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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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?

vrtuljakwb · Accepted Answer

Saying &quot;any         b     ∗     satisfying    (   2   ) is optimal given that         a     ∗     is chosen according to (1)&quot; is a bit ambiguous, as it suggests that there is a closer relationship between         a     ∗     and         b     ∗     than simply &quot;if an         a     ∗     with    (   1   ) exists, then a         b     ∗     with    (   2   ) also exists, but         a     ∗     and         b     ∗     are linked in no other way&quot;; that is, it suggests that given         a     ∗     with (1), there might exist         b     ∗     with    (   2   ) which is not optimal, but then that         b     ∗     is completely &quot;separated&quot; from         a     ∗    . I can assume that what is most likely meant is that the existence of an         a     ∗     satisfying    (   1   ) is enough to prove that all         b     ∗     satisfying    (   2   ) are optimal, but obviously this depends on the actual scenario.This ambiguity is mirrored in saying &quot;Assume next that         b     ∗     is chosen such that    (   2   ) holds. Then, one shows that         a     ∗     can only be optimal parameters if    (   1   ) holds&quot;. Firstly, do you mean that if    (   1   ) holds,         a     ∗     is optimal, or merely that         a     ∗     cannot be optimal if    (   1   ) doesn&#039;t hold? Secondly, and more importantly, does the entire statement mean that the authors prove that it is enough for a         b     ∗     satisfying    (   2   ) to exist in order for either all         a     ∗     satisfying    (   1   ) (even those completely independent from         b     ∗    ) to also be optimal or in order to have as a fact that         a     ∗     cannot be optimal if they don&#039;t satisfy (1), depending on the answer to the previous question?In the best case when both issues amount to it there being enough for the corresponding parameters to exist, there remains only to be shown that there exists         a     ∗     such that    (   1   ) holds, as you noted.

The proof to follow is taken from a research paper in risk analysis which is in reviewing. I have be

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