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    It is not the case that Continuous probability distributions require a restriction to countable additivity rather than full additivity

    ?Set your confidence on the premises below to see your aggregate.

    Reasons For

    2 perspectives
    Reason for 1 of 2
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    • 1.de Finetti's coherence framework demonstrates that full additivity can be preserved by treating probability as a finitely additive measure without requiring countable additivity at all.
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    • 2.Restricting to countable additivity is itself a conventional stipulation, not a mathematical necessity, as Dubins and Savage showed in 'How to Gamble If You Must' (1965).
      ?

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    • 3.The alleged violation of full additivity presupposes that uncountable sums must behave like countable ones, but this conflates two distinct mathematical regimes with different convergence properties.
      ?

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    Reason for 2 of 2
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    • 1.Nonstandard analysis, developed by Abraham Robinson, assigns infinitesimal but non-zero probabilities to individual points in continuous distributions, dissolving the tension without restricting additivity.
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    • 2.If each point receives a positive infinitesimal probability and hyperfinite summation is employed, full additivity is recoverable within a well-defined mathematical framework endorsed by Bernstein and Wattenberg (1969).
      ?

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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.In continuous distributions there are uncountably many states
      ?

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    • 2.Each individual state in a continuous distribution has probability 0
      ?

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    • 3.Events containing uncountably many states often have non-zero probability
      ?

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