Let ( Ω , F , P ) be a probability space and X : ( Ω , F ) → ( R , F E ) a random variable that admits a density function f X : R → R .I have some questions about the information given by the cumulative distribution function of X ( F X : R → [ 0 , 1 ]).i) Does the CDF determine f X uniquely ?ii) Does the CDF determine the law of X uniquely ?iii) Can we find a subset of events that uniquely determine the CDF ?For i), I would say that as F X ( t ) = ∫ − ∞ t f X ( x ) d x, the density function is the derivative of the CDF and therefore the CDF can only define one density function. But I feel like maybe we could find a counterexample by considering Lebesgue measure and taking sets of the form (a,b) and [a,b].For ii), as ∀ x ∈ R , P ( X = x ) = F X ( x ) − F X ( x − ) and P ( X ≤ x ) = F X ( x ), I am tempted to say that the law of X is indeed by definition determined uniquely by the CDF.For iii), maybe we could take all events of the form { X ∈ ( a , b ) : a , b ∈ R } but I'm not quite sure about that, and particularly regarding the uniqueness.

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Let   (  Ω  ,      F    ,      P    ) be a probability space and   X  :  (  Ω  ,      F    )  →  (      R    ,            F        E    ) a random variable that admits a density function       f    X    :      R    →      R  .I have some questions about the information given by the cumulative distribution function of X (      F    X    :      R    →  [  0  ,  1  ]).i) Does the CDF determine       f    X   uniquely ?ii) Does the CDF determine the law of X uniquely ?iii) Can we find a subset of events that uniquely determine the CDF ?For i), I would say that as       F    X    (  t  )  =      ∫          −      ∞              t            f    X    (  x  )  d  x, the density function is the derivative of the CDF and therefore the CDF can only define one density function. But I feel like maybe we could find a counterexample by considering Lebesgue measure and taking sets of the form (a,b) and [a,b].For ii), as   ∀  x  ∈      R  ,       P    (  X  =  x  )  =      F    X    (  x  )  −      F    X    (  x  −  ) and       P    (  X  ≤  x  )  =      F    X    (  x  ), I am tempted to say that the law of X is indeed by definition determined uniquely by the CDF.For iii), maybe we could take all events of the form   {  X  ∈  (  a  ,  b  )  :  a  ,  b  ∈      R    } but I&#039;m not quite sure about that, and particularly regarding the uniqueness.

razdiralem · Accepted Answer

By subtraction, the CDF   F  (  x  )  :=  P  (  X  ≤  x  ) determines the probabilities   P  (  X  ∈  [  a  ,  b  ]  ) for every   a  ,  b  ∈      R  . By an approximation theorem, such as Caratheodory&#039;s theorem, this information determines   P  (  X  ∈  E  ) for every   E  ∈  B  (      R    ). So (ii) is true.(i) follows (though, as always, the density is unique only up to almost everywhere equality).

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