Unit of time and normalization of time preference rates
$begingroup$
Consider an infinite horizon cake eating differential game described by
begin{align}
&max_{u_1(t)} int_0^infty{e^{-r_1 t}ln(u_1(t))dt}\
&max_{u_2(t)} int_0^infty{e^{-r_2 t}ln(u_2(t))dt}\
text{s.t.} quad &dot x(t) = -u_1(t)-u_2(t)
end{align}
where $t in [0,infty)$ refers to the current time, $u_i(t)$ to the consumption rate of player $i in {1,2}$ and $x(t)$ to the size of the cake.
I'm interested in the time preference rates $r_i > 0$.
Without further explanations I found in Clemhout and Wan, 1989, On Games of cake-eating, p. 131
Since one can always choose a unit for time without losing generality, we suppose that the length of a unit of time is such that the preference rates of the two players sum up to unity.
That is, we can normalize $r_1 + r_2 = 1$.
But what does this mean actually? I'm particular puzzled about the without loss of generality part.
dynamical-systems optimal-control differential-games
asked Dec 11 '18 at 23:45
cluelessclueless
298111
$endgroup$
add a comment |
$begingroup$
Consider an infinite horizon cake eating differential game described by
begin{align}
&max_{u_1(t)} int_0^infty{e^{-r_1 t}ln(u_1(t))dt}\
&max_{u_2(t)} int_0^infty{e^{-r_2 t}ln(u_2(t))dt}\
text{s.t.} quad &dot x(t) = -u_1(t)-u_2(t)
end{align}
where $t in [0,infty)$ refers to the current time, $u_i(t)$ to the consumption rate of player $i in {1,2}$ and $x(t)$ to the size of the cake.
I'm interested in the time preference rates $r_i > 0$.
Without further explanations I found in Clemhout and Wan, 1989, On Games of cake-eating, p. 131
Since one can always choose a unit for time without losing generality, we suppose that the length of a unit of time is such that the preference rates of the two players sum up to unity.
That is, we can normalize $r_1 + r_2 = 1$.
But what does this mean actually? I'm particular puzzled about the without loss of generality part.
dynamical-systems optimal-control differential-games
asked Dec 11 '18 at 23:45
cluelessclueless
298111
$endgroup$
add a comment |
$begingroup$
Consider an infinite horizon cake eating differential game described by
begin{align}
&max_{u_1(t)} int_0^infty{e^{-r_1 t}ln(u_1(t))dt}\
&max_{u_2(t)} int_0^infty{e^{-r_2 t}ln(u_2(t))dt}\
text{s.t.} quad &dot x(t) = -u_1(t)-u_2(t)
end{align}
where $t in [0,infty)$ refers to the current time, $u_i(t)$ to the consumption rate of player $i in {1,2}$ and $x(t)$ to the size of the cake.
I'm interested in the time preference rates $r_i > 0$.
Without further explanations I found in Clemhout and Wan, 1989, On Games of cake-eating, p. 131
Since one can always choose a unit for time without losing generality, we suppose that the length of a unit of time is such that the preference rates of the two players sum up to unity.
That is, we can normalize $r_1 + r_2 = 1$.
But what does this mean actually? I'm particular puzzled about the without loss of generality part.
dynamical-systems optimal-control differential-games
asked Dec 11 '18 at 23:45
cluelessclueless
298111
$endgroup$
Consider an infinite horizon cake eating differential game described by
begin{align}
&max_{u_1(t)} int_0^infty{e^{-r_1 t}ln(u_1(t))dt}\
&max_{u_2(t)} int_0^infty{e^{-r_2 t}ln(u_2(t))dt}\
text{s.t.} quad &dot x(t) = -u_1(t)-u_2(t)
end{align}
where $t in [0,infty)$ refers to the current time, $u_i(t)$ to the consumption rate of player $i in {1,2}$ and $x(t)$ to the size of the cake.
I'm interested in the time preference rates $r_i > 0$.
Without further explanations I found in Clemhout and Wan, 1989, On Games of cake-eating, p. 131
Since one can always choose a unit for time without losing generality, we suppose that the length of a unit of time is such that the preference rates of the two players sum up to unity.
That is, we can normalize $r_1 + r_2 = 1$.
But what does this mean actually? I'm particular puzzled about the without loss of generality part.
dynamical-systems optimal-control differential-games
dynamical-systems optimal-control differential-games
asked Dec 11 '18 at 23:45
cluelessclueless
298111
asked Dec 11 '18 at 23:45
cluelessclueless
298111
asked Dec 11 '18 at 23:45
cluelessclueless
298111
asked Dec 11 '18 at 23:45
cluelessclueless
298111
asked Dec 11 '18 at 23:45
cluelessclueless
298111
298111
add a comment |
add a comment |
1 Answer
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$begingroup$
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=rho$. By introducing a new independent variable $tau=rho t$ we get
$$max_{u_1(tau)} frac{1}{rho}int_0^infty{e^{-frac{r_1}{rho}tau}ln(u_1(tau))dtau}.$$
The factor $frac{1}{rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $frac{r_i}{rho}$ now sum to 1.
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=rho$. By introducing a new independent variable $tau=rho t$ we get
$$max_{u_1(tau)} frac{1}{rho}int_0^infty{e^{-frac{r_1}{rho}tau}ln(u_1(tau))dtau}.$$
The factor $frac{1}{rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $frac{r_i}{rho}$ now sum to 1.
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
$endgroup$
add a comment |
$begingroup$
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=rho$. By introducing a new independent variable $tau=rho t$ we get
$$max_{u_1(tau)} frac{1}{rho}int_0^infty{e^{-frac{r_1}{rho}tau}ln(u_1(tau))dtau}.$$
The factor $frac{1}{rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $frac{r_i}{rho}$ now sum to 1.
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
$endgroup$
add a comment |
$begingroup$
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=rho$. By introducing a new independent variable $tau=rho t$ we get
$$max_{u_1(tau)} frac{1}{rho}int_0^infty{e^{-frac{r_1}{rho}tau}ln(u_1(tau))dtau}.$$
The factor $frac{1}{rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $frac{r_i}{rho}$ now sum to 1.
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
$endgroup$
First note that $r_1>0$ and $r_2>0$. Let $r_i$ be such that $r_1+r_2=rho$. By introducing a new independent variable $tau=rho t$ we get
$$max_{u_1(tau)} frac{1}{rho}int_0^infty{e^{-frac{r_1}{rho}tau}ln(u_1(tau))dtau}.$$
The factor $frac{1}{rho}$ in front of the integral can be dropped as it doesn't influence the result. And the new discount rates $frac{r_i}{rho}$ now sum to 1.
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
answered Dec 14 '18 at 9:47
DmitryDmitry
671618
671618
add a comment |
add a comment |
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