Dear Mr. "Typo = I'm right."
e. f(x)>.5[f(x+W)+f(x-L)]
f. f(x) > .5[f(x+W)+f(x-L)]
Thank you for not calling me a retard.
Is Behavioral Economics Doomed?
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"Is there a difference between the equations in e and f? I'm missing it..."
e. f(x)>.5[f(x+G)+f(x-L)] for ALL x
This is the assumption in:
1. first sentence of first full paragraph on page 1282 of Rabin (2000)
2. Rabin (2000) corollary (p. 1291)
3. Rabin and Thaler (2001)f. f(x)>.5[f(x+G)+f(x-L)] for SOME x (e.g., x<$300,000)
This is the assumption in second sentence of Rabin's theorem (p. 1289)“I can give you a non-expected utility formulation that rejects all such gambles but isn't cara. Would you shut up if I do so? “
I read Safra and Segal's “generalization” of Rabin's theorem to non-EU models. Non-EU models drop the independence axiom (and others?) but retain the assumptions of transitivity and dominance. Rabin's assumption of concavity goes beyond the necessary assumptions of EU and non-EU models which obviously can handle someone with a linear utility function. U(x)=x is consistent with EU and non-EU models but violates concavity assumption. Rabin's theorem has absolutely nothing to do with the independence axiom. It only depends on transitivity, dominance and concavity.
I promise to attempt to rebut your “non-EU, non-CARA” argument once and let you have the last word re: that argument. This is my counteroffer.
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Let L={x,p; y,1-p} denote the lottery that pays $x with probability p and $y with probability 1-p and consider the utility function
$$U(L)=min{x,y}$$
This utility function prefers a known payment x to any lottery that pays x+G with probability p and x-L with probability 1-p. Yet, this is not equivalent to CARA.
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Let L=[$0,0; $1000, 1]
L=100% chance of $1000
U(L)=Min [x,y]=$0
Person prefers $0 to 100% chance of $1000? This violates the assumption that the function is monotonically increasing.
If 1-p>0 and y>0 and x=0, it violates monotonicity assumption."This utility function prefers a known payment x to any lottery that pays x+G with probability p and x-L with probability 1-p."
I don't understand how this sentence can be derived from your utility function since the utility function involves the terms x, y and p, but not G and L. Also, in the utility function L represents a gamble. In the conclusion, L represents the amount of the loss.
But let x=0, G=1000, L=0 and p=1. Once again, I think your example says a person prefers $0 to a 100% chance of $1000.
Please clarify.
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Are you playing dumb, or have you really never seen maximin preferences before?
If so, undergrad detected.
Let L=[$0,0; $1000, 1]
L=100% chance of $1000
U(L)=Min [x,y]=$0
Person prefers $0 to 100% chance of $1000? This violates the assumption that the function is monotonically increasing.
If 1-p>0 and y>0 and x=0, it violates monotonicity assumption.
"This utility function prefers a known payment x to any lottery that pays x+G with probability p and x-L with probability 1-p."
I don't understand how this sentence can be derived from your utility function since the utility function involves the terms x, y and p, but not G and L. Also, in the utility function L represents a gamble. In the conclusion, L represents the amount of the loss.
But let x=0, G=1000, L=0 and p=1. Once again, I think your example says a person prefers $0 to a 100% chance of $1000.
Please clarify. -
$$L=(x,p; y,1-p)$$ denotes the gamble that pays $x with probability p and $y with probability 1-p.
The utility function U(x,p,y)=min{x,y} for p strictly between 0 and 1 is the well-known maximin criterion. The decision maker values a lottery according to its worst possible outcome. The preferences represented by U are (weakly) monotonic. Of course, for p=0, the utility function is defined as U(x,y,0)=y and for p=1 it is defined as U(x,y,1)=y.
When facing a lottery between W dollars for sure and W+G with probability p>0 and W-L with probability 1-p>0, where G>0 and L>0, the decision maker always prefers the sure lottery.
By the way, these are standard preferences in an ambiguity world but the axiomatization for a risk world is obvious. They satisfy first-order stochastic dominance (but are discontinuous). The preferences represented by U are not equivalent to CARA. In particular, they violate the independence axiom so they are not even equivalent to any EU preferences.
Let L=[$0,0; $1000, 1]
L=100% chance of $1000
U(L)=Min [x,y]=$0
Person prefers $0 to 100% chance of $1000? This violates the assumption that the function is monotonically increasing.
If 1-p>0 and y>0 and x=0, it violates monotonicity assumption.
"This utility function prefers a known payment x to any lottery that pays x+G with probability p and x-L with probability 1-p."
I don't understand how this sentence can be derived from your utility function since the utility function involves the terms x, y and p, but not G and L. Also, in the utility function L represents a gamble. In the conclusion, L represents the amount of the loss.
But let x=0, G=1000, L=0 and p=1. Once again, I think your example says a person prefers $0 to a 100% chance of $1000.
Please clarify. -
I gave you a very concrete and explicit counterexample to your example using your terminology. Your utility function implies violations of monotonicity for an infinite number of cases. In response, you do not identify a single flaw in my analysis but merely opine about minimax rules. Please specify the flaw in my analysis.
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He actually did respond to your analysis. He assumed you were familiar with minimax preferences, since they're common core material. Thus he didn't originally tell you what happens when p=0. You pointed out a counter example that relies on a different treatment of p=0 than is used in minimax. He corrected that, and explained the way corners are treated.
Dont be dumb.
I gave you a very concrete and explicit counterexample to your example using your terminology. Your utility function implies violations of monotonicity for an infinite number of cases. In response, you do not identify a single flaw in my analysis but merely opine about minimax rules. Please specify the flaw in my analysis.
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"relies on a different treatment of p=0"
I assumed p=0 means p=0. I interpreted the algebra presented as algebra. I didn't realize we were talking about "corners." My bad.
Let L=[$0, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers $0 to a greater than 99.9999999999999999999999999% chance of $1 trillion? Have I got it now?Oops. I guess if p=0 doesn't mean p=0, then x=$0 doesn't mean x=$0. My bad AGAIN. I MUST be an undergraduate who suffered a traumatic head injury recently. I'm sorry.
Hopefully, the third time's the charm.
Let L=[$0.01, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers a penny to a greater than 99.9999999999999999999999999% chance of $1 trillion?
Is that it? -
Yes, the stuff you wrote follows from maximin preferences.
"relies on a different treatment of p=0"
I assumed p=0 means p=0. I interpreted the algebra presented as algebra. I didn't realize we were talking about "corners." My bad.
Let L=[$0, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers $0 to a greater than 99.9999999999999999999999999% chance of $1 trillion? Have I got it now?
Oops. I guess if p=0 doesn't mean p=0, then x=$0 doesn't mean x=$0. My bad AGAIN. I MUST be an undergraduate who suffered a traumatic head injury recently. I'm sorry.
Hopefully, the third time's the charm.
Let L=[$0.01, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers a penny to a greater than 99.9999999999999999999999999% chance of $1 trillion?
Is that it? -
So since you understand now, do you agree that this is a rebuttal of your "CARA only" interpretation of Rabin's results?
Yes, the stuff you wrote follows from maximin preferences.
"relies on a different treatment of p=0"
I assumed p=0 means p=0. I interpreted the algebra presented as algebra. I didn't realize we were talking about "corners." My bad.
Let L=[$0, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers $0 to a greater than 99.9999999999999999999999999% chance of $1 trillion? Have I got it now?
Oops. I guess if p=0 doesn't mean p=0, then x=$0 doesn't mean x=$0. My bad AGAIN. I MUST be an undergraduate who suffered a traumatic head injury recently. I'm sorry.
Hopefully, the third time's the charm.
Let L=[$0.01, 10^-1000; $1 trillion, 1-10^-1000]
So person prefers a penny to a greater than 99.9999999999999999999999999% chance of $1 trillion?
Is that it? -
Ok. Counterexample given. Can we just move on now?
Obviously, choosing W for sure over a 50-50 lottery that pays either W+G or W-L for all wealth levels below some threshold does not imply in expected utility with CARA preferences. This is entirely obvious and it didn't take me very long to come up with a counterexample. I'm sure others can come up with other counterexamples as well. Safra and Segal's trick to obtain calibration theorems for non-expected utility is the introduction of background risk. I'm surprised you claimed to have read the paper and not to have noticed its main conceptual innovation.
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You've got to be kidding. You don't specify a utility function of the form f(x)= blah blah blah. You specify a function f(L)=blah blah blah. If you specified the function like a normal function in algebra where you plot x on the x axis and f(x) on the y axis, you will see that your function is ever more risk averse than CARA. So sure, I'll modify my claim to say that Rabin's theorem and Safra and Segal's generalization shows that someone with CARA or even more risk aversion shows this effect. But you were trying to say the phenomenon occurs if we are LESS restrictive than CARA. So no, your example does not convince me that Rabin's theorem and Safra and Segal's theorem are "non-parametric."
Also, your function appears to violate weak ordering. Let L= [.5, x; .5, x] That is, p=1-p and and y=x. [NOTE: if this assumption violates any assumptions of your theory, please forgive me.]
So U(L)=min[x,y]=x=y. So x is preferred to 100% chance of x. Last time I checked, this was a no no in non-EU models, just like in EU. -
Lol.
You've got to be kidding. You don't specify a utility function of the form f(x)= blah blah blah. You specify a function f(L)=blah blah blah. If you specified the function like a normal function in algebra where you plot x on the x axis and f(x) on the y axis, you will see that your function is ever more risk averse than CARA. So sure, I'll modify my claim to say that Rabin's theorem and Safra and Segal's generalization shows that someone with CARA or even more risk aversion shows this effect. But you were trying to say the phenomenon occurs if we are LESS restrictive than CARA. So no, your example does not convince me that Rabin's theorem and Safra and Segal's theorem are "non-parametric."
Also, your function appears to violate weak ordering. Let L= [.5, x; .5, x] That is, p=1-p and and y=x. [NOTE: if this assumption violates any assumptions of your theory, please forgive me.]
So U(L)=min[x,y]=x=y. So x is preferred to 100% chance of x. Last time I checked, this was a no no in non-EU models, just like in EU. -
Ok, a2e4. You outed yourself as a troll. I'm actually embarrassed I spent time arguing with you. Good luck on your future trolling endeavors, hopefully in another topic because we're tired of this one.
You've got to be kidding. You don't specify a utility function of the form f(x)= blah blah blah. You specify a function f(L)=blah blah blah. If you specified the function like a normal function in algebra where you plot x on the x axis and f(x) on the y axis, you will see that your function is ever more risk averse than CARA. So sure, I'll modify my claim to say that Rabin's theorem and Safra and Segal's generalization shows that someone with CARA or even more risk aversion shows this effect. But you were trying to say the phenomenon occurs if we are LESS restrictive than CARA. So no, your example does not convince me that Rabin's theorem and Safra and Segal's theorem are "non-parametric."
Also, your function appears to violate weak ordering. Let L= [.5, x; .5, x] That is, p=1-p and and y=x. [NOTE: if this assumption violates any assumptions of your theory, please forgive me.]
So U(L)=min[x,y]=x=y. So x is preferred to 100% chance of x. Last time I checked, this was a no no in non-EU models, just like in EU.