gamma \right ) \left ( \beta _{L}+\beta _{H}\right ) +\left ( 2+\gamma \right ) \beta _{L}\beta _{H}\right ] R \\ &&+\left ( 1+\gamma +\beta _{L}+\beta _{H}+\beta _{L}\beta _{H} + \gamma \left [ \left ( 1-\pi \right ) \beta _{H}+\pi \beta _{L} \right ] \right ) = 0 \end{eqnarray*}

Noticing that $\gamma =\left ( 1-\alpha \right ) /\alpha$ , we obtain

\begin{equation*} R_{1}^{\pm }\ =\ \frac {1}{2}\,\left ( 1+\alpha +\frac {1}{\beta _{L}}+\frac {1}{\beta _{H}}\pm \sqrt {\left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right ) }\right ) \end{equation*}

Let $D\left ( \pi \right ) \equiv \left ( \delta +\alpha -1\right ) ^{2} +4\delta \pi \left ( 1-\alpha \right )$ . It is easily seen that $D(\pi )$ increases with $\pi$ , and

(A3) \begin{equation} D\left ( 0\right ) = \left ( \delta +\alpha -1\right )^{2} \,, \qquad D\left ( \overline {\pi }\right ) = \left ( 1-\delta +\alpha \right )^{2} \,, \qquad D\left ( 1\right ) = \left ( \delta +1-\alpha \right )^{2} \end{equation}

where $\overline {\pi }$ is given by (17).

Since for all $\pi$ , $D\left ( \pi \right ) \geq \left ( \delta +\alpha -1 \right ) ^{2}$ , it follows that

\begin{equation*} R_{1}^{-} \ \leq \ 1+\frac {1}{\beta _{H}} \ \leq R_{1}^{+} \end{equation*}

Denote $R_{1}^{*} \equiv R_{1}^{-}$ . Then the corresponding steady-state capital stock is $k^{*}_{1} = \left ( \alpha A / R_{1}^{*} \right )^{1/(1-\alpha )}$ . Since $1/\beta _{L}\lt R_{1}^{*}\lt 1+1/\beta _{H}$ , at the steady state we have for $i=L,H$ ,

\begin{equation*} b_{1}^{i*}=\gamma k^{*}_{1}\frac {\beta _{i} R_{1}^{*}-1}{1+\beta _{i}-\beta _{i} R_{1}^{*}} \end{equation*}

and

\begin{eqnarray*} c_{1}^{i*} &=&\left ( 1-\alpha \right ) A(k^{*}_{1})^{\alpha }+\left ( R_{1}^{*}-1\right ) b_{1}^{i*}=\gamma k^{*}_{1}\alpha A(k^{*}_{1})^{\alpha -1}+\gamma k^{*}_{1}\frac {\left ( R_{1}^{*}-1\right ) \left ( \beta _{i}R_{1}^{*}-1\right ) }{\beta _{i}+1-\beta _{i}R_{1}^{*}} \\ &=&\gamma k^{*}_{1}\left [ R_{1}^{*}+\frac {\left ( R_{1}^{*}-1\right ) \left ( \beta _{i}R_{1}^{*}-1\right ) }{\beta _{i}+1-\beta _{i}R_{1}^{*}}\right ] =\frac {\gamma k^{*}_{1}}{\beta _{i}+1-\beta _{i}R_{1}^{*}} \end{eqnarray*}

A.4 Proof of Proposition 4

(0) Consider a sequence $\left ( b_{t}^{L},b_{t}^{H},k_{t}\right ) _{t=0}^{\infty }$ . Fix $t$ . Let $\lambda _{t}^{i} \equiv b_{t}^{i}/k_{t}$ . Let $\lambda _{t+1}^{L},\lambda _{t+1}^{H}$ be such that for $i=L,H$ ,

\begin{equation*} \lambda _{t+1}^{i}k_{t+1}=\frac {\beta _{i}R\left ( k_{t}\right ) k_{t}\left ( \lambda _{t}^{i}+\gamma \right ) -\gamma k_{t+1}}{1+\beta _{i}} \end{equation*}

We observe that, at this stage of the proof, $\lambda _{t+1}^{i}$ may lie outside of the interval $\left ( 0,1/\pi _{i}\right )$ . We have

\begin{equation*} \left ( \lambda _{t+1}^{i}+\frac {\gamma }{1+\beta _{i}}\right ) k_{t+1}=\frac {\beta _{i}R\left ( k_{t}\right ) k_{t}\left ( \lambda _{t}^{i}+\gamma \right ) }{1+\beta _{i}} \end{equation*}

and, therefore,

\begin{equation*} \frac {\lambda _{t+1}^{H}+\frac {\gamma }{1+\beta _{H}}}{\lambda _{t+1}^{L}+\frac {\gamma }{1+\beta _{L}}}=\frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}}\frac {\lambda _{t}^{H}+\gamma }{\lambda _{t}^{L}+\gamma } \end{equation*}

Since we need $\pi _{L}\lambda _{t}^{L}+\pi _{H}\lambda _{t}^{H}=1$ , we study the following equation

(A4) \begin{equation} \zeta \left ( \lambda _{t+1}^{H}\right ) \equiv \frac {\lambda _{t+1}^{H}+\frac {\gamma }{1+\beta _{H}}}{\frac {1-\pi _{H}\lambda _{t+1}^{H}}{\pi _{L}}+\frac {\gamma }{1+\beta _{L}}}=\frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}}\frac {\lambda _{t}^{H}+\gamma }{\frac {1-\pi _{H}\lambda _{t}^{H}}{\pi _{L}}+\gamma }\equiv \xi \left ( \lambda _{t}^{H}\right ) \end{equation}

Functions $\zeta$ and $\xi$ are increasing in the interval $\left ( 0,1/\pi _{H}\right )$ .

Notice that, given $\lambda _{t}^{H}$ , equation (A4) has a solution $\lambda _{t+1}^{H}\in \left ( 0,1/\pi _{H}\right )$ if and only if $b_{t+1}^{L}, b_{t+1}^{H}\gt 0$ . More precisely, let $\lambda _{t+1}^{H}$ be solution to (A4) in $\left ( 0,1/\pi _{H}\right )$ , $b_{t+1}^{i}=\lambda _{t+1}^{i}k_{t+1}$ and

\begin{equation*} \lambda _{t+1}^{L}=\frac {1}{\pi _{L}}-\frac {\pi _{H}}{\pi _{L}}\lambda _{t+1}^{H} \end{equation*}

with

\begin{equation*} k_{t+1} = \sum \nolimits _{i}\frac {\pi _{i}\left [ \beta _{i}R\left ( k_{t}\right ) \left ( b_{t}^{i}+\gamma k_{t}\right ) -\gamma k_{t+1}\right ] }{1+\beta _{i}} = k_{t}R\left ( k_{t}\right ) \sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \frac {b_{t}^{i}}{k_{t}}+\gamma \right ) }{1+\beta _{i}}-\gamma k_{t+1}\sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}} \end{equation*}

that is

(A5) \begin{equation} k_{t+1}=k_{t}R\left ( k_{t}\right ) \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda _{t}^{i}+\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}} \end{equation}

Let us prove that each solution to the equation $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ in $\left ( 0,1/\pi _{H}\right )$ corresponds to a steady state where both bequests are strictly positive. Indeed, let $\lambda ^{H*}$ be a solution to equation $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ in this interval and

\begin{equation*} \lambda ^{L*}=\frac {1}{\pi _{L}}-\frac {\pi _{H}}{\pi _{L}}\lambda ^{H*} \end{equation*}

Let the sequence $\left ( \hat {b}_{t}^{L},\hat {b}_{t}^{H},\hat {k}_{t}\right )$ be such that $\hat {b}_{0}^{L}=\lambda ^{L*}k_{0}$ and $\hat {b}_{0}^{L}=\lambda ^{H* }k_{0}$ . Let $\lambda _{t}^{i}=\hat {b}_{t}^{i}/\hat {k}_{t}$ , for every $i$ and $t \geq 0$ . By induction, we obtain $\lambda _{t}^{i}=\lambda ^{i*}$ and $\hat {b}_{t}^{i}=\lambda ^{i*}\hat {k}_{t}$ , for every $t \geq 0$ .

From (A5), we have

\begin{equation*} \left ( 1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}\right ) \hat {k}_{t+1}=\alpha f\left ( \hat {k}_{t}\right ) \sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda ^{i*}+\gamma \right ) }{1+\beta _{i}} \end{equation*}

since $\hat {k}_{t}R\left ( \hat {k}_{t}\right ) =\alpha A\hat {k}_{t}^{\alpha }=\alpha f\left ( \hat {k}_{t}\right )$ .

Therefore,

\begin{equation*} \hat {k}_{t+1}=\alpha f\left ( \hat {k}_{t}\right ) \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda ^{i*}+\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}} = \alpha S f\left ( \hat {k}_{t}\right ) \end{equation*}

where

\begin{equation*} S \equiv \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda ^{i*} +\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}} \end{equation*}

is a constant.

Since $f\left ( \hat {k}_{t}\right ) =A\hat {k}_{t}^{\alpha }$ with $\alpha \in \left ( 0,1\right )$ and $\hat {k}_{t+1}=\alpha Sf\left ( \hat {k}_{t}\right )$ with $\alpha S$ constant, it is known that $\hat {k}_{t}$ monotonically converges to some capital level $k^{*}$ .

(1) Consider part (1) of Proposition3. We want to prove that:

(1.1) For any $0\leq \lambda _{t}^{H}\leq 1/\pi _{H}$ , there exists a unique $\lambda _{t+1}^{H}\in \left ( 0,1/\pi _{H}\right )$ such that $\zeta \left ( \lambda _{t+1}^{H}\right ) =\xi \left ( \lambda _{t}^{H}\right )$ .

(1.2) There exists a unique $\lambda ^{H*} \in \left ( 0,1/\pi _{H}\right )$ which solves $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ .

(1.1) To prove the first claim, we show that $\zeta \left ( 0\right ) \lt \xi \left ( 0\right ) \leq \xi \left ( \lambda _{t}^{H}\right )$ , and $\zeta \left ( 1/\pi _{H}\right ) \gt \xi \left ( 1/\pi _{H}\right ) \geq \xi \left ( \lambda _{t}^{H}\right )$ . Indeed, inequality $\zeta \left ( 0\right ) \lt \xi \left ( 0\right )$ is equivalent to

\begin{equation*} \frac {\frac {\gamma }{1+\beta _{H}}}{\frac {1}{\pi _{L}}+\frac {\gamma }{1+\beta _{L}}}\lt \frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}}\frac {\gamma }{\frac {1}{\pi _{L}}+\gamma } \end{equation*}

that is to

\begin{equation*} 1+\gamma \pi _{L}\lt \frac {\beta _{H}}{\beta _{L}}\left ( 1+\beta _{L}+\gamma \pi _{L}\right ) \end{equation*}

which is always true since $\beta _{H}\gt \beta _{L}\gt 0$ .

The inequality $\zeta \left ( 1/\pi _{H}\right ) \gt \xi \left ( 1/\pi _{H}\right )$ is equivalent to

\begin{equation*} \frac {\frac {1}{\pi _{H}}+\frac {\gamma }{1+\beta _{H}}}{\frac {\gamma }{1+\beta _{L}}}\gt \frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}}\frac {\frac {1}{\pi _{H}}+\gamma }{\gamma } \end{equation*}

that is to

\begin{equation*} \frac {1}{1+\gamma \pi _{H}}\gt \frac {1}{\beta _{L}}-\frac {1}{\beta _{H}} \end{equation*}

which is true because, in this case, $\delta \lt \overline {\delta }$ . Therefore, $\zeta \left ( 0\right ) \lt \xi \left ( 0\right ) \leq \xi \left ( \lambda _{t}^{H}\right )$ and $\zeta \left ( 1/\pi _{H}\right ) \gt \xi \left ( 1/\pi _{H}\right ) \geq \xi \left ( \lambda _{t}^{H}\right )$ .

Hence, there exists $0\lt \lambda _{t+1}^{H}\lt 1/\pi _{H}$ such that $\zeta \left ( \lambda _{t+1}^{H}\right ) =\xi \left ( \lambda _{t}^{H}\right )$ . The strict monotonicity of function $\zeta$ ensures the uniqueness. Let $\lambda _{t+1}^{H}=\varphi \left ( \lambda _{t}^{H}\right )$ be the unique solution to $\zeta \left ( \lambda _{t+1}^{H}\right ) =\xi \left ( \lambda _{t}^{H}\right )$ . The function $\varphi$ is continuous in the interval $\left ( 0,1/\pi _{H}\right )$ and is strictly increasing, with $\varphi \left ( 0\right ) \gt 0$ and $\varphi \left ( 1/\pi _{H}\right ) \lt 1/\pi _{H}$ . This means that starting from any initial pair $\left ( b_{0}^{L},b_{0}^{H}\right )$ with at least one positive bequest, bequests $b_{t}^{L}$ and $b_{t}^{H}$ are both strictly positive for any $t\geq 1$ .

(1.2) Let us focus on the second claim, which is determinant in the proof of convergence. As a preliminary step, we observe that any solution in $\left ( 0,1/\pi _{H}\right )$ to equation $\lambda =\varphi \left ( \lambda \right )$ corresponds to a steady state, which, according to Proposition3, is unique. We obtain also the uniqueness of $\lambda ^{H*}$ , a solution to $\lambda =\varphi \left ( \lambda \right )$ .

The uniqueness of the solution ensures that we have $\varphi \left ( \lambda \right ) \gt \lambda$ on $\left ( 0,\lambda ^{H*}\right )$ and $\varphi \left ( \lambda \right ) \lt \lambda$ on $\left ( \lambda ^{H*}, 1/\pi _{H}\right )$ . Then, if $0\leq \lambda _{0}^{H}\lt \lambda ^{H*}$ , the sequence $\left ( \lambda _{t}^{H}\right )_{t=0}^{\infty }$ is increasing and converges to $\lambda ^{H*}$ , and, in the opposite case $\lambda _{0}^{H}\gt \lambda ^{H*}$ , this sequence is decreasing and converges to $\lambda ^{H*}$ . We can therefore ensure the convergence of $\left ( \lambda _{t}^{L},\lambda _{t}^{H}\right )$ to $\left ( \lambda ^{L*}, \lambda ^{H*}\right )$ .

This implies also the convergence of $\left (b_{t}^{L},b_{t}^{H}\right )$ to $\left ( b_{1}^{L*},b_{1}^{H*} \right )$ . Indeed, we observe that

\begin{equation*} k_{t+1}=k_{t}R\left ( k_{t}\right ) \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda _{t}^{i}+\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}}=\alpha f\left ( k_{t}\right ) \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda _{t}^{i}+\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}}=\alpha S_{t}f\left ( k_{t}\right ) \end{equation*}

where

\begin{equation*} S_{t}\equiv \frac {\sum \nolimits _{i}\frac {\pi _{i}\beta _{i}\left ( \lambda _{t}^{i}+\gamma \right ) }{1+\beta _{i}}}{1+\gamma \sum \nolimits _{i}\frac {\pi _{i}}{1+\beta _{i}}} \end{equation*}

By the convergence of $\lambda _{t}^{i}$ to $\lambda ^{i*}$ , we obtain $S_{t} \rightarrow S$ .

Fix any $0\lt \varepsilon \lt S$ . There exists $T$ such that $S-\varepsilon \lt S_{t}\lt S+\varepsilon$ for any $t\geq T$ .

Let $\left ( \overline {k}_{t}\right )_{t=T}^{\infty }$ and $\left ( \underline {k}_{t}\right )_{t=T}^{\infty }$ be defined as

\begin{equation*} \overline {k}_{T} = \underline {k}_{T}=k_{T}\,, \qquad \overline {k}_{t+1} = \alpha A\left ( S+\varepsilon \right ) \overline {k}_{t}^{\alpha }\,, \qquad \underline {k}_{t+1} = \alpha A\left ( S-\varepsilon \right ) \underline {k}_{t}^{\alpha } \end{equation*}

By induction, we have $\underline {k}_{t}\leq k_{t}\leq \overline {k}_{t}$ , for every $t\geq T$ . Clearly,

\begin{equation*} \lim _{t\rightarrow \infty }\overline {k}_{t}=\left [ \alpha A\left ( S+\varepsilon \right ) \right ] ^{\frac {1}{1-\alpha }} \qquad \text { and } \qquad \lim _{t\rightarrow \infty }\underline {k}_{t}=\left [ \alpha A\left ( S-\varepsilon \right ) \right ] ^{\frac {1}{1-\alpha }} \end{equation*}

Hence we obtain

\begin{equation*} \lim _{t\rightarrow \infty }\sup k_{t}\leq \left [ \alpha A\left ( S+\varepsilon \right ) \right ] ^{\frac {1}{1-\alpha }} \qquad \text { and } \qquad \lim _{t\rightarrow \infty }\inf k_{t}\geq \left [ \alpha A\left ( S-\varepsilon \right ) \right ] ^{\frac {1}{1-\alpha }} \end{equation*}

Since $\varepsilon$ is arbitrary, we have $\lim _{t \rightarrow \infty } k_{t} = \left ( \alpha AS \right )^{1/(1-\alpha )}$ . The convergence of $k_{t}$ implies the convergence of $b_{t}^{i}$ and $c_{t}^{i}$ . Therefore, the sequence $\left ( c_{t}^{L},c_{t}^{H},b_{t}^{L},b_{t}^{H},k_{t}\right ) _{t=0}^{\infty }$ converges to the values defined in part (1) of Proposition3.

(2) Consider part (2) of Proposition 3 and suppose that $\delta \gt \overline {\delta }$ .

(2.1) First, we prove the existence of some $t$ such that $b_{t}^{H}\geq b_{t}^{L}$ . Assume the contrary: $b_{t}^{L}\gt b_{t}^{H}$ for every $t \geq 0$ .

(2.1.1) We prove that $b_{t+1}^{H}\gt 0$ for every $t\geq 0$ . Indeed, assume the contrary, $b_{t+1}^{H}=0$ for some  $t$ . Then, from $b_{t+1}^{L}\gt b_{t+1}^{H}=0$ , we have

\begin{equation*} \beta _{L}R(k_{t})(b_{t}^{L}+\gamma k_{t})\geq \gamma k_{t+1}\geq \beta _{H}R(k_{t})(b_{t}^{H}+\gamma k_{t}) \end{equation*}

for some $t$ , that is

\begin{equation*} b_{t}^{H}+\gamma k_{t}\leq \frac {\beta _{L}}{\beta _{H}}\left ( b_{t}^{L}+\gamma k_{t}\right ) \end{equation*}

or, equivalently,

\begin{equation*} \frac {\beta _{L}}{\beta _{H}}\gt \frac {\lambda _{t}^{H}+\gamma }{\lambda _{t}^{L}+\gamma } \end{equation*}

We know that $b_{t+1}^{H}\gt 0$ if the equation $\zeta \left ( \lambda _{t+1}^{H}\right ) =\xi \left ( \lambda _{t}^{H}\right )$ has a solution in the interval $\left ( 0,1/\pi _{H}\right )$ . We already have $\zeta \left ( 0\right ) \lt \xi \left ( 0\right ) \lt \xi \left ( \lambda _{t}^{H}\right )$ . We will verify that $\zeta \left ( 1/\pi _{H}\right ) \gt \xi \left ( \lambda _{t}^{H}\right )$ . Indeed, we have

\begin{equation*} \zeta \left ( \frac {1}{\pi _{H}}\right ) = \frac {\frac {1}{\pi _{H}}+\frac {\gamma }{1+\beta _{H}}}{\frac {\gamma }{1+\beta _{L}}}\gt \frac {1+\beta _{L}}{1+\beta _{H}}=\frac {\beta _{L}}{\beta _{H}}\frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}} \gt \frac {\lambda _{t}^{H}+\gamma }{\lambda _{t}^{L}+\gamma }\frac {\beta _{H}}{\beta _{L}}\frac {1+\beta _{L}}{1+\beta _{H}}=\xi \left ( \lambda _{t}^{H}\right ) \end{equation*}

Hence, equation $\zeta \left ( \lambda _{t+1}^{H}\right ) =\xi \left ( \lambda _{t}^{H}\right )$ has a solution in $\left ( 0,1/\pi _{H}\right )$ . Therefore, $\lambda _{t+1}^{H}\gt 0$ and $b_{t+1}^{H}\gt 0$ , a contradiction. Then, $b_{t+1}^{H}\gt 0$ for every $t\geq 0$ under the assumption that $b_{t}^{L}\gt b_{t}^{H}$ for every $t\geq 0$ .

(2.1.2) Since $b_{t+1}^{L}\gt b_{t+1}^{H}\gt 0$ for every $t \geq 0$ , using the same arguments as in the preliminary part of the proof, we have $0\lt \lambda _{t}^{i}\lt 1/\pi _{H}$ for any $t\geq 0$ . Moreover, this sequence is monotonic and converges to a solution to equation $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ . Hence, equation $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ has a solution in the interval $\left ( 0,1/\pi _{H}\right )$ . As proven above, this implies the existence of a steady state with positive bequests: a contradiction with the second part of Proposition3. Hence, there exists some $t_{0}$ such that $b_{t_{0}}^{H}\geq b_{t_{0}}^{L}$ .

(2.2) We will prove that $b_{t}^{H}\geq b_{t}^{L}$ for every $t\geq t_{0}$ . Indeed, since $\beta /\left ( 1+\beta \right )$ increases with $\beta$ and $1/\left ( 1+\beta \right )$ decreases, we have

\begin{equation*} \frac {\beta _{H}R\left ( k_{t_{0}}\right ) \left ( b_{t_{0}}^{H}+\gamma k_{t_{0}}\right ) -\gamma k_{{t_{0}}+1}}{1+\beta _{H}} \ \geq \ \frac {\beta _{L}R\left ( k_{t_{0}}\right ) \left ( b_{t_{0}}^{L}+\gamma k_{t_{0}}\right ) -\gamma k_{{t_{0}}+1}}{1+\beta _{L}} \end{equation*}

This implies $b_{{t_{0}}+1}^{H}\geq b_{{t_{0}}+1}^{L}$ . By induction, we have $b_{t}^{H}\geq b_{t}^{L}$ for every $t\geq {t_{0}}$ .

(2.3) We prove the existence of some $t_{1}\geq t_{0}$ such that $b_{t_{1}}^{L}=0$ . Assume the contrary: $b_{t}^{H}\geq b_{t}^{L}\gt 0$ for any $t\geq t_{0}$ . This implies the existence of a steady state with strictly positives bequests: a contradiction. Therefore, there exists $t_{1}\geq t_{0}$ such that $b_{t_{1}}=0$ .

(2.4) Now, we prove that $b_{t}^{L}=0$ for every $t\geq t_{1}$ . Assume the contrary: there is some $t\geq t_{1}$ such that $b_{t}^{L}=0$ and $b_{t+1}^{L}\gt 0$ . In this case, both $b_{t+1}^{L}$ and $b_{t+1}^{H}$ are strictly positive. Since $b_{t}^{L}=0$ , we have $\lambda _{t}^{H}=1/\pi _{H}$ . Using the same arguments as in part (1) with $b_{t+1}^{L},b_{t+1}^{H}\gt 0$ , we find that $\lambda _{t+1}^{H}=\varphi \left ( \lambda _{t}^{H}\right ) \gt \lambda _{t}^{H}=1/\pi _{H}$ , a contradiction.

Hence, $b_{t}^{L}=0$ for every $t\geq t_{1}$ . Therefore, the sequence $\left ( b_{t}^{L},b_{t}^{H}\right )$ converges to $\left ( 0, b_{2}^{H*}\right )$ in Proposition3.

(3) Consider the cutting-edge case where $\delta = \overline {\delta }$ . Consider functions $\zeta$ and $\xi$ defined as in part (1) of the proof. We observe that, for any $\lambda \in \left [ 0,1/\pi _{H}\right ]$ , we have $\zeta \left ( \lambda \right ) \leq \xi \left ( \lambda \right )$ , with equality if and only if $\lambda =1/\pi _{H}$ . Using the same arguments as in part (2) of the proof, we have $b_{t}^{H}\gt 0$ for every $t\geq 1$ . Now, we consider two cases: either $b_{0}^{L}=0$ or $b_{0}^{L}\gt 0$ .

In the first case, following the same line of arguments as in part (2) of the proof, we have $b_{t}^{L}=0$ for any $t\geq 1$ and the solution converges to the one described in Proposition3.

In the second case, we have $0\lt \lambda _{0}^{H}\lt 1/\pi _{H}.$ Using arguments in part (1) of the proof, we have $0\lt \lambda _{0}^{L}\lt \lambda _{2}^{L}\lt 1/\pi _{H}$ . By induction we obtain that the sequence $\left ( \lambda _{t}^{H}\right ) _{t\geq 1}$ is strictly increasing and converges to the unique solution to $\zeta \left ( \lambda \right ) =\xi \left ( \lambda \right )$ , that is $1/\pi _{H}$ . A direct consequence of this is that $\lambda _{t}^{L}$ converges to $0$ . As in part (2), the convergence of $\lambda _{t}^{L}$ to $0$ and of $\lambda _{t}^{H}$ to $1/\pi _{H}$ implies the convergence of $\left ( b_{t}^{L},b_{t}^{H},k_{t}\right )$ . It is easy to compute that they converge to the values defined in part (1) of Proposition3.

A.5 Proof of Proposition 5

Denote

\begin{eqnarray*} R_{1}^{*} \left ( \pi \right ) &\equiv &\frac {1+\alpha }{2}+\frac {1}{2}\left ( \frac {1}{\beta _{L}}+\frac {1}{\beta _{H}}\right ) -\frac {1}{2}\sqrt {\left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right ) } \\ R_{2}^{*} \left ( \pi \right ) &\equiv &\frac {1}{\beta _{H}}+\frac {1}{1+\gamma \pi } \end{eqnarray*}

and consider the impact of $\pi$ on the steady-state interest rate.

Condition $\delta \lt \overline {\delta }$ is equivalent to $\pi \lt \overline {\pi }$ . Thus, there are three cases: (1) if $\overline {\pi }\gt 1$ , which is equivalent to $\delta \lt \alpha$ , then $R^{*}\left ( \pi \right ) =R_{1}^{*}\left ( \pi \right )$ ; (2) if $0\lt \overline {\pi }\leq 1$ , then $R^{*}\left ( \pi \right ) =R_{1}^{*}\left ( \pi \right )$ for $\pi \lt \overline {\pi }$ and $R^{*}\left ( \pi \right ) = R_{2}^{*}\left ( \pi \right )$ for $\pi \geq \overline {\pi }$ ; and (3) if $\overline {\pi }\leq 0$ , which is equivalent to $\delta \geq 1$ , then $R^{*} \left ( \pi \right ) = R_{2}^{*}\left ( \pi \right )$ .

When $0\lt \overline {\pi }\leq 1$ and $\pi =\overline {\pi }$ , by (A3) and the fact that $\delta \lt 1$ we have

\begin{equation*} R_{1}^{*} \left ( \overline {\pi }\right ) = \frac {1}{2}\left [ 1+\alpha +\frac {1}{\beta _{L}}+\frac {1}{\beta _{H}}-\sqrt {D(\overline {\pi })}\right ] =\frac {1}{2}\left ( \frac {1}{\beta _{L}}+\frac {1}{\beta _{H}}+\delta \right ) =\frac {1}{\beta _{L}} = \frac {1}{\beta _{H}}+\delta = R_{2}^{*}\left ( \overline {\pi }\right )\end{equation*}

Therefore, $R^{*} \left ( \pi \right )$ is continuous. It follows that $k^{*}$ is also continuous in $\pi$ . Since both $R_{1}^{*} \left ( \pi \right )$ and $R_{2}^{*} \left ( \pi \right )$ decrease with $\pi$ , $k^{*}$ increases with $\pi$ .

A.6 Proof of Proposition 6

Let $D\left ( \delta \right ) \equiv \left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right )$ . Denote

\begin{eqnarray*} R_{1}^{*} \left ( \delta \right ) &\equiv &\frac {1+\alpha }{2}+\frac {1}{\overline {\beta }}+\frac {2\pi -1}{2}\delta -\frac {1}{2}\sqrt {D\left ( \delta \right ) } \\ R_{2}^{*} \left ( \delta \right ) &\equiv &\frac {1}{\overline {\beta }}-\left ( 1-\pi \right ) \delta +\frac {1}{1+\gamma \pi } \end{eqnarray*}

and consider the impact of $\delta$ on the steady-state interest rate.

Since $\overline {\delta }$ given by (11) corresponds to $\overline {\pi }$ given by (17), it is easily seen from (A3) that

\begin{equation*} D\left ( \overline {\delta }\right ) =\left ( 1+\alpha -\frac {1}{1+\gamma \pi }\right )^{2} \end{equation*}

Therefore, $R^{*} \left ( \delta \right )$ is continuous, since

\begin{equation*} R_{1}^{*} \left ( \overline {\delta }\right ) = \frac {1}{\overline {\beta }}+\frac {1}{2}\left [ 1+\alpha +\frac {2\pi -1}{1+\gamma \pi }-\sqrt {D\left ( \overline {\delta }\right ) }\right ] =\frac {1}{\overline {\beta }}+\frac {\pi }{1+\gamma \pi } = R_{2}^{*} \left ( \overline {\delta }\right ) \end{equation*}

Notice that

\begin{eqnarray*} (R_{2}^{*})^{\prime }\left ( \delta \right ) &\equiv &\pi -1\lt 0 \\ (R_{1}^{*})^{\prime } \left ( \delta \right ) &\equiv &\frac {2\pi -1}{2}-\frac {\delta +\alpha -1+2\pi \left ( 1-\alpha \right ) }{2\sqrt {D(\delta )}}\lt 0 \end{eqnarray*}

Indeed, $(R_{1}^{*})^{\prime } \left ( \delta \right ) \lt 0$ if $\delta \gt \left ( 2\pi -1\right ) \left ( \sqrt {D(\delta )} - 1 + \alpha \right )$ . When $2\pi -1\gt 0$ , we have $\delta \gt (2\pi - 1) \delta$ , and the above inequality holds if $\sqrt {D(\delta )} \lt \delta + 1 - \alpha$ , which follows from (A3). When $2\pi -1\lt 0$ , we have $\delta \gt (1 - 2\pi ) \delta$ , and the above inequality holds if $\sqrt {D(\delta )} \gt 1 - \alpha - \delta$ , which also follows from (A3).

Since $k^{*}$ is inversely related to $R^{*}$ , it follows that $k^{*}$ increases with $\delta$ .

A.7 Proof of Proposition 7

The level of social inequality in terms of income in the steady-state equilibrium is represented by the Gini index:

\begin{equation*} G=2\int _{0}^{1}\left [ x-g\left ( x\right ) \right ] dx \end{equation*}

where $g:\left [ 0,1\right ] \rightarrow \left [ 0,1\right ]$ is the Lorenz curve.

Consider a share $\pi _{L}$ of less altruistic agents with income $y^{L}$ , and a share $\pi _{H}$ of more altruistic agents with income $y^{H}\geq y^{L}$ . The continuous Lorenz curve is given by

\begin{eqnarray*} g\left ( x\right ) &=&\frac {y^{L}}{\pi _{L}y^{L}+\pi _{H}y^{H}}x \qquad \text {if } \qquad 0\leq x\leq \pi _{L} \\ g\left ( x\right ) &=&\frac {\pi _{L}y^{L}+\left ( x-\pi _{L}\right ) y^{H}}{\pi _{L}y^{L}+\pi _{H}y^{H}} \qquad \text {if } \qquad \pi _{L} \lt x \leq 1 \end{eqnarray*}

We obtain

\begin{eqnarray*} G &=&2\int _{0}^{1}\left [ x-g\left ( x\right ) \right ] dx=2\left ( \int _{0}^{\pi _{L}}\left [ x-g\left ( x\right ) \right ] dx+\int _{\pi _{L}}^{1}\left [ x-g\left ( x\right ) \right ] dx\right ) \\ &=&1-\pi _{L}^{2}\frac {y^{L}}{\pi _{L}y^{L}+\pi _{H}y^{H}}-2\left ( 1-\pi _{L}\right ) \frac {\pi _{L}y^{L}-\pi _{L}y^{H}}{\pi _{L}y^{L}+\pi _{H}y^{H}}-\left ( 1-\pi _{L}^{2}\right ) \frac {y^{H}}{\pi _{L}y^{L}+\pi _{H}y^{H}} \end{eqnarray*}

and hence,

(A6) \begin{equation} G=\frac {\pi _{L}\pi _{H}\left ( y^{H}-y^{L}\right ) }{\pi _{L}y^{L}+\pi _{H}y^{H}} \end{equation}

Consider two parts of Proposition 3.

(1) For $i=L,H$ , we have $y_{1}^{i}=R_{1}^{*}b_{1}^{i*}+\left ( 1-\alpha \right ) A(k^{*}_{1})^{\alpha }$ . Observing that $R_{1}^{*} = \alpha A(k^{*}_{1})^{\alpha -1}$ and using (12), it follows from (A6) that

\begin{equation*} G_{1}^{*}=\pi _{L}\pi _{H}\frac {R_{1}^{*}\frac {b_{1}^{H*}}{\gamma k^{*}_{1}}-R_{1}^{*}\frac {b_{1}^{L*}}{\gamma k^{*}_{1}}}{\alpha A(k^{*}_{1})^{\alpha -1}+\pi _{L}R_{1}^{*}\frac {b_{1}^{L*}}{\gamma k^{*}_{1}}+\pi _{H}R_{1}^{*}\frac {b_{1}^{H*}}{\gamma k^{*}_{1}}}=\pi _{L}\pi _{H}\frac {\frac {1}{\beta _{L}}-\frac {1}{\beta _{H}}}{1-R_{1}^{*}+\frac {\pi _{H}}{\beta _{L}}+\frac {\pi _{L}}{\beta _{H}}} \end{equation*}

(2) We have $y_{2}^{L}=\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }$ and $y_{2}^{H}=\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }+R_{2}^{*} k^{*}_{2}/\pi _{H}$ , and hence,

\begin{equation*} G_{2}^{*}=\frac {\pi _{L}\pi _{H}\left ( y_{2}^{H}-y_{2}^{L}\right ) }{\pi _{L}y_{2}^{L}+\pi _{H}y_{2}^{H}}=\frac {\pi _{L}R_{2}^{*}k^{*}_{2}}{\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }+R_{2}^{*}k^{*}_{2}}=\frac {\pi _{L} \alpha A(k^{*}_{2})^{\alpha }} {\left ( 1-\alpha \right ) A(k^{*}_{2})^{\alpha }+\alpha A(k^{*}_{2})^{\alpha }}=\alpha \pi _{L} \end{equation*}

A.8 Proof of Proposition 8

Denote

\begin{eqnarray*} G_{1}^{*}\left ( \pi \right ) &\equiv &\frac {2\delta \pi (1-\pi )}{1-\alpha +\left ( 2\pi -1\right ) \delta +\sqrt {\left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right ) }} \\ G_{2}^{*}\left ( \pi \right ) &\equiv &\alpha \left ( 1-\pi \right ) \end{eqnarray*}

Using (A3), we have

\begin{eqnarray*} G_{1}^{*}\left ( \overline {\pi }\right ) &=&\frac {2\delta \overline {\pi }\left ( 1- \overline {\pi } \right ) }{1-\alpha +\left ( 2\overline {\pi }-1\right ) \delta +\sqrt {D(\overline {\pi })}}=\frac {2\delta \overline {\pi }\left ( 1-\overline {\pi }\right ) }{1-\alpha +\left ( 2\overline {\pi }-1\right ) \delta +1-\delta +\alpha } \\ &=&\frac {\delta \overline {\pi }\left ( 1-\overline {\pi }\right ) }{1-\delta +\delta \overline {\pi }}=\frac {1}{1+\gamma }\left ( 1-\overline {\pi }\right ) =\alpha \left ( 1-\overline {\pi }\right ) = G_{2}^{*} \left ( \overline {\pi }\right ) \end{eqnarray*}

since $\delta \overline {\pi }=\left ( 1-\delta \right ) /\gamma$ . Therefore, $G^{*} \left ( \pi \right )$ is continuous.

To analyze the shape of $G_{1}^{*}\left ( \pi \right )$ , denote $G_{1}^{*}\left ( \pi \right ) =n \left ( \pi \right ) / d \left ( \pi \right )$ , where

\begin{equation*} n\left ( \pi \right ) \equiv 2\delta \pi \left ( 1-\pi \right ) \qquad \text {and} \qquad d\left ( \pi \right ) \equiv 1-\alpha +\delta \left ( 2\pi -1\right ) +\sqrt {D\left ( \pi \right )} \end{equation*}

Since

\begin{equation*} n^{\prime }\left ( \pi \right ) = 2\delta \left ( 1-2\pi \right ) \qquad \text {and} \qquad d^{\prime }\left ( \pi \right ) = 2\delta \left [ 1+\frac {1-\alpha }{\sqrt {D\left ( \pi \right ) }}\right ] \end{equation*}

we have

(A7) \begin{equation} (G_{1}^{*})^{\prime }\left ( \pi \right ) =2\delta \frac {\left ( 1-2\pi \right ) d\left ( \pi \right ) -2\delta \pi \left ( 1-\pi \right ) \left [ 1+\frac {1-\alpha }{\sqrt {D\left ( \pi \right ) }}\right ] }{d\left ( \pi \right )^{2}} \end{equation}

Now we have to distinguish between the two cases.

(1) Suppose that $\delta \lt 1-\alpha$ . Then $G_{1}^{*}\left ( 0\right ) =G_{1}^{*}\left ( 1\right ) =0$ . We show that there exists a unique $\hat {\pi }$ such that for $\pi \lt \hat {\pi }$ , $G_{1}^{*}\left ( \pi \right )$ strictly increases with $\pi$ , while for $\pi \geq \hat {\pi }$ , $G_{1}^{*}\left ( \pi \right )$ strictly decreases with $\pi$ .

Indeed, after some algebra, equation $(G_{1}^{*})^{\prime }\left ( \pi \right ) =0$ can be written as $\mathcal {L} \left ( \pi \right ) = \mathcal {R}\left ( \pi \right )$ , where

\begin{eqnarray*} \mathcal {L} \left ( \pi \right ) &\equiv &\left ( 1-\alpha -\delta \right ) \frac {1-2\pi }{2\delta \pi } \\ \mathcal {R} \left ( \pi \right ) &\equiv &\frac {\pi \sqrt {D\left ( \pi \right ) }+\left ( 1-\alpha \right ) \left ( 3\pi -1\right ) }{\sqrt {D\left ( \pi \right ) }+1-\alpha -\delta }=\frac {\sqrt {D\left ( \pi \right ) }+\left ( 1-\alpha \right ) \left ( 3-\frac {1}{\pi }\right ) }{\frac {\sqrt {D\left ( \pi \right ) }}{\pi }+\frac {1-\alpha -\delta }{\pi }} \end{eqnarray*}

Since in this case $1-\alpha -\delta \gt 0$ , $\mathcal {L} \left ( \pi \right )$ is convex and decreasing with $\mathcal {L} \left ( 0^{+} \right ) = \infty$ , $\mathcal {L} \left ( 1/2 \right ) = 0$ and $\mathcal {L} (1) = -\left ( 1-\delta -\alpha \right ) /\left ( 2\delta \right ) \lt 0$ .

Further, $\mathcal {R} \left ( \pi \right )$ increases with $\pi$ . Indeed, $\sqrt {D \left ( \pi \right )} + \left ( 1-\alpha \right ) \left ( 3-1/\pi \right )$ increases with $\pi$ , while $\sqrt {D\left ( \pi \right ) }/\pi +\left ( 1-\alpha -\delta \right ) / \pi$ decreases with $\pi$ . By (A3),

\begin{equation*} \mathcal {R} \left ( 0 \right ) =-\frac {1}{2}\frac {1-\alpha }{1-\alpha -\delta }\lt 0 \qquad \text {and} \qquad \mathcal {R} \left ( \frac {1}{2}\right ) =\frac {1}{2}\frac {\sqrt {D\left ( 1/2\right )} +1-\alpha }{\sqrt {D\left ( 1/2\right ) }+1-\alpha -\delta }\gt 0 \end{equation*}

Therefore, in this case there exists a unique value $0\lt \hat {\pi }\lt 1/2$ such that $\mathcal {L} \left ( \hat {\pi }\right ) = \mathcal {R} \left ( \hat {\pi }\right )$ , or $(G_{1}^{*})^{\prime }\left ( \hat {\pi }\right ) = 0$ . Moreover, for $\pi \lt \hat {\pi }$ , $\mathcal {L} \left ( \pi \right ) \gt \mathcal {R} \left ( \pi \right )$ , so that $G_{1}^{*}\left ( \pi \right )$ is increasing, while for $\pi \gt \hat {\pi }$ , $\mathcal {L} \left ( \pi \right ) \lt \mathcal {R} \left ( \pi \right )$ , and $G_{1}^{*} \left ( \pi \right )$ is decreasing.

When $\delta \lt \alpha$ , we have $\overline {\pi }\gt 1$ , and hence, as in the proof of Proposition5, we have $G^{*} \left ( \pi \right ) =G_{1}^{*}\left ( \pi \right )$ . In this case the threshold level of $\pi$ , up to which $G^{*}$ is increasing and after which $G^{*}$ is decreasing, is $\hat {\pi }$ . Similarly, when $\alpha \lt \delta \leq 1$ , for $\pi \lt \overline {\pi }$ we have $G^{*}\left ( \pi \right ) =G_{1}^{*}\left ( \pi \right )$ , while for $\pi \geq \overline {\pi }$ , we have $G^{*}\left ( \pi \right ) =G_{2}^{*}\left ( \pi \right )$ . In this case, the threshold level of $\pi$ is $\tilde {\pi } \equiv \min \left \{ \hat {\pi },\overline {\pi }\right \}$ .

(2) Suppose that $\delta \gt 1-\alpha$ . By Bernoulli’s rule,

\begin{equation*} G_{1}^{*}\left ( 0\right ) =\frac {n^{\prime }\left ( 0\right ) }{d^{\prime }\left ( 0\right ) }=\frac {\delta +\alpha -1}{\delta }\gt 0=G_{1}^{*}\left ( 1\right ) \end{equation*}

Moreover, in this case $G_{1}^{*}\left ( 0\right ) \lt \alpha =G_{2}^{*}\left ( 0\right )$ if and only if $\delta \lt 1$ .

The second case has two subcases.

(2.1) Suppose that $1-\alpha \lt \delta \lt 2\left ( 1-\alpha \right )$ . Then we show that $(G_{1}^{*})^{\prime }\left ( 0^{+}\right ) \gt 0$ and $(G_{1}^{*})^{\prime }\left ( 1^{-}\right ) \lt 0$ , so there exists an interior $\hat {\pi }=\arg \max _{0\leq \pi \leq 1}G_{1}^{*}\left ( \pi \right )$ .

Indeed, using (A3) and (A7), we obtain

\begin{equation*} (G_{1}^{*})^{\prime }\left ( 1\right ) =-\frac {\delta }{1-\alpha +\delta }\lt 0 \end{equation*}

Applying Bernoulli’s rule to (A7), we get

\begin{eqnarray*} (G_{1}^{*})^{\prime }\left ( 0^{+}\right ) &=&2\delta\!\! \lim _{\pi \rightarrow 0^{+}}\!\!\frac {-2d\left ( \pi \right ) +\left ( 1-2\pi \right ) d^{\prime }\!\left ( \pi \right ) -2\delta \!\left ( 1-2\pi \right )\! \left [ 1+\frac {1-\alpha }{\sqrt {D\left ( \pi \right ) }}\right ] \!+4\delta ^{2}\left ( 1-\alpha \right ) ^{2}\frac {\pi \left ( 1-\pi \right ) }{D\left ( \pi \right ) ^{\frac {3}{2}}}}{2d\left ( \pi \right ) d^{\prime }\left ( \pi \right ) } \\ &=&\lim _{\pi \rightarrow 0^{+}}\frac {-d\left ( \pi \right ) \sqrt {D\left ( \pi \right ) }+2\delta ^{2}\left ( 1-\alpha \right ) ^{2}\frac {\pi \left ( 1-\pi \right ) }{D\left ( \pi \right ) }}{d\left ( \pi \right ) \left [ 1-\alpha +\sqrt {D\left ( \pi \right ) }\right ] } \end{eqnarray*}

In this case, by (A3), $\sqrt {D\left ( 0\right ) }=\delta +\alpha -1\gt 0$ . Applying again Bernoulli’s rule, we get

\begin{eqnarray*} (G_{1}^{*})^{\prime }\left ( 0^{+}\right ) &=&\lim _{\pi \rightarrow 0^{+}}\frac {-d^{\prime }\left ( \pi \right ) \sqrt {D\left ( \pi \right ) }-d\left ( \pi \right ) \frac {D^{\prime }\left ( \pi \right ) }{2\sqrt {D\left ( \pi \right ) }}+2\delta ^{2}\left ( 1-\alpha \right ) ^{2}\frac {\left ( 1-2\pi \right ) D\left ( \pi \right ) -\pi \left ( 1-\pi \right ) D^{\prime }\left ( \pi \right ) }{D\left ( \pi \right ) ^{2}}}{d^{\prime }\left ( \pi \right ) \left [ 1-\alpha +\sqrt {D\left ( \pi \right ) }\right ] +d\left ( \pi \right ) \frac {D^{\prime }\left ( \pi \right ) }{2\sqrt {D\left ( \pi \right ) }}} \\ &=&\frac {2\left ( 1-\alpha \right ) -\delta }{\delta +\alpha -1} \end{eqnarray*}

Thus, for $1-\alpha \lt \delta \lt 2\left ( 1-\alpha \right )$ , $(G_{1}^{*})^{\prime }\left ( 0^{+}\right ) \gt 0$ .

(2.2) Suppose that $\delta \geq 2\left ( 1-\alpha \right )$ . We have just seen that in this case $(G_{1}^{*})^{\prime }\left ( 0^{+}\right ) \leq 0$ . Let us show that $(G_{1}^{*})^{\prime }\left ( \pi \right ) \lt 0$ for all $\pi \gt 0$ . By (A7), this is equivalent to

\begin{equation*} \pi \left ( 1-\pi \right ) d^{\prime }\left ( \pi \right ) \gt \left ( 1-2\pi \right ) d\left ( \pi \right ) \end{equation*}

Since $d\left ( 0\right ) =0$ , it is sufficient to check that for all $\pi \gt 0$ ,

\begin{equation*} \left [ \pi \left ( 1-\pi \right ) d^{\prime }\left ( \pi \right ) \right ] ^{\prime }\gt \left [ \left ( 1-2\pi \right ) d\left ( \pi \right ) \right ] ^{\prime } \end{equation*}

that is $2d\left ( \pi \right ) \gt -\pi \left ( 1-\pi \right ) d^{\prime \prime }\left ( \pi \right )$ or $2d\left ( \pi \right ) \gt \pi \left ( 1-\pi \right ) \left \vert d^{\prime \prime }\left ( \pi \right ) \right \vert$ , since $d^{\prime \prime }\left ( \pi \right ) \lt 0$ . Or, equivalently, it is sufficient to show that

\begin{equation*} \left [ \sqrt {D\left ( \pi \right ) }-\left ( \delta +\alpha -1\right ) +2\delta \pi \right ] \left [ D\left ( \pi \right ) \right ] ^{\frac {3}{2}}\gt 2\pi \left ( 1-\pi \right ) \delta ^{2}\left ( 1-\alpha \right ) ^{2} \end{equation*}

It is easily seen that

\begin{equation*} \sqrt {D\left ( \pi \right ) }-\left ( \delta +\alpha -1\right ) \gt \frac {2\delta \pi \left ( 1-\alpha \right ) }{\sqrt {D\left ( \pi \right ) }} \end{equation*}

and, since $\delta \geq 2\left ( 1-\alpha \right ) \gt 1-\alpha$ ,

\begin{equation*} \sqrt {D\left ( \pi \right ) }=\sqrt {\left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right ) }\gt \delta -\left ( 1-\alpha \right ) \geq 1-\alpha \end{equation*}

Therefore,

\begin{eqnarray*} &&\left [ \sqrt {D\left ( \pi \right ) }-\left ( \delta +\alpha -1\right ) +2\delta \pi \right ] D\left ( \pi \right ) ^{\frac {3}{2}} \ = \ \left [ \sqrt {D\left ( \pi \right ) }-\left ( \delta +\alpha -1\right ) \right ] D\left ( \pi \right ) ^{\frac {3}{2}}+2\delta \pi D\left ( \pi \right )^{\frac {3}{2}} \\ && \gt \ \frac {2\delta \pi \left ( 1-\alpha \right ) }{\sqrt {D\left ( \pi \right )}} D\left ( \pi \right ) ^{\frac {3}{2}}+2\delta \pi D\left ( \pi \right )^{\frac {3}{2}} \ = \ 2\delta \pi D\left ( \pi \right ) \left [ 1-\alpha +\sqrt {D\left ( \pi \right ) }\right ] \\ && \gt \ 2\delta \pi \left ( 1-\alpha \right ) ^{2}\left [ 1-\alpha +\delta - \left ( 1-\alpha \right ) \right ] \ = \ 2 \pi \delta ^{2}\left ( 1-\alpha \right )^{2} \geq 2\pi \left ( 1-\pi \right ) \delta ^{2}\left ( 1-\alpha \right )^{2} \end{eqnarray*}

Again, as in the proof of Proposition5, in both cases (2.1) and (2.2), when $\delta \lt \alpha$ , we have $G^{*}\left ( \pi \right ) =G_{1}^{*}\left ( \pi \right )$ . When $\alpha \lt \delta \leq 1$ , for $\pi \lt \overline {\pi }$ we have $G^{*}\left ( \pi \right ) =G_{1}^{*}\left ( \pi \right )$ , while for $\pi \geq \overline {\pi }$ , we have $G^{*} \left ( \pi \right ) =G_{2}^{*}\left ( \pi \right )$ . When $\delta \gt 1$ , we have $\overline {\pi }\leq 0$ , and hence $G^{*}\left ( \pi \right ) =G_{2}^{*}\left ( \pi \right )$ . Thus, if $1-\alpha \lt \delta \lt \min \left \{ 1,2\left ( 1-\alpha \right ) \right \}$ , then $G^{*}\left ( \pi \right )$ has an inverted-U shape. If $\delta \geq \min \left \{ 1,2\left ( 1-\alpha \right ) \right \}$ , then $G^{*}\left ( \pi \right )$ decreases with $\pi$ .

A.9 Proof of Proposition 9

Denote

\begin{equation*} G_{1}^{*}\left ( \delta \right ) \equiv \frac {2\delta \pi (1-\pi )}{1-\alpha +\left ( 2\pi -1\right ) \delta +\sqrt {\left ( \delta +\alpha -1\right ) ^{2}+4\delta \pi \left ( 1-\alpha \right )}} \end{equation*}

We have

\begin{eqnarray*} G_{1}^{*}\left ( \overline {\delta }\right ) &=&\frac {2\overline {\delta }\pi \left ( 1-\pi \right ) }{1-\alpha +(2\pi -1)\overline {\delta }+\sqrt {D(\overline {\delta })}}=\frac {2\overline {\delta }\pi \left ( 1-\pi \right ) }{1-\alpha +(2\pi -1) \overline {\delta }+1+\alpha -\overline {\delta }} \\ &=&\frac {\overline {\delta }\pi \left ( 1-\pi \right ) }{1-(1-\pi )\overline {\delta }}=\frac {\pi \left ( 1-\pi \right ) }{1+\gamma \pi -1+\pi }=\alpha \left ( 1-\pi \right ) = G_{2}^{*}\left ( \overline {\delta }\right ) \end{eqnarray*}

since $\gamma =\left ( 1-\alpha \right ) / \alpha$ . Therefore, $G^{*} \left ( \delta \right )$ is continuous.

Let us show that $(G_{1}^{*})^{\prime }\left ( \delta \right ) \gt 0$ . Indeed,

\begin{equation*} (G_{1}^{*})^{\prime }\left ( \delta \right ) =\frac {2\pi \left ( 1-\pi \right ) }{\left [ 1-\alpha +\delta \left ( 2\pi -1\right ) +\sqrt {D\left ( \delta \right ) }\right ] ^{2}}\left [ 1-\alpha +\sqrt {D\left ( \delta \right ) }-\frac {\delta D^{\prime }\left ( \delta \right ) }{2\sqrt {D\left ( \delta \right ) }}\right ] \end{equation*}

and, since by (A3), $\sqrt {D\left ( \delta \right )} \geq \delta +\alpha -1$ , we have

\begin{eqnarray*} 1-\alpha +\sqrt {D\left ( \delta \right ) }-\frac {\delta D^{\prime }\left ( \delta \right ) }{2\sqrt {D\left ( \delta \right ) }} &=&\frac {1-\alpha }{\sqrt {D\left ( \delta \right ) }}\left [ \sqrt {D\left ( \delta \right ) }+\frac {D\left ( \delta \right ) -\delta \left ( \delta +\alpha -1\right ) }{1-\alpha }-2\delta \pi \right ] \\ &=&\frac {1-\alpha }{\sqrt {D(\delta )}}\left [ 2\delta \pi +\sqrt {D\left ( \delta \right ) }-\left ( \delta +\alpha -1\right ) \right ] \gt 0 \end{eqnarray*}

A.10 Proof of Proposition 10

In the steady state $j=1,2$ , for agent $i=L,H$ , we obtain

(A8) \begin{equation} (U_{j}^{i*})^{\prime }\left ( \pi \right ) =\frac {(c_{j}^{i*})^{\prime }\left ( \pi \right ) }{c_{j}^{i*}\left ( \pi \right ) }+\beta _{i}\frac {(c_{j}^{i*})^{\prime }\left ( \pi \right ) +(b_{j}^{i*})^{\prime }\left ( \pi \right ) }{c_{j}^{i*}\left ( \pi \right ) +b_{j}^{i*}\left ( \pi \right ) } \end{equation}

In the following, for simplicity, we omit the argument $\pi$ . We have $(U_{j}^{i*})^{\prime }=0$ if

(A9) \begin{equation} \frac {(c_{j}^{i*})^{\prime }}{c_{j}^{i*}}\left [ 1+\left ( 1+\beta _{i}\right ) \frac {c_{j}^{i*}}{b_{j}^{i*}}\right ] +\beta _{i}\frac {(b_{j}^{i*})^{\prime }}{b_{j}^{i*}}=0 \end{equation}

(1) Consider the steady-state equilibrium $\left (c_{1}^{L*},b_{1}^{L*}, c_{1}^{H*},b_{1}^{H*},k^{*}_{1}\right )$ where both types of agents leave bequests. For $\pi \lt \overline {\pi }$ , the utilities of the more and the less altruistic agents are given by $U_{1}^{H*}$ and $U_{1}^{L*}$ respectively. Recall that $(R_{1}^{*})^{\prime } \lt 0$ and

\begin{equation*} \frac {(k^{*}_{1})^{\prime }}{k^{*}_{1}} = -\frac {1}{1-\alpha }\frac {(R_{1}^{*})^{\prime }}{R_{1}^{*}} \end{equation*}

Furthermore, by part (1) of Proposition3,

(A10) \begin{equation} \frac {c_{1}^{i*}}{b_{1}^{i*}}\ =\ \frac {1}{\beta _{i}R_{1}^{*}-1} \end{equation}

and hence

(A11) \begin{equation} \frac {(b_{1}^{i*})^{\prime }}{b_{1}^{i*}}=\frac {(c_{1}^{i*})^{\prime }}{c_{1}^{i*}}\ +\frac {\beta _{i}(R_{1}^{*})^{\prime }}{\beta _{i}R^{*}_{1}-1} \end{equation}

On the other hand,

(A12) \begin{equation} \frac {(c_{1}^{i*})^{\prime }}{c_{1}^{i*}}\ =\ \frac {(k^{*}_{1})^{\prime }}{k^{*}_{1}}+\frac {\beta _{i} (R_{1}^{*})^{\prime }}{\beta _{i}+1-\beta _{i}R_{1}^{*}}=\frac {(R_{1}^{*})^{\prime }}{R_{1}^{*}}\left ( \frac {\beta _{i}R_{1}^{*}}{\beta _{i}+1-\beta _{i}R_{1}^{*}}-\frac {1}{1-\alpha }\right ) \end{equation}

Using (A8)–(A12), after some algebra we have

\begin{equation*} (U_{1}^{i*})^{\prime } = \frac {(R_{1}^{*})^{\prime }}{R_{1}^{*}}\left [ \frac {\beta _{i}\left ( 1+\beta _{i}\right ) R_{1}^{*}}{1+\beta _{i}-\beta _{i}R_{1}^{*}}-\frac {1+\alpha \beta _{i}}{1-\alpha }\right ] \end{equation*}

Since $(R_{1}^{*})^{\prime } \lt 0$ , we obtain that $(U_{1}^{i*})^{\prime } \lt 0$ if and only if

\begin{equation*} \frac {\beta _{i}\left ( 1+\beta _{i}\right ) R_{1}^{*}}{1+\beta _{i}-\beta _{i}R_{1}^{*}}-\frac {1+\alpha \beta _{i}}{1-\alpha }\gt 0 \end{equation*}

or, equivalently,

(A13) \begin{equation} R_{1}^{*} \left ( \pi \right ) \gt \frac {1+\beta _{i}}{\beta _{i}}\frac {1+\alpha \beta _{i}}{2-\alpha +\beta _{i}} \end{equation}

Note that the right-hand side of this inequality depends on $\beta _{i}$ and hence is different for different types of agents.

(1.1) Consider the utility of the less altruistic agents. Denote

\begin{equation*} \overline {R} \equiv \frac {1+\beta _{L}}{\beta _{L}} \frac {1+\alpha \beta _{L}}{2-\alpha +\beta _{L}} = 1 + \frac {1}{\beta _{L}} - \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \end{equation*}

Let the critical value of the level of altruism $\hat {\pi }$ be a solution to the equation $R_{1}^{*} \left ( \pi \right ) = \overline {R}$ . This solution exists only if $R_{1}^{*} \left ( 1 \right ) \leq \overline {R} \leq R_{1}^{*} \left ( 0 \right )$ . The first inequality holds when

\begin{equation*} \frac {1}{\beta _{H}} + \alpha \leq 1 + \frac {1}{\beta _{L}} - \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \iff \frac {1}{\beta _{L}} - \frac {1}{\beta _{H}} \geq \frac {(1-\alpha )(1+\alpha \beta _{L})}{(2-\alpha +\beta _{L}) \beta _{L}} \equiv \check {\delta } \end{equation*}

For the second inequality, recall that $R_{1}^{*} \left ( 0 \right ) = 1/\beta _{L} + \alpha$ when $\delta \lt 1 - \alpha$ , and $R_{1}^{*} \left ( 0 \right ) = 1/\beta _{H} + 1$ when $\delta \gt 1 - \alpha$ . While it is always true that

\begin{equation*} 1 + \frac {1}{\beta _{L}} - \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \leq \frac {1}{\beta _{L}} + \alpha \iff \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \geq 1-\alpha \iff 1+\alpha \beta _{L} \gt 0 \end{equation*}

the second inequality holds when

\begin{equation*} 1 + \frac {1}{\beta _{L}} - \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \leq \frac {1}{\beta _{H}} + 1 \iff \frac {1}{\beta _{L}} - \frac {1}{\beta _{H}} \leq \frac {(1-\alpha )(1+\beta _{L})^2}{(2-\alpha +\beta _{L}) \beta _{L}} \equiv \hat {\delta } \end{equation*}

Since $R_{1}^{*} \left ( \pi \right )$ strictly decreases with $\pi$ , it follows that when $\check {\delta } \leq \delta \leq \hat {\delta }$ , there exists a unique solution $\hat {\pi }$ to the equation $R_{1}^{*} \left ( \pi \right ) = \overline {R}$ , and $0 \leq \hat {\pi } \leq 1$ .

Define also a critical value for the capital share in total income:

\begin{equation*} \overline {\alpha } \equiv \frac {1}{1+\beta _{L}+\beta _{L}^{2}} \end{equation*}

There are three cases.

(1.1.1) Suppose that $\delta \lt \check {\delta }$ . Then we have $R_{1}^{*} \left ( \pi \right ) \geq R_{1}^{*} \left ( 1 \right ) \gt \overline {R}$ for all $\pi$ , and hence, according to (A13), $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \lt 0$ .

(1.1.2) Suppose that $\check {\delta } \leq \delta \leq \hat {\delta }$ . Note that we have

\begin{equation*} \overline {\pi } \lessgtr \hat {\pi } \ \iff \ R_{1}^{*} \left ( \overline {\pi } \right ) \gtrless \overline {R} \ \iff \ \frac {1}{\beta _{L}} \gtrless \frac {1+\beta _{L}}{\beta _{L}}\frac {1+\alpha \beta _{L}}{2-\alpha +\beta _{L}} \ \iff \ \alpha \lessgtr \overline {\alpha } \end{equation*}

Therefore, since $R_{1}^{*} \left ( \pi \right )$ strictly decreases, when $\alpha \lt \overline {\alpha }$ , we have $R_{1}^{*} \left ( \pi \right ) \gt R_{1}^{*} \left ( \overline {\pi } \right ) \gt \overline {R}$ for all $\pi \lt \overline {\pi }$ , and hence $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \lt 0$ for all $\pi \lt \overline {\pi }$ .

Alternatively, when $\alpha \gt \overline {\alpha }$ , we have $\hat {\pi } \lt \overline {\pi }$ . Therefore, for all $\pi \leq \hat {\pi }$ , we have $R_{1}^{*} \left ( \pi \right ) \gt \overline {R}$ , and hence $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \lt 0$ for all $\pi \leq \hat {\pi }$ . At the same time, for all $\pi \geq \hat {\pi }$ , we have $R_{1}^{*} \left ( \pi \right ) \lt \overline {R}$ , and hence $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \gt 0$ for all $\hat {\pi } \lt \pi \lt \overline {\pi }$ .

(1.1.3) Suppose that $\delta \gt \hat {\delta }$ . Then we have $R_{1}^{*} \left ( \pi \right ) \leq R_{1}^{*} \left ( 0 \right ) \lt \overline {R}$ for all $\pi$ , and hence, according to (A13), $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \gt 0$ .

(1.2) Consider now the more altruistic agents. Clearly, for all $\pi$ ,

\begin{equation*} R_{1}^{*} \left ( \pi \right ) \geq R_{1}^{*} \left ( 1 \right ) = \frac {1}{\beta _{H}} + \alpha = \frac {1+\alpha \beta _{H}}{\beta _{H}} \gt \frac {1+\alpha \beta _{H}}{\beta _{H}}\frac {1+\beta _{H}}{1+\beta _{H}+1-\alpha } = \frac {1+\beta _{H}}{\beta _{H}}\frac {1+\alpha \beta _{H}}{2-\alpha +\beta _{H}} \end{equation*}

Therefore, by (A13), $(U_{1}^{H*})^{\prime }\left ( \pi \right ) \lt 0$ .

Let us also show that bequests of the more altruistic agents decrease with $\pi$ . Substituting (A12) into (A11), we obtain

\begin{equation*} \frac {(b_{1}^{i*})^{\prime }}{b_{1}^{i*}} = \frac {(R_{1}^{*})^{\prime }}{R_{1}^{*}} \left [ \frac {\beta _{i}^2 R_{1}^{*}}{(\beta _{i}+1-\beta _{i}R_{1}^{*})(\beta _{i}R_{1}^{*} - 1)} -\frac {1}{1-\alpha } \right ] \end{equation*}

Since $(R_{1}^{*})^{\prime } \lt 0$ , we have $(b_{1}^{i*})^{\prime } \lt 0$ if and only if $(\beta _{i}R_{1}^{*}-1)(\beta _{i}+1-\beta _{i}R_{1}^{*}) \lt (1-\alpha ) \beta _{i}^2 R_{1}^{*}$ , which can be rewritten as

(A14) \begin{equation} (R_{1}^{*})^2 - 2 R_{1}^{*} \left ( \frac {\alpha }{2} + \frac {1}{\beta _{i}} \right ) + \frac {1}{\beta _{i}} + \frac {1}{\beta _{i}^2} \gt 0 \end{equation}

The largest root of the quadratic equation $\beta _{i}^2 x^2 - (\alpha \beta _{i}^2 + 2\beta _{i}) x + 1 + \beta _{i} = 0$ is given by

\begin{equation*} x^{+}_{i} = \frac {\alpha }{2} + \frac {1}{\beta _{i}} + \sqrt { \frac {\alpha ^2}{4} - \frac {1-\alpha }{\beta _{i}}} \lt \alpha + \frac {1}{\beta _{i}} \end{equation*}

For $i = H$ , we have $R_{1}^{*} (\pi ) \geq R_{1}^{*} (1) = 1/\beta _{H} + \alpha \gt x^{+}_{H}$ , and hence (A14) holds, meaning that $(b_{1}^{H*})^{\prime } \lt 0$ for all $\pi$ .

(2) Consider now the steady-state equilibrium $\left (c_{2}^{L*},b_{2}^{L*}, c_{2}^{H*},b_{2}^{H*},k^{*}_{2}\right )$ . For $\pi \geq \overline {\pi }$ , the utilities of the less and the more altruistic agents are given by $U_{2}^{L*}$ and $U_{2}^{H*}$ respectively. Recall also that

\begin{equation*} (R_{2}^{*})^{\prime } = -\frac {\gamma }{\left ( 1+\gamma \pi \right ) ^{2}} \qquad \text { and } \qquad \frac {(k^{*}_{2})^{\prime }}{k^{*}_{2}} = \frac {1}{\left ( 1+\gamma \pi \right ) ^{2}}\frac {1}{\alpha R_{2}^{*}} \gt 0 \end{equation*}

(2.1) Less altruistic agents leave no bequests and consume their wages, so that

\begin{equation*} U_{2}^{L*}=\ln c_{2}^{L*}+\beta _{L}\ln \left ( c_{2}^{L*} + b_{2}^{L*} \right ) = \left ( 1+\beta _{L}\right ) \ln c_{2}^{L*} = \left ( 1+\beta _{L}\right ) \ln \left [ (1-\alpha ) A(k^{*}_{2})^{\alpha } \right ] \end{equation*}

Since $k^{*}_{2}$ strictly increases with $\pi$ , $U_{2}^{L*}\left ( \pi \right )$ also strictly increases with $\pi$ : $(U_{2}^{L*})^{\prime }\left ( \pi \right ) \gt 0$ for all $\pi \geq \overline {\pi }$ .

(2.2) For the more altruistic agents, we have

(A15) \begin{equation} \frac {(b_{2}^{H*})^{\prime }}{b_{2}^{H*}}=\frac {(k^{*}_{2})^{\prime }}{k^{*}_{2}}-\frac {1}{\pi }\ =-\frac {1}{\pi R_{2}^{*}}\left [ R_{2}^{*}-\frac {\pi }{\alpha \left ( 1+\gamma \pi \right ) ^{2}}\right ] \end{equation}

Note that $b_{2}^{H*}$ strictly decreases with $\pi$ . Indeed, $(b_{2}^{H*})^{\prime }\lt 0$ because

\begin{equation*} R_{2}^{*}-\frac {\pi }{\alpha \left ( 1+\gamma \pi \right ) ^{2}}=\frac {1}{\beta _{H}}+\frac {1}{1+\gamma \pi }-\frac {\pi \left ( 1+\gamma \right ) }{\left ( 1+\gamma \pi \right ) ^{2}}=\frac {1}{\beta _{H}}+\frac {1-\pi }{\left ( 1+\gamma \pi \right ) ^{2}}\gt 0 \end{equation*}

Moreover,

(A16) \begin{equation} \frac {c_{2}^{H*}}{b_{2}^{H*}}=\frac {1+\gamma \pi }{\beta _{H}} \end{equation}

and hence

(A17) \begin{equation} \frac {(c_{2}^{H*})^{\prime }}{c_{2}^{H*}} = \frac {(b_{2}^{H*})^{\prime }}{b_{2}^{H*}}+\frac {\gamma }{1+\gamma \pi } = -\frac {1}{\pi \left ( 1+\gamma \pi \right ) R_{2}^{*}}\left [ R_{2}^{*}-\frac {\pi }{\alpha \left ( 1+\gamma \pi \right ) }\right ] \end{equation}

Therefore, using (A8), (A15), (A16) and (A17), and taking into account (10), we obtain

\begin{eqnarray*} (U_{2}^{H*})^{\prime } &=&\frac {(c_{2}^{H*})^{\prime }}{c_{2}^{H*}}+\beta _{H}\frac {(c_{2}^{H*})^{\prime }+(b_{2}^{H*})^{\prime }}{c_{2}^{H*}+b_{2}^{H*}}=\frac {b_{2}^{H*}}{c_{2}^{H*}+b_{2}^{H*}}\left ( \frac {(c_{2}^{H*})^{\prime }}{c_{2}^{H*}}\left [ 1+\left ( 1+\beta _{H}\right ) \frac {c_{2}^{H*}}{b_{2}^{H*}}\right ] +\beta _{H}\frac {(b_{2}^{H*})^{\prime }}{b_{2}^{H*}}\right ) \\ &=&-\frac {1}{\pi }\frac {b_{2}^{H*}}{c_{2}^{H*}+b_{2}^{H*}}\left [ \frac {1}{\beta _{H}}+\frac {1+\left ( 1-\pi \right ) \left ( 1+\beta _{H}\right ) }{1+\pi \gamma }\right ] \lt 0 \end{eqnarray*}

Hence $(U_{2}^{H*})^{\prime } \left ( \pi \right ) \lt 0$ for any $\pi \geq \overline {\pi }$ .

Summing up the results for both steady states, we conclude the following.

(1) Less altruistic agents.

(1.1) Suppose that $\delta \lt \min \{ \alpha, \check {\delta } \}$ . Since $\delta \lt \alpha$ , we have $\overline {\pi } \gt 1$ , and hence only the steady state where both types of agents leave positive bequests is possible, so the utility is given by $U^{L*} = U_{1}^{L*}$ . Since $\delta \lt \check {\delta }$ , $(U_{1}^{L*})^{\prime }\left ( \pi \right ) \lt 0$ for all $\pi$ , and therefore $U^{L*}$ strictly decreases with $\pi$ for all $\pi$ .

(1.2) Suppose that $\min \{ \alpha, \check {\delta } \} \lt \delta \lt \min \{ 1, \hat {\delta } \}$ . When $\check {\delta } \lt \delta \lt \alpha$ , the utility is given by $U^{L*} = U_{1}^{L*}$ for all $\pi$ , while when $\alpha \lt \delta \lt 1$ , the utility is given by $U^{L*} = U_{1}^{L*}$ for $\pi \lt \overline {\pi }$ , and $U^{L*} = U_{2}^{L*}$ for $\pi \geq \overline {\pi }$ . In both cases, there is a threshold $\check {\pi }$ defined as follows:

\begin{equation*} \check {\pi } \equiv \overline {\pi } \quad \text {if} \quad \alpha \leq \overline {\alpha } \qquad \text {and} \qquad \check {\pi } \equiv \hat {\pi } \lt \overline {\pi } \quad \text {if} \quad \alpha \gt \overline {\alpha } \end{equation*}

such that $U^{L*}$ is continuous at $\pi = \check {\pi }$ , strictly decreases with $\pi$ for all $\pi \lt \check {\pi }$ , and strictly increases with $\pi$ for all $\pi \gt \check {\pi }$ .

(1.3) Suppose that $\delta \gt \min \{ 1, \hat {\delta } \}$ . When $\hat {\delta } \lt \delta \lt 1$ , the utility is given by $U^{L*} = U_{1}^{L*}$ for $\pi \lt \overline {\pi }$ , and $U^{L*} = U_{2}^{L*}$ for $\pi \geq \overline {\pi }$ , while when $\delta \gt 1$ , the utility is given by $U^{L*} = U_{2}^{L*}$ for all $\pi$ . Since $\delta \gt \hat {\delta }$ , we have $(U_{1}^{L*})^{\prime } \left ( \pi \right ) \gt 0$ , and it is always the case that $(U_{2}^{L*})^{\prime } \left ( \pi \right ) \gt 0$ . Therefore, $U^{L*}$ strictly increases with $\pi$ for all $\pi$ .

(2) More altruistic agents.

In all possible cases, we have $(U_{1}^{H*})^{\prime } \left ( \pi \right ) \lt 0$ , and $(U_{2}^{H*})^{\prime } \left ( \pi \right ) \lt 0$ , and, therefore, $U^{H*}$ strictly decreases with $\pi$ for all $\pi$ .