Euler–Lagrange equations of stochastic differential games: application to a game of a productive asset

Abstract

This paper analyzes a noncooperative and symmetric dynamic game where players have free access to a productive asset whose evolution is a diffusion process with Brownian uncertainty. A Euler–Lagrange equation is found and used to provide necessary and sufficient conditions for the existence and uniqueness of a smooth Markov Perfect Nash Equilibrium. The Euler–Lagrange equation also provides a stochastic Keynes–Ramsey rule, which has the form of a forward–backward stochastic differential equation. It is used to study the properties of the equilibrium and to make some comparative statics exercises.

Appendix: Proofs

In this appendix, we collect the proofs of the results that are not shown in the body text. It also include some auxiliary results that are used in the proofs.

1.1 Proof of Theorem 2

Let $R_T=[0,T]\times (0,\infty )$. We first state an auxiliary result.

Existence of solutions to quasilinear parabolic PDEs.

Theorem 13

There exists at least one bounded solution in $R_T$ of the Cauchy problem

$$\begin{aligned} u_{\tau }-\frac{\partial }{\partial x} a(\tau ,x,u,u_x) = b(\tau ,x,u,u_x), \end{aligned}$$

with initial condition

$$\begin{aligned} u(0,x)=u_0(x),\quad x>0. \end{aligned}$$

if all of the following conditions are satisfied.

C1.:

$u_0$ is of class $C^4$ and bounded.

C2.:

Functions $a$ and $b$ are of class $C^3$ and $C^2$, respectively.

C3.:

There are non-negative constants $b_1$ and $b_2$ such that for all $x$ and $u$

$$\begin{aligned} \left( b(\tau ,x,u,0)+\frac{\partial }{\partial x} a(\tau ,x,u,0)\right) u\le b_1 u^2+b_2. \end{aligned}$$

C4.:

For all $M>0$, there are constants $\mu _2(M)\ge \mu _1(M)>0$ such that, if $\tau , x$ and $u$ are bounded in modulus by $M$, then for arbitrary $p$

$$\begin{aligned} \mu _1(M)\le \frac{\partial }{\partial p} a(\tau ,x,u,p) \le \mu _2(M) \end{aligned}$$

and

$$\begin{aligned} \left( |a|+\left| \frac{\partial a}{\partial u}\right| \right) (1+|p|)+\left| \frac{\partial a}{\partial x}\right| +|b|\le \mu _2(M)(1+|p|)^2. \end{aligned}$$

The problem admits no more than a classical solution in $R_T$ that is bounded together with its derivatives of first and second orders if the following additional conditions hold.

C5.:

For all $M>0$ there are non-negative constants $\nu _1(M)$ and $\nu _2(M)$ such that

$$\begin{aligned}&\max _{\begin{array}{c} (t,x)\in R_T\\ |u|, |p|\le M \end{array}}\left| \frac{\partial ^2 a}{\partial p\partial u}, \frac{\partial ^2 a}{\partial p^2}, \frac{\partial A}{\partial p}\right| \le \nu _1(M),\\&\min _{\begin{array}{c} (t,x)\in R_T\\ |u|, |p|\le M \end{array}} \frac{\partial A}{\partial u}\ge -\nu _2(M), \end{aligned}$$

where

$$\begin{aligned} A=a-\frac{\partial a}{\partial u} p - \frac{\partial a}{\partial x}. \end{aligned}$$

Proof

The result is a consequence of Theorem 8.1 of Ladyzhenskaya et al (1969), the only difference being that we have set the problem in $[0,T]\times (0,\infty )$ instead of $[0,T]\times {\mathbb {R}}$ and that we require more smoothness. The method of proof of Theorem 8.1 in Ladyzhenskaya et al consists in considering truncated problems on the strip $[0,T]\times [1/n,n]$ with boundary conditions $u_n(0,x)=u_0(x)$ for all $x\in [1/n,n]$ and¹² $u_n(t,1/n)=u_0(1/n), u_n(t,n)=u_0(n)$ for all $t\ge 0$. These solutions converge smoothly to a solution $u$ of the original equation as $n\rightarrow \infty$. $\square$

Motivated by the necessity to drop the boundedness of the data defining the game, we will consider the EL Eq. (10) for the function $u=\phi /f$ written in the divergence form

$$\begin{aligned} u_{\tau } - \frac{{\partial }}{{\partial }x} a(x,u_x) = b(x, u, u_x), \end{aligned}$$

(39)

where $\tau =T-t$ and

$$\begin{aligned} a(x,p)&=\frac{\sigma (x)^2}{2} p,\\ b(x, u, p)&= \Big ( F -Nu f +(N-1) {R}(u f )+\frac{f' }{f } \sigma ^2 \Big )p\\&\quad + \Big ( F -Nu f +(N-1) {R}(u f )+ \sigma ' \sigma \Big )u \frac{f' }{f }\\&\quad +\frac{1}{2} \sigma ^2 u \frac{f'' }{f } -\frac{\sigma ^2}{2} \frac{ {P}(u f )}{f} (pf+u f')^2+\frac{r-F'}{f} {R(u f)}. \end{aligned}$$

The initial condition is

$$\begin{aligned} u(0,x)=\varphi _0 (x)=\frac{\varphi (x)}{f(x)}. \end{aligned}$$

(40)

Note that even if the initial condition $\varphi _0$ is now bounded by assumption, still we cannot apply the above theorem directly. The difficulties are two: (i) the function $\sigma$ vanishes at $x=0$, thus it is not uniformly bounded away from zero; and (ii) the function $P$ is in general not defined at $0$ and in fact $\lim _{c\rightarrow 0^+} P(c) = \infty$ for problems with CRRA utility, where $P(c)=(1+\delta )/c$. To deal with (i) we consider the PDE (39) on bounded subintervals $I_n=[1/n,n], n=1,2, \ldots$, and then we take a limit as $n\rightarrow \infty$, while for (ii) we will prove that the solutions $u_n$ found in the subintervals above remain uniformly bounded away from 0, in the sense that there exists a lower bound $l_m>0$ such that $u_n\ge l_m$ for all $n\ge m$ for all $x\in I_m$. As a byproduct of the proof, we obtain the estimates claimed in the theorem.

• C.1. is fulfilled, since $\varphi _0(x)=\varphi (x)/f(x)$ is bounded and smooth on $(0,\infty )$, by assumption.

• C.2. holds, as the function $a$ has the required smoothness. As explained above, we will prove below that $u$ never vanishes on $(0,\infty )$, thus the term $P(uf)$ does not pose any problem at all for the smoothness of function $b$.

• C.3. There are constants $b_1$ and $b_2$ such that

  $$\begin{aligned} b(x,u,0) u&= \Big (F - Nu f +(N-1) {R}(u f )+ \sigma ' \sigma \Big )u^2 \frac{f' }{f } \\&\quad +\frac{1}{2} \sigma ^2 u^2 \frac{f'' }{f } -\frac{\sigma ^2}{2} \frac{ {P}(u f )}{f} (u f')^2 u+\frac{r-F'}{f} {R(u f)}u \le b_1u^2+b_2 \end{aligned}$$

  (since $a(x,0)=0$). To see this, note that, thanks to our assumptions,

  $$\begin{aligned} b(x,u,0)u&\le \alpha ^+u^3f' + \left( (F +\sigma ' \sigma )\frac{f'}{f} +\frac{1}{2} \sigma ^2\frac{f''}{f} - \frac{\pi ^{-}}{2}\sigma ^2\left( \frac{f'}{f}\right) ^2 + \gamma ^+\right) u^2\\&< \alpha ^+(\min _{x\in (0,\infty )}f') u^3 + \beta ^+u^2\le \beta ^+u^2, \end{aligned}$$

  since $\alpha ^+< 0$ and the solution $u>0$ (this will be proved below).

• C.4. first part, is also fulfilled, as

  $$\begin{aligned} \frac{\partial a(x,p)}{\partial p} = \frac{\sigma ^2(x)}{2} \end{aligned}$$

  is positive for $x>0$ and continuous, thus it is bounded away from 0, as well as bounded above in any compact subset $[1/M,M]$ of $(0,\infty )$.¹³ The second part of C.4 is a local assumption, that also holds because function $|b|$ is quadratic in $p$, with continuous coefficients, and thus bounded on compact subsets of $(0,\infty )$; the same is true for

  $$\begin{aligned} \left( |a|+\left| \frac{\partial a}{\partial u}\right| \right) (1+|p|)+\left| \frac{\partial a}{\partial x}\right| = \frac{\sigma ^2(x)}{2} |p| (1+|p|) + |\sigma (x)\sigma '(x)||p|, \end{aligned}$$

  which is also quadratic in $p$, and since both $\sigma$ and $\sigma '$ are continuous and thus they are bounded in compact subsets of $[0,\infty )$.

• C.5. Both $\sigma$ and $\sigma '$ are continuous, and therefore bounded over any compact interval of the state space, achieving uniqueness of the bounded solution

We now show that we can construct a solution $u>0$ of the PDE (39) as limit of positive truncated solutions $u_n$ in $[0,T]\times [1/n,n]$ as $n\rightarrow \infty$. Then, C.2 holds, as $P$ is smooth in $(0,\infty )$. In fact we prove more than that, as we obtain upper and lower estimates for the solution. The latter will imply in particular that the solution is positive for $x>0$.

Let the PDE (39) with initial condition (40) and boundary conditions at the extreme points of the interval $I_n$ given by

$$\begin{aligned} \begin{aligned} u_n(\tau ,1/n)&= \frac{\beta ^+e^{\tau \beta ^+}\varphi _0(1/n)}{\alpha ^+\varphi _0(1/n) f'(1/n)(1-e^{\tau \beta ^+})+\beta ^+},\\ u_n(\tau ,n)&= \frac{\beta ^+e^{\tau \beta ^+}\varphi _0(n)}{\alpha ^+\varphi _0(n) f'(n)(1-e^{\tau \beta ^+})+\beta ^+}. \end{aligned} \end{aligned}$$

(41)

The reason for this particular selection of the boundary conditions is shown next. Formulating these approximating problems, we have eliminated the degeneration in the truncated equation. It is easy to check that all conditions in Theorem 8.1 of Ladyzhenskaya et al. (1968) are fulfilled in a small neighborhood of $\tau = 0$, since $\varphi _0(x)$ is bounded away from zero in $I_n$ and hence a solution $u_n$ of the truncated problem exists which does not vanish. From this, our aim is to extend this truncated solution to all $[0,T]$ and after this, to get the solution $\phi (t,x)$ with initial condition $\phi (0,x)=\varphi (x)$ as the smooth limit of $u_n(\tau ,x) f(x)$ as $n\rightarrow \infty$. As explained above, this is the procedure used in Ladyzhenskaya et al. (1968). We show next that the local solution can be extended in time and space. To this end, we define

$$\begin{aligned} M_n(\tau )=\max _{y\in [1/n,n]}u_n(\tau ,y). \end{aligned}$$

By Danskin’s Theorem, function $M_n$ is almost everywhere differentiable, and at points of differentiability, the derivative $\dot{M}_n(\tau )=u_{n,\tau }(\tau ,x_n(\tau ))$, where $u_n(\tau ,x_n(\tau ))=M_n(\tau )$.

We prove that for any $\tau$

$$\begin{aligned} M_n(\tau ) \le \frac{\beta ^+e^{\tau \beta ^+}M_n(0)}{\alpha ^+(\min _{I_n}f') M_n(0)(1-e^{\tau \beta ^+})+\beta ^+}. \end{aligned}$$

(42)

where $M_n(0)=\sup _{x\in I_n}\varphi _0(x)$. Suppose, by way of contradiction, that $M_n(\tau _0)$ is greater that the right-hand side of (42) for some $\tau _0>0$. Let $\overline{ \tau }_0$ be the inferior of all the $\tau _0$s satisfying this property, and hence, by continuity of $M_n, M_n(\overline{\tau }_0)$ equals the right-hand side of (42). Then $x_n(\tau )$ is interior to $I_n$ for every $\tau \in [\overline{\tau }_0, \tau _0]$, due to the boundary conditions (41). In consequence, $u_{n,x}(\tau ,x_n(\tau ))=0$ and $u_{n,xx}(\tau ,x_n(\tau )) \le 0$ for any $\tau \in [\overline{\tau }_0, \tau _0]$. Hence, $(\partial /\partial x) (\sigma ^2 u_{n,x})|_{(\tau , x_n(\tau ))} \le 0$. This information, used in the Eq. (39) for $\nu _n$, provides the following chain of inequalities

$$\begin{aligned} \begin{aligned} \dot{M}_n(\tau )&\le b( x_n(\tau ), M_n (\tau ),0)\\&= \Big ( F-NfM_n (\tau ) +(N-1) {R}(fM_n (\tau )) +\sigma ' \sigma \Big )M_n (\tau )\frac{f'}{f}\\&\quad +\frac{1}{2} \sigma ^2 M_n(\tau )\frac{f''}{f} -\frac{1}{2} \sigma ^2 \frac{ {P}(fM_n(\tau ) )}{f} (M_n (\tau )f')^2 + \frac{r-F'}{f} {R}(fM_n (\tau )) \\&\le \alpha ^+M_n^2(\tau )f' + \Big ( (F +\sigma ' \sigma )\frac{f'}{f} \!+\!\frac{1}{2} \sigma ^2\frac{f''}{f} - \frac{\pi ^{-}}{2}\sigma ^2\left( \frac{f'}{f}\right) ^2\!+\!\gamma ^+\Big )M_n (\tau )\\&< \alpha ^+(\min _{I_n}f') M_n^2(\tau ) + \beta ^+M_n(\tau ), \quad \text {a.e. }\tau . \end{aligned} \end{aligned}$$

(43)

We have used the definitions of $\alpha ^+$ and $\beta ^+$ done in Assumption (A3) and (A5), respectively, and the fact that $\alpha ^+< 0$. From this differential inequality of Ricatti we get the estimate

$$\begin{aligned} M_n(\tau ) \le \frac{\beta ^+e^{(\tau -\overline{\tau }_0) \beta ^+}M_n(\overline{\tau }_0)}{\alpha ^+(\min _{I_n}f') M_n(\overline{\tau }_0)(1-e^{(\tau -\overline{\tau }_0) \beta ^+})+\beta ^+}. \end{aligned}$$

Once the expression for $M_n(\overline{\tau }_0)=\frac{\beta ^+e^{\overline{\tau }_0 \beta ^+}M_n(0)}{\alpha ^+(\min _{I_n}f') M_n(0)(1-e^{\overline{\tau }_0\beta ^+})+\beta ^+}$, which holds by our contradiction argument is substituted into (A.1) inequality (42) easily follows.

Following the same technique, we get the lower estimate. Consider the solution $v_n$ of the PDE with the boundary conditions

$$\begin{aligned} v_n(\tau ,1/n)&= \frac{\beta ^{-}e^{\tau \beta ^{-}}\varphi _0(1/n)}{\alpha ^-\varphi _0(1/n)f'(1/n)(1-e^{\tau \beta ^{-}})+\beta ^{-}},\\ v_n(\tau ,n)&= \frac{\beta ^{-}e^{\tau \beta ^{-}}\varphi _0(n)}{\alpha ^-\varphi _0(n) f'(n)(1-e^{\tau \beta ^{-}})+\beta ^{-}}. \end{aligned}$$

and

$$\begin{aligned} m_n(\tau )=\min _{y\in [1/n,n]}v_n(\tau ,y), \end{aligned}$$

with the minimum attained at some $y_n(\tau )$. Similar arguments as done above for the maximum lead to the estimate

$$\begin{aligned} m_n(\tau ) \ge \frac{\beta ^{-}e^{\tau \beta ^{-}}m_n(0)}{\alpha ^-m_n(0) (\max _{I_n} f') (1-e^{\tau \beta ^{-}})+\beta ^{-}}. \end{aligned}$$

(44)

as follows. By way of contradiction one finds

$$\begin{aligned} \dot{m}_n(\tau )\ge g(y_n(\tau ), m_n (\tau ),0)\ge \alpha ^-(\max _{I_n} f') m_n^2(\tau ) +\beta ^{-}m_n(\tau ), \quad \text {a.e. }\tau . \end{aligned}$$

Reasoning much as for the case of $M_n$, one gets the estimate (44) easily. Thus, we have shown that the local solution is strictly uniformly bounded away from zero in the intervals $I_n$ and that it is also bounded above. Since this fact is independent of $\tau$, as well as the upper bound obtained above, the solution can be extended up to the whole $[0,T]$, for any $T$.

Hence, since $v_n\rightarrow v$, we have that $v$ is bounded away from zero, and hence $v$ is a solution of the Cauchy problem, since all the conditions of the Theorem of Ladyzhenskaya et al. (1968) are fulfilled. Taking limits as $n\rightarrow \infty$ one has

$$\begin{aligned} \frac{\phi (t,x)}{f(x)} = u(T-t,x) \ge m(T-t) \ge \frac{\beta ^{-}e^{(T-t) \beta ^{-}}m(0)}{\alpha ^-m(0)(\sup _{(0,\infty )} f') (1-e^{(T-t)\beta ^{-}})+\beta ^{-}}. \end{aligned}$$

By the above estimates the limit $\phi (t,x)= u(T-t,x)f(x)$ is a solution of the Cauchy problem (10) satisfying

$$\begin{aligned} \frac{\phi (t,x)}{f(x)}= v(T-t,x) \le M(T-t) \le \frac{\beta ^+e^{(T-t) \beta ^+}M}{\alpha ^+(\inf _{(0,\infty )}f') M(1-e^{(T-t)\beta ^+})+\beta ^+}, \end{aligned}$$

since $M_n(0)\rightarrow M$ as $n\rightarrow \infty$, and $\inf _{(0,\infty )} f'\le \min _{I_n} f'$.

1.2 Proof of Theorem 7

With CRRA preferences, $\rho ^{+}=\rho ^{-}=\rho =1/\delta , \alpha ^+=\alpha ^-=\alpha$ and $\pi ^{+}=\pi ^{-}=b=1+\delta$. See the definition of these constants in Sect. 3, Assumption (A3) and in (12). See also the computations done after Definition 4. We will follow the proof of Theorem 2, using the same notation. Retaking inequality (43) in the aforementioned proof, in the case of CRRA preferences we have

$$\begin{aligned} \begin{aligned} \dot{M}_n(\tau )&\le b( x_n(\tau ), M_n (\tau ),0)\\&= \Big ( F-NfM_n (\tau ) +(N-1) afM_n (\tau ) +\sigma ' \sigma \Big )M_n (\tau )\frac{f'}{f}\\&\quad +\frac{1}{2} \sigma ^2 M_n(\tau )\frac{f''}{f} -\frac{1}{2} \sigma ^2 \frac{b}{f^2M_n(\tau )} (M_n (\tau )f')^2 + \frac{r-F'}{f}M_n (\tau )a\\&\le \alpha f' M_n^2(\tau )\! +\! \Big ( (F +\sigma ' \sigma ) f' \!+\!\frac{1}{2} \sigma ^2 {f''} \!-\!\frac{1}{2} \sigma ^2 \frac{b}{f} f'^2 \!+\! a (r\!-\!F')f\Big ) \frac{M_n (\tau )}{f} \end{aligned} \end{aligned}$$

Since $f$ is solution of the stationary EL Eq. (10) with CRRA preferences, then

$$\begin{aligned} ( F +\sigma ' \sigma ) {f'} +\frac{1}{2} \sigma ^2 {f''} -\frac{1}{2} \sigma ^2 \frac{b}{f} f'^2 + (r-F')af= -\alpha f f'. \end{aligned}$$

Plugging this into the inequality above it simplifies to

$$\begin{aligned} \dot{M}_n(\tau ) \le \alpha f' M_n^2(\tau )-\alpha f' M_n(\tau )\le \alpha (\inf _{I_n} f') M_n^2(\tau ) - \alpha (\sup _{I_n} f') M_n(\tau ). \end{aligned}$$

Hence

$$\begin{aligned} M_n(\tau ) \le \frac{M_n(0) e^{-\tau \alpha (\sup _{I_n}f')}}{- M_n(0) (1-e^{-\tau \alpha (\sup _{I_n}f')})+1}. \end{aligned}$$

A similar computation for the minimum shows

$$\begin{aligned} m_n(\tau ) \ge \frac{m_n(0) e^{-\tau \alpha (\inf _{I_n}f')}}{- m_n(0) (1-e^{-\tau \alpha (\inf _{I_n}f')})+1}. \end{aligned}$$

Notice that in the former case $f'$ is evaluated at the point where $u_n$ attains a maximum in $I_n=[1/n,n]$, say $x_n$, whereas in the latter case it is at the point where $u_n$ attains a minimum, say $y_n$. Thus,

$$\begin{aligned} \frac{m_n(0) e^{-\tau \alpha (\inf _{I_n}f')}}{- m_n(0) (1-e^{-\tau \alpha (\inf _{I_n}f')})+1}\le u_n(t,x)\le \frac{M_n(0) e^{-\tau \alpha (\sup _{I_n}f') }}{- M_n(0) (1-e^{-\tau \alpha (\sup _{I_n}f')})+1}. \end{aligned}$$

Taking the limit as $n$ tends to $\infty$ and since $u_n(t,x)\rightarrow \phi (t,x)/f(x)$ as $n\rightarrow \infty$, we find

$$\begin{aligned} \frac{m e^{-\tau \alpha (\inf _{I_n} f')}}{- m (1-e^{-\tau \alpha (\inf _{I_n} f')})+1}\le \frac{\phi (\tau ,x)}{f(x)}\le \frac{M e^{-\tau \alpha (\sup _{I_n} f')}}{- M (1-e^{-\tau \alpha (\sup _{I_n} f')})+1}. \end{aligned}$$

As $T \rightarrow \infty , \tau \rightarrow \infty$ and one finally find that $\phi (t,x)$ converges to $f(x)$.

1.3 Proof of Theorem 8

We follow the same scheme of proof as in Theorems 2 and 7. Notice that the hypotheses of the theorem imply $\varphi '(x) = S''(x)/L''(S'(x)) \ge 0$. Now, derive the EL Eq. (10) with respect to $x$ and let $v=\phi _x$. Then $w$ solves the Cauchy problem

$$\begin{aligned} v_{\tau } -\frac{1}{2} \frac{{\partial }}{{\partial }x} (\sigma (x)^2 v_x)= & {} g_x(x,\phi ,v) + v g_c(x,\phi ,v) + v_x g_v(x,\phi ,v)\\&\quad + v \sigma '(x)^2 + v\sigma (x) \sigma ''(x) + v_x \sigma (x)\sigma '(x). \\ v(0,x)= & {} \varphi '(x)\ge 0. \end{aligned}$$

Function $g$ is

$$\begin{aligned} g(x,c,v) = (F(x) - N c + (N-1) {R}(c))v - \frac{1}{2} \sigma (x)^2 {P} (c) v^2 + (r - F'(x)) {R}(c). \end{aligned}$$

We will follow a similar strategy of proof as in Theorem 2, considering truncated intervals for the variable $x, I_n=[1/n,n]$ and the solution $v_n(\tau ,x)$ in $[0,T]\times I_n$ satisfying the boundary conditions

$$\begin{aligned} v_n(\tau ,1/n) = \varphi '(1/n)>0 ,\quad v_n(\tau ,n) = \varphi '(n)>0. \end{aligned}$$

Let $\nu _n(\tau ) = \min _{x\in I_n} v_n(\tau ,x)$. A reasoning by contradiction, assuming the existence of $\tau _0$ satisfying $\nu _n(\tau _0) < \nu _n(0)$, will lead to a contradiction as follows. Let $\overline{ \tau }_0$ be the inferior of all the $\tau$s satisfying this property; by continuity of $\nu _n, \nu _n(\overline{\tau }_0)=\nu _n(0)$. Then, we obtain the inequality

$$\begin{aligned} \dot{\nu }_n \ge g_x(x,\phi ,\nu _n)+\nu _n \big (g_c(x,\phi _n,\nu _n)+\sigma '(x)^2+\sigma (x)\sigma ''(x)\big ). \end{aligned}$$

We have used the same arguments as those used in the proof of Theorem 2, hence we do not repeat it here. The notation $\phi _n$ is used for the solution of the EL Eq. (10) in $[0,T]\times I_n$. The term $\phi _n(\tau ,\zeta (\tau ))$, where $\zeta (\tau )$ minimizes $v_n$ over $I_n$, does not pose any problem at all. Given that $g_x(x,c,v) = v\big (F'(x)-v\sigma (x)\sigma '(x) {P}(c)\big )-F''(x) {R}(c),$ we have

$$\begin{aligned} \dot{\nu }_n&\ge - F''(x) {R}(\phi _n)\\&\quad + \nu _n \big (F'(x)-\nu _n\sigma (x)\sigma '(x) {P}(\phi _n) + g_c(x,\phi _n,\nu _n)+\sigma '(x)^2+\sigma (x)\sigma ''(x)\big ), \end{aligned}$$

with $\nu _n(\overline{\tau })=0$. Since $F''$ is concave and ${R}$ is non-negative, $-F''(x) {R}(c)\ge 0$, thus

$$\begin{aligned} \nu _n(\tau ) \ge \nu _n (\overline{\tau }_0) e^{\int _{\overline{\tau }_0}^{\tau } \{\cdots \} \, d\tau } = 0\quad \forall \tau \in [\overline{\tau }_0,\tau _0], \end{aligned}$$

in contradiction with $\nu _n(\tau _0) < 0$. Hence $0\le \nu _n(\tau )\le v_n(\tau ,x)$ for all $\tau , x\in I_n$, and then the limit function $\phi _x(\tau ,x)\ge 0$.

1.4 Proof of Theorem 9

We first establish a lemma about the concavity of the consumption rate at the final time $T$. It is established for general $L$, not only in the class CRRA. In the lemma, $\rho _{\{\}}$ stands for the elasticity of the marginal utility and $\pi _{\{\}}$ for the relative prudence index of a given utility function. For the linear game, the lemma implies that $\varphi$ is concave if and only if the bequest function $S$ satisfies

$$\begin{aligned} \frac{S'(x) S'''(x)}{{S''}^2(x)} \ge 1+\frac{1}{\delta }. \end{aligned}$$

Lemma 3

$\varphi ''\le 0$ if and only if for all $x> 0$

$$\begin{aligned} \rho _S(x)\pi _S(x) \ge \rho _L(\varphi (x))\pi _L(\varphi (x)). \end{aligned}$$

Proof

Deriving twice in $L'(\varphi (x)) = S'(x)$ we get

$$\begin{aligned}&L''(\varphi (x))\varphi '(x) = S''(x),\\&L'''(\varphi (x))\varphi '(x)^2 + L''(\varphi (x))\varphi ''(x) = S'''(x). \end{aligned}$$

Solving for $\varphi ''(x)$ and imposing $\varphi ''(x)\le 0$ we obtain the inequality (we eliminate arguments)

$$\begin{aligned} \frac{S'''}{S''^2} \ge \frac{L'''}{L''^2} \end{aligned}$$

or equivalently, multiplying both sides of the inequality by $S'>0$

$$\begin{aligned} \left( \frac{-xS'''}{S''}\right) \left( \frac{-S'}{xS''}\right) \ge \left( \frac{-\varphi L'''}{L''} \right) \left( \frac{-S'}{\varphi L''}\right) . \end{aligned}$$

Noting that $S'=L'(\varphi )$, and plugging this equality into the right-hand side of the inequality above, we obtain the claim of the lemma. $\square$

We now proceed with the proof of Theorem 9 with the same techniques as those used in the proofs of Theorems 2 and 8, deriving twice in the EL Eq. (10) to find a PDE for $\phi _{xx}$. We refer the reader to the proofs of those theorems for filling in the missing details. Deriving twice in (10) we get

$$\begin{aligned} (\phi _{xx})_{\tau }-\frac{\partial }{\partial x} \left( \frac{\sigma ^2 x^2}{2} \phi _{xxx}\right) = j(x,\phi ,\phi _x,\phi _{xx},\phi _{xxx}) \end{aligned}$$

where the function $j$ is defined as

$$\begin{aligned} j(x,c,v,w,z)&=w\left( 3\sigma ^2+2F'(x)+3\alpha v +\frac{r-F'(x) }{\delta } - 4\sigma ^2 x \frac{1+\delta }{c} v \right. \\&\quad \left. +\frac{5}{2} \sigma ^2x^2\frac{1+\delta }{c^2} v^2-\sigma ^2x^2\frac{1+\delta }{c}w\right) \\&\quad - \sigma ^2 v^2 \frac{1+\delta }{c} \left( 1-\frac{xv}{c}\right) ^2\\&\quad -F'''(x)\frac{c}{\delta }+F''(x)\left( 1-\frac{2}{\delta }\right) v\\&\quad + \left( 2\sigma ^2 x + F(x) +g_N(c) - \sigma ^2x^2 \frac{1+\delta }{c} v\right) z. \end{aligned}$$

Recall that in the CRRA case, $\alpha =-N+(N-1)/\delta$. Following the same method of proof as in the above referenced theorems, and defining $w_n$ as the solution of the PDE above in the interval $I_n$, we get

$$\begin{aligned} w_{n,\tau } - \frac{\partial }{\partial x} \left( \frac{\sigma ^2 x^2}{2} w_{n,x}\right) = j(x,\phi _n,\phi _{n,x},w_{n},w_{n,x}). \end{aligned}$$

(45)

Here, $\phi _n$ denotes the restriction of $\phi$ to $[0,T]\times I_n$. Let $\omega _n(\tau ) = \max _{I_n} w_n (\tau ,x)$. Reasoning by contradiction supposing that $\omega _n(\tau ) >0$ at some $\overline{\tau }>0$, one has that $w_{n,x}(\overline{\tau })= 0$ and $\frac{\partial }{\partial x} \left( \frac{\sigma ^2 x^2}{2} w_{n,x}\right) \le 0$, thus plugging this into Eq. (45) we get

$$\begin{aligned} \dot{\omega }_n \le \omega _n\left\{ \cdots \right\} - \sigma ^2 \phi _{n,x}^2 \frac{1+\delta }{\phi _n} \left( 1-\frac{x\phi _{n,x}}{\phi _n}\right) ^2 -F'''(x)\frac{\phi _n}{\delta }+F''(x)\left( 1-\frac{2}{\delta }\right) \phi _{n,x}. \end{aligned}$$

In the linear game $F''=F'''=0$, hence the second summand in the above expression is non-positive. Given that $\omega _n(0)=\sup _{[0,\infty )} \varphi ''(x) \le 0$ is also non-positive, we arrive to a contradiction, because it is never possible to have $\omega _n(\overline{\tau }) >0$ from the above estimate for $\dot{\omega }_n$.

In the following proofs, recall that $g_N(c)=-Nc+(N-1)R(c)$.

1.5 Proof of Theorem 10

Let us show that if $\sigma _i\in \varSigma , i=1,2$ and $\sigma _1\le \sigma _2$, then $Y^{\sigma _1}\le Y^{\sigma _2}$ a.s. Clearly, for $\sigma \in \varSigma , \mu (Y(s))$ satisfies the Novikov condition and then

$$\begin{aligned} M(s) = \exp {\left\{ \int _t^s \mu (Y(a)) d(a) - \frac{1}{2} \int _t^s \mu ^2(Y(a))da\right\} }, \quad s\in [t,T] \end{aligned}$$

is a $\mathbf {P}$-martingale, where $\mathbf {P}$ is the objective probability measure. Define now the probability measure $\mathbf {\widetilde{P}}$ by

$$\begin{aligned} \frac{d \mathbf {\widetilde{P}}}{d \mathbf {P}} = M(T). \end{aligned}$$

It is known that $\mathbf {\widetilde{P}}$ is absolutely continuous with respect to $\mathbf {P}$. By Girsanov’s Theorem, $\widetilde{\omega }(s) = \omega (s) - \int _t^s \mu (Y(a)) d(a)$ is a $\mathbf {\widetilde{P}}$-Brownian motion and, in the new measure, $Y$ satisfies

$$\begin{aligned} dY (s) = \frac{1}{2}\sigma (Y(s))\sigma '(Y(s)) ds + \sigma (Y(s)) dw (s). \end{aligned}$$

Now, given that $0<\sigma _1\le \sigma _2$, it holds that $\int _x ^y \frac{dz}{\sigma _1(z)} \ge \int _x ^y \frac{dz}{\sigma _2(z)}$ for all $y\ge x$, which is the sufficient condition of Example 2 of Zhiyuan (1984) assuring $Y^{\sigma _1}\le Y^{\sigma _2}$ for every $s\in [t,T], \mathbf {\widetilde{P}}$-a.s., whence $\mathbf {P}$-a.s. Then $\varphi (Y^{\sigma _1}(T))\le \varphi (Y^{\sigma _2}(T))\,\mathbf {P}$-a.s., since $\varphi$ is non-decreasing by hypothesis.

Now, let us define $f^i(\omega , s, C,Z)$ as the drift term in the backward SDE for $C$ in (36) when the forward process is $Y^{\sigma _i}(s), i=1,2$, that is

$$\begin{aligned} f^i(\omega , s, C,Z) = C \left( \rho (C) (F'(Y^{\sigma _i})-r) + \frac{1}{2} \pi (C) Z^2 - \frac{g_N(C)Z}{\sigma (Y^{\sigma _i})} \right) , \end{aligned}$$

as well as $\varphi ^i(T) = \varphi (Y^{\sigma _i}(T))$. In the above, $\omega$ denotes an element of the sample space $\varOmega$ where the probability $\mathbf {P}$ is defined, in order to stress the dependence of the drift term with respect to the stochastic process $Y^{\sigma _i}$. Let also $(C^1,Z^ 1), (C^2,Z^ 2)$ be solutions of the BSDE (36) when $\sigma = \sigma _i, i=1,2$. We can write (36) in integral form

$$\begin{aligned} C^i(s) = \varphi ^i - \int _s^T f^i(\omega ,v, C(v),Z(v))dv - \int _s^T Z^ i(v) dw(v). \end{aligned}$$

Let us check that $-f^1(\omega , s, C^ 2,Z^ 2)\le -f^2 (\omega , s, C^ 1,Z^ 1)$. Note that there are two terms that depend on $i$ in the definition of $f^i$: one is $- \frac{g_N(C)Z}{\sigma (Y^{\sigma _i})}$; we have already proved that ${\sigma (Y^{\sigma _1})}\le {\sigma (Y^{\sigma _2})}$. Since $\sigma$ is increasing, that $Z\ge 0$ for all $s \, \mathbf {P}$-a.s., and that $g_N(C)\le 0$ (see Sect. 5), then $- \frac{g_N(C)}{\sigma (Y^{\sigma _2})} \le - \frac{g_N(C)}{\sigma (Y^{\sigma _1})}$. The other term is $F'(Y^ {\sigma _i})$; since $F$ is concave, $F'$ is non-increasing, thus $-F'(Y^{\sigma _1}) \le -F'(Y^{\sigma _2})$. Hence, $-f^1(\omega , s, C^ 2,Z^ 2)\le -f^2 (\omega , s, C^ 1,Z^ 1)$, as claimed. Now, the Comparison Theorem 2.2 in Karoui et al. (1997) ensures that $C^1(s) \le C^2(s)$ for all $s, \mathbf {P}$-a.s. Since the process $C^ i$ is deterministic at $(t,x)$ and $C^ i(t)=\phi ^ {\sigma ^ i}(t,x)$, we have that $\phi ^{\sigma _1}(t,x)\le \phi ^{\sigma _2}(t,x)$ and the proof is finished.

1.6 Proof of Theorem 11

To show the result, we will use a representation of the MPNE by means of an FBSDE, alternative to the KR rule introduced in Sect. 5, more amenable for our purposes. Let the process $Y$ that satisfies

$$\begin{aligned} dY =(F(Y) + (\sigma {\sigma }')(Y))\, ds+\sigma (Y)\, d\omega ,\quad Y(t)=x, \end{aligned}$$

where we have omitted the argument $s$ to shorten notation. Then, for $C(s)=\phi (s,Y(s))$ and the square-integrable adapted process ($Z=\sigma (Y(s))\phi _x(s,Y(s))$), where $\phi$ is a solution of the EL Eq. (10) we have, as in Sect. 6.4

$$\begin{aligned} dC (s)= & {} \left( R(C(s)) (F'(Y(s))-r) - \frac{1}{2} {P}({C}(s)) Z^2(s) + \frac{g_N(C(s))}{\sigma (Y(s))}Z(s) \right) ds \\&+ Z(s) dw (s), \end{aligned}$$

and $C(T) = \varphi (Y(T))$. Note that the process $Z$ does not depend on the number of players, $N$. Let

$$\begin{aligned} f^ N (\omega ,s,C,Z) = R(C) (F'(Y(s))-r) - \frac{1}{2} {P}({C}) Z^2 + \frac{g_N(C)}{\sigma (Y(s))}Z \end{aligned}$$

be the drift term of the SDE of $C$ when the number of players is $N$ and let $(C^ {N},Z^ N)$ be the corresponding solution. We have included $\omega \in \varOmega$ into the notation to stress the dependence with respect to the process $Y$. We know that $g_N$ is negative thanks to Assumption (A3) (a) (see also (12)). On the other hand, it is easy to see that $g_N$, as a function of $N$, is monotonous increasing if $\rho (c)>1$ and decreasing if $\rho (c)<1$. Let us suppose first that $\rho (c)>1$. Given that $Z\ge 0$ a.s. by Theorem 5, then $\frac{g_{N_1}(C)}{\sigma }Z \le \frac{g_{N_2}(C)}{\sigma }Z$ a.s. if $N_1\le N_2$. Hence $-f^ {N_1} (\omega ,s,C^ {N_2},Z^ {N_2}) \le -f^{N_2}(\omega , s, C^ {N_2},\xi ^ {N_2})$, as well as $C^{N_1}(T) = C^{N_2}(T)$. According to the Comparison Theorem 2.2 of Karoui et al. (1997), $C^{N_1}(s)\le C^{N_2}(s)$ for all $t\le s\le T$, a.s. The case $\rho (c) <1$ is analyzed analogously. Since $\phi ^ {N_i}(t,x)=C^{N_i}(t)$ is deterministic, the theorem is proved.

1.7 Proof of Theorem 12

We follow the same steps as in the proof of Theorem 11, using the same FBSDE representation. The process $Y$ is independent of the preference rate and the drift term of the SDE of $C$ in (A.6), $f^ {r}$, is decreasing in the preference rate. Hence $r_1\le r_2$ implies $-f^{r_1}\le -f^{r_2}$ and, by the Comparison Theorem 2.2 of Karoui et al. (1997), we have the result.