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I'm looking at the following definition

The random variables $X_{1}, \ldots, X_{d}$ are said to be comonotonic if they admit as copula the Frechet upper bound.

I am however not quite sure how to read this (my background is in Economics rather than Mathematics). Does this mean

a) the copula of the random variables is the Frechet upper bound $=>$ they are comonotonic


b) the copula of the random variables is the Frechet upper bound $<=>$ they are comonotonic

In other words: Is the word "if" semantically equivalent to "if and only if" in the context of a definition like this?


marked as duplicate by MJD, user147263, PhoemueX, Community Mar 12 '15 at 16:46

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  • $\begingroup$ The correct option is a). I believe that your confusion stems from the location of the "if" within this statement. Typically, a logical statement starts with an "if", and is followed by a "then", which makes it easy to infer which part should be on each side of the implication arrow ($\implies$). My suggestion is that you always try to convert the given statement to an "if/then" structure (if it's not already given as such), before attempting to understand it logically. $\endgroup$ – barak manos Mar 8 '15 at 15:08
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    $\begingroup$ @barakmanos I beg to differ. In general, “X is Y if Z”, where Y is the term being defined, means that you are allowed to call X an Y when Z is true and in no other case. $\endgroup$ – egreg Mar 8 '15 at 15:08
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    $\begingroup$ yes in the definitions if means if and only if $\endgroup$ – ali Mar 8 '15 at 15:10
  • $\begingroup$ Of course, everything I mentioned on the previous comment does not apply to "if and only if" statements, where the logic implication goes in both directions ($\iff$). $\endgroup$ – barak manos Mar 8 '15 at 15:14
  • $\begingroup$ A definition is an equivalence : roughly speaking, it licences us to replace the "long" condition : the definiens, with the "short" term : the definiendum. $\endgroup$ – Mauro ALLEGRANZA Mar 8 '15 at 15:15

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