If F is a continuous distribution function on ( R , B , μ L ) with distribution μ F , use Fubini's theorem to show that1. ∫ R F ( x ) d μ F ( x ) = 1 2 2. If X 1 , X 2 are i.i.d random variables with common distribution F, then P ( { X 1 ≤ X 2 } ) = 1 / 2 and E ( F ( X 1 ) ) = 1 / 2.My Attempt:I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.For example, I don't understand what's going on in line 4 of bs_math's answer.

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If   F is a continuous distribution function on   (      R    ,      B    ,      μ                  L              ) with distribution       μ    F  , use Fubini&#039;s theorem to show that1.       ∫                  R              F  (  x  )    d      μ    F    (  x  )  =      1    2  2. If       X    1    ,      X    2   are i.i.d random variables with common distribution   F, then   P  (  {      X    1    ≤      X    2    }  )  =  1      /    2 and   E  (  F  (      X    1    )  )  =  1      /    2.My Attempt:I don&#039;t really understand bs_math&#039;s answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.For example, I don&#039;t understand what&#039;s going on in line 4 of bs_math&#039;s answer.

pressacvt · Accepted Answer

The distribution function is generally defined as  F  :  x  ↦      μ    F    (  (  −  ∞  ,  x  ]  )  ,The OP relies on another definition stating that            F      ~        (  x  )  =      {                                        μ            F                    (          (          0          ,          x          ]          ,                          x          &amp;gt;          0          ,                                      0          ,                          x          =          0          ,                                      −                      μ            F                    (          (          −          x          ,          0          ]          ,                          x          &amp;lt;          0.                        Then we have             F      ~        =  F  +  c for the constant summand   c  =  −      μ    F    (  (  −  ∞  ,  0  ]  ). I have never seen this as a definition of the distribution function. The fact that one can show the claim                    (1)                              ∫                      R                          F        (        x        )                d                  μ          F                (        x        )        =                  1          2                    for   F shows that (1) is actually false as soon as   c  ≠  0. So I assume there is a misunderstanding regarding the intended definition of   F.Now, in order to show (1) use that by the very definition of the Lebesgue integral it is  F  (  x  )  =  μ  (  (  −  ∞  ,  x  ]  )  =      ∫          R            1          (      −      ∞      ,      x      ]        (  y  )    d      μ    F    (  y  )  .Then, consider the following transformations:                              ∫                      R                          F        (        x        )        d                  μ          F                (        x        )                            =                  ∫                                    R                                                ∫                      R                                    1                      (            −            ∞            ,            x            ]                          (        y        )                d                  μ          F                (        y        )                d                  μ          F                (        x        )                                          =                  ∫                      R                                    ∫                      R                                    1                      (            −            ∞            ,            x            ]                          (        y        )                d                  μ          F                (        x        )                d                  μ          F                (        y        )                                          =                  ∫                      R                                    ∫                                    R                                                1                      [            y            ,            ∞            )                          (        x        )                d                  μ          F                (        x        )                d                  μ          F                (        y        )                                          =                  ∫                      R                          1        −        F        (        y        )                d                  μ          F                (        y        )                            (2)                                  =        1        −                  ∫                      R                          F        (        x        )                d                  μ          F                (        x        )        ,            where we use Fubinis&#039;s theorem in the second line. To get from the second to the third line we observe for every   x  ,  y  ∈      R        1          (      −      ∞      ,      x      ]        (  y  )  =  1  ⇔  y  ≤  x  ⇔      1          [      y      ,      ∞      )        (  x  )  =  1.In the fourth line we use that for fixed   y one has                              ∫                      R                                    1                      [            y            ,            ∞            )                          (        x        )                d                  μ          F                (        x        )                            =                  ∫                      R                          1        −                  1                      (            −            ∞            ,            y            )                          (        x        )                d                  μ          F                (        x        )                                          =        1        −                  ∫                      R                                    1                      (            −            ∞            ,            y            )                          (        x        )                d                  μ          F                (        x        )                                          =        1        −                  lim                      h            ↘            0                                    ∫                      R                                    1                      (            −            ∞            ,            y            −            h            ]                          (        x        )                d                  μ          F                (        x        )                                          =        1        −                  lim                      h            ↘            0                          F        (        y        −        h        )                                          =        1        −        F        (        y        )        ,            where we use Beppo-Levi&#039;s theorem in the third line.The claim (1) follows immediately from (2) by rearranging terms.Here all intgrals are to be understood as Lebesgue integrals.

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