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Whats the difference between local non-satiation and monotonicity?
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Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?
microeconomics consumer-theory
asked 4 hours ago
EconJohn♦
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Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?
microeconomics consumer-theory
asked 4 hours ago
EconJohn♦
3,2771836
Do you mean weak or strong monotonicity?
– denesp
2 hours ago
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up vote
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down vote
favorite
Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?
microeconomics consumer-theory
asked 4 hours ago
EconJohn♦
3,2771836
Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?
microeconomics consumer-theory
microeconomics consumer-theory
asked 4 hours ago
EconJohn♦
3,2771836
asked 4 hours ago
EconJohn♦
3,2771836
asked 4 hours ago
EconJohn♦
3,2771836
asked 4 hours ago
EconJohn♦
3,2771836
asked 4 hours ago
EconJohn♦
3,2771836
3,2771836
Do you mean weak or strong monotonicity?
– denesp
2 hours ago
add a comment |Â
Do you mean weak or strong monotonicity?
– denesp
2 hours ago
Do you mean weak or strong monotonicity?
– denesp
2 hours ago
Do you mean weak or strong monotonicity?
– denesp
2 hours ago
add a comment |Â
1 Answer
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Monotonicity of preferences is a stronger condition that local nonsatiation. Monotonicity implies local nonsatiation, but not the other way around.
To see this:
Claim: Let $succsim$ be a monotonic preference relation over $mathbbR^n_+$. Therefore, $succsim$ is locally nonsatiated.
Proof: Fix some $varepsilon > 0$. Let there be an arbitrary $x in mathbbR^n_+$ and let $mathbf1 in mathbbR^n_+$ be the unit vector. For any $lambda > 0$, we also have $x + lambda mathbf1 in mathbbR^n_+$. Since clearly $x + lambda mathbf1 gg x,$ $x + lambda mathbf1 succ x$ by monotonicity. Consider the following metric over $mathbbR^n_+$:
$$
d(x+lambda mathbf1, x) = ||x+lambda mathbf1 -x|| = lambda||mathbf1|| = lambda sqrtn.
$$
Thus for $lambda < fracvarepsilonsqrtn$, $d(x+lambda mathbf1, x) < varepsilon$ yet $x + lambda mathbf1 succ x$. Since $x$ was arbitrary, the existence of such a point implies that $succsim$ is locally nonsatiated. $blacksquare$
To show that locally nonsatiated preferences do not imply monotonic preferences, you can come up with a utility function $u(cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a satiation point for at least one of the others. For example, in $mathbbR^2_+$:
$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$
A consumer with such preferences is satiated with respect to $x_2$ at $x_2=1$.
answered 3 hours ago
Kenneth Rios
656111
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Monotonicity of preferences is a stronger condition that local nonsatiation. Monotonicity implies local nonsatiation, but not the other way around.
To see this:
Claim: Let $succsim$ be a monotonic preference relation over $mathbbR^n_+$. Therefore, $succsim$ is locally nonsatiated.
Proof: Fix some $varepsilon > 0$. Let there be an arbitrary $x in mathbbR^n_+$ and let $mathbf1 in mathbbR^n_+$ be the unit vector. For any $lambda > 0$, we also have $x + lambda mathbf1 in mathbbR^n_+$. Since clearly $x + lambda mathbf1 gg x,$ $x + lambda mathbf1 succ x$ by monotonicity. Consider the following metric over $mathbbR^n_+$:
$$
d(x+lambda mathbf1, x) = ||x+lambda mathbf1 -x|| = lambda||mathbf1|| = lambda sqrtn.
$$
Thus for $lambda < fracvarepsilonsqrtn$, $d(x+lambda mathbf1, x) < varepsilon$ yet $x + lambda mathbf1 succ x$. Since $x$ was arbitrary, the existence of such a point implies that $succsim$ is locally nonsatiated. $blacksquare$
To show that locally nonsatiated preferences do not imply monotonic preferences, you can come up with a utility function $u(cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a satiation point for at least one of the others. For example, in $mathbbR^2_+$:
$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$
A consumer with such preferences is satiated with respect to $x_2$ at $x_2=1$.
answered 3 hours ago
Kenneth Rios
656111
add a comment |Â
up vote
3
down vote
Monotonicity of preferences is a stronger condition that local nonsatiation. Monotonicity implies local nonsatiation, but not the other way around.
To see this:
Claim: Let $succsim$ be a monotonic preference relation over $mathbbR^n_+$. Therefore, $succsim$ is locally nonsatiated.
Proof: Fix some $varepsilon > 0$. Let there be an arbitrary $x in mathbbR^n_+$ and let $mathbf1 in mathbbR^n_+$ be the unit vector. For any $lambda > 0$, we also have $x + lambda mathbf1 in mathbbR^n_+$. Since clearly $x + lambda mathbf1 gg x,$ $x + lambda mathbf1 succ x$ by monotonicity. Consider the following metric over $mathbbR^n_+$:
$$
d(x+lambda mathbf1, x) = ||x+lambda mathbf1 -x|| = lambda||mathbf1|| = lambda sqrtn.
$$
Thus for $lambda < fracvarepsilonsqrtn$, $d(x+lambda mathbf1, x) < varepsilon$ yet $x + lambda mathbf1 succ x$. Since $x$ was arbitrary, the existence of such a point implies that $succsim$ is locally nonsatiated. $blacksquare$
To show that locally nonsatiated preferences do not imply monotonic preferences, you can come up with a utility function $u(cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a satiation point for at least one of the others. For example, in $mathbbR^2_+$:
$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$
A consumer with such preferences is satiated with respect to $x_2$ at $x_2=1$.
answered 3 hours ago
Kenneth Rios
656111
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Monotonicity of preferences is a stronger condition that local nonsatiation. Monotonicity implies local nonsatiation, but not the other way around.
To see this:
Claim: Let $succsim$ be a monotonic preference relation over $mathbbR^n_+$. Therefore, $succsim$ is locally nonsatiated.
Proof: Fix some $varepsilon > 0$. Let there be an arbitrary $x in mathbbR^n_+$ and let $mathbf1 in mathbbR^n_+$ be the unit vector. For any $lambda > 0$, we also have $x + lambda mathbf1 in mathbbR^n_+$. Since clearly $x + lambda mathbf1 gg x,$ $x + lambda mathbf1 succ x$ by monotonicity. Consider the following metric over $mathbbR^n_+$:
$$
d(x+lambda mathbf1, x) = ||x+lambda mathbf1 -x|| = lambda||mathbf1|| = lambda sqrtn.
$$
Thus for $lambda < fracvarepsilonsqrtn$, $d(x+lambda mathbf1, x) < varepsilon$ yet $x + lambda mathbf1 succ x$. Since $x$ was arbitrary, the existence of such a point implies that $succsim$ is locally nonsatiated. $blacksquare$
To show that locally nonsatiated preferences do not imply monotonic preferences, you can come up with a utility function $u(cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a satiation point for at least one of the others. For example, in $mathbbR^2_+$:
$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$
A consumer with such preferences is satiated with respect to $x_2$ at $x_2=1$.
answered 3 hours ago
Kenneth Rios
656111
Monotonicity of preferences is a stronger condition that local nonsatiation. Monotonicity implies local nonsatiation, but not the other way around.
To see this:
Claim: Let $succsim$ be a monotonic preference relation over $mathbbR^n_+$. Therefore, $succsim$ is locally nonsatiated.
Proof: Fix some $varepsilon > 0$. Let there be an arbitrary $x in mathbbR^n_+$ and let $mathbf1 in mathbbR^n_+$ be the unit vector. For any $lambda > 0$, we also have $x + lambda mathbf1 in mathbbR^n_+$. Since clearly $x + lambda mathbf1 gg x,$ $x + lambda mathbf1 succ x$ by monotonicity. Consider the following metric over $mathbbR^n_+$:
$$
d(x+lambda mathbf1, x) = ||x+lambda mathbf1 -x|| = lambda||mathbf1|| = lambda sqrtn.
$$
Thus for $lambda < fracvarepsilonsqrtn$, $d(x+lambda mathbf1, x) < varepsilon$ yet $x + lambda mathbf1 succ x$. Since $x$ was arbitrary, the existence of such a point implies that $succsim$ is locally nonsatiated. $blacksquare$
To show that locally nonsatiated preferences do not imply monotonic preferences, you can come up with a utility function $u(cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a satiation point for at least one of the others. For example, in $mathbbR^2_+$:
$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$
A consumer with such preferences is satiated with respect to $x_2$ at $x_2=1$.
answered 3 hours ago
Kenneth Rios
656111
edited 1 hour ago
answered 3 hours ago
Kenneth Rios
656111
answered 3 hours ago
Kenneth Rios
656111
answered 3 hours ago
Kenneth Rios
656111
656111
add a comment |Â
add a comment |Â
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Do you mean weak or strong monotonicity?
– denesp
2 hours ago