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College StatisticsAnswered question
Willow Pratt Willow Pratt 2022-07-11

The proof to follow is taken from a research paper in risk analysis which is in reviewing. I have been tasked with repeating the results in the paper, under a different risk measure, but I am not sure if the main argument in the paper is correct as it is presented, although I believe the main result should hold.
The context is the following. We are optimizing a risk function over a set of parameters ( a i ) i = 1 m , ( b i ) i = 1 m , with particular choices of ( a i ) i = 1 m , ( b i ) i = 1 m denoted as a , b respectively. The aim is to establish a criterion for optimality of the parameters.
The argument in the paper is as follows:
Begin by assuming that a is chosen such that ( 1 ) holds, where ( 1 ) is a condition on a . It is then possible to show that there exists some b such that a condition ( 2 ) on b holds. We then show that any b satisfying ( 2 ) is optimal given that a is chosen according to ( 1 ) . Assume next that b is chosen such that ( 2 ) holds. Then, one shows that a can only be optimal parameters if ( 1 ) holds. The conclusion is that choosing b such that ( 2 ) holds is sufficient for obtaining optimal ( a , b ).
There are two things that make me uneasy about this argument. Firstly, as we begin by assuming a condition on a , my instinct is to look next at what happens if a does not satisfy ( 1 ) .
Secondly, it is easy to find b such that ( 2 ) does not hold. While existence of b satisfying ( 2 ) given a such that ( 1 ) holds, it seems to me like we assume existance in the second part of the proof. Then, we seemingly use the assumed existence of b such that ( 2 ) to show that we must choose a such that ( 1 ) , and hence b will exists such that ( 2 ) . Is this not a circular argument?

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