tag:blogger.com,1999:blog-7905515.post3719994446059682702..comments2023-03-26T09:28:59.359-05:00Comments on Falkenblog: Why Ambiguity Aversion is RationalEric Falkensteinhttp://www.blogger.com/profile/07243687157322033496noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-7905515.post-23705842377398581052011-10-01T16:02:45.311-05:002011-10-01T16:02:45.311-05:00In one sense, what I am doing is assuming that the...In one sense, what I am doing is assuming that the ambiguity can be converted into a probability distribution that approximates the ambiguity.<br /><br />Some more points:<br /><br />1) I'm not disputing that if you only have preferences over expected values than the preference for A and D would be paradoxical.<br /><br />2) I don't know what you mean by 1MM times is different than 10,000 times. I don't think 1MM is any different than 10,000. I can go and check it, but I doubt I would have any different results. Again, happy to send over some Matlab code to show what I mean.<br /><br />3) Here's what I don't think I'm explaining well enough. I think we're talking about different things. <br />You seem to be focusing on the fact that if I think there is some x% probability of a black ball, then I will switch. Indeed, that is what most focus on. I think x is a random variable. For each x, I can compute the expected profit, exactly. No dispute there. However, I need to sample many x's in order to get the actual distribution of profits. This seems to be where I'm not explaining things well enough. I'm implicitly assuming that people make some assumption about how x is chosen. I think in practice that is reasonable. If they think, well the allocation between yellow and black could be anything (and I assume that means they assume it is a uniform random varible), then the black gamble and the red and yellow gambles need to be sampled from to actually get the distribution of profits. The red gamble and the not red gamble are known in advance.<br /><br />Again, if people only care about expected values, then it doesn't change anything. However, most people care about the possible downside. Hence, most people switch.John Hallnoreply@blogger.comtag:blogger.com,1999:blog-7905515.post-50745498703223375872011-09-30T08:43:51.504-05:002011-09-30T08:43:51.504-05:00They key is that if you do this 1MM times, they ar...They key is that if you do this 1MM times, they are identical:(same probability of winning in both gambles-1/3 in the first, 2/3 in the second), which is why the preference towards A and D is paradoxical. You don't need to sample to assess the true ex ante probabilities, because the are presented without ambiguity.Eric Falkensteinhttps://www.blogger.com/profile/07243687157322033496noreply@blogger.comtag:blogger.com,1999:blog-7905515.post-45342441628155406422011-09-28T14:49:51.958-05:002011-09-28T14:49:51.958-05:00I'm not sure I explained well enough. I'd ...I'm not sure I explained well enough. I'd be happy to send along some code that may make it more clear.<br /><br />Even in Ellsberg, you never know the urn distribution with certainty. You merely know that if you prefer A to B that implies something about what you think the distribution is and it influences what your bet in C vs D should be.<br /><br />So 30 red balls, no matter what. Let's say black=floor(70*p) and yellow=70-black. Let's assume p is a random variable (say, uniform between 0 and 1) and you simulate it some 10,000 times. So you have the breakdown of the marbles if you had run this experiment 10,000 times. The person could then determine what the probability distribution is for each gamble. Of course, after they choose the gamble, they will draw from one jar and have one marble and one profit. However, they can still come up with an ex ante distribution of expected profits. For instance, A and D will look like dirac deltas, but B and C will have different distributions depending on how p is simulated.<br /><br />In this sense, you could compute something like a mean to expected shortfall ratio (assuming you have to pay to play the game) or some utility function like mean-lambda*expected shortfall and I think this would readily explain the way most people behave in these games. They don't like the uncertainty with gambles B and C.John Hallnoreply@blogger.comtag:blogger.com,1999:blog-7905515.post-82213919460108248702011-09-27T11:40:37.554-05:002011-09-27T11:40:37.554-05:00So we should invest in AIG and BAC because if we l...So we should invest in AIG and BAC because if we lose we'd look like fools and thus they are potentially undervalued now because no fund manager wants to deal with this 'career risk'?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7905515.post-51435146502857764402011-09-27T10:06:26.212-05:002011-09-27T10:06:26.212-05:00Well, considering the Ellsberg Paradox is still co...Well, considering the Ellsberg Paradox is still considered an unsolved paradox, I don't think your solution works. <br /><br />I don't see what the 10k times does. Once you draw one black or yellow ball, you know the urn distribution with certainty, which would be either 30 red and 60 black, or 30 read and 60 yellow. But the bet is for drawing once.Eric Falkensteinhttps://www.blogger.com/profile/07243687157322033496noreply@blogger.comtag:blogger.com,1999:blog-7905515.post-7194196318899160902011-09-27T09:23:12.607-05:002011-09-27T09:23:12.607-05:00I see ambiguity aversion as more tightly joined wi...I see ambiguity aversion as more tightly joined with risk aversion than you do. For instance, consider the Ellsberg Paradox (I'm using the 1 urn example from wikipedia). It would be impossible to estimate what the probability distribution is in advance. However, if you assume that the split between black and yellow balls is a uniform random variable (and I believe most people implicitly believe this is the case when they face this game), then it is possible to do the draw 10,000 times and determine the probability distribution of the gambles. The gambles with "ambiguity" are clearly riskier and the paradox melts away.John Hallnoreply@blogger.com