How does the equation come up: ∫ 0 ∞ θ e θ x Q ( x , ∞ ) λ ( d x ) = ∫ 0 ∞ ( e θ x − 1 ) Q ( d x )with Q is a measure and λ is the Lebesgue measure.I calculated: ∫ 0 ∞ θ e θ x ∫ R 1 ( x , ∞ ) ( y ) Q ( d y ) λ ( d x ) = ∫ R ∫ 0 ∞ θ e θ x 1 ( x , ∞ ) ( y ) λ ( d y ) Q ( d x ) = ∫ R ∫ x ∞ θ e θ x λ ( d y ) Q ( d x )but don’t know how to come to ∫ 0 ∞ ( e θ x − 1 ) Q ( d x ).KR, toni

Question

How does the equation come up:      ∫          0              ∞        θ      e          θ      x        Q  (  x  ,  ∞  )  λ  (  d  x  )  =      ∫          0              ∞        (      e          θ      x        −  1  )  Q  (  d  x  )with   Q is a measure and   λ is the Lebesgue measure.I calculated:      ∫          0              ∞        θ      e          θ      x            ∫                  R                        1              (      x      ,      ∞      )        (  y  )  Q  (  d  y  )  λ  (  d  x  )  =      ∫                  R                  ∫          0              ∞        θ      e          θ      x                  1              (      x      ,      ∞      )        (  y  )  λ  (  d  y  )  Q  (  d  x  )  =      ∫                  R                  ∫          x              ∞        θ      e          θ      x        λ  (  d  y  )  Q  (  d  x  )but don’t know how to come to       ∫          0              ∞        (      e          θ      x        −  1  )  Q  (  d  x  ).KR, toni

Lisbonaid · Accepted Answer

It comes from a known result (*). Let   (  X  ,      A    ,  μ  ) be a   σ-finite measure space. Let   u  :  X  →  [  0  ,  ∞  ) be a measurable function. Let   ϕ  :  [  0  ,  ∞  )  →  [  0  ,  ∞  ) be increasing and s.t.   ϕ  (  0  )  =  0 and continuously differentiable. Then      ∫    X    ϕ  (  u  (  x  )  )  μ  (  d  x  )  =      ∫          (      0      ,      ∞      )            ϕ    ′    (  t  )  μ  (  {  x  :  u  (  x  )  ≥  t  }  )  d  twhere the rhs is an improper Riemann integral. In our case,   (  X  ,      A    ,  μ  )  =  (            R        +    ,      B    (            R        +    )  ,  Q  ),   u  (  x  )  =  x,   ϕ  (  y  )  =      e          θ      y        −  1. Thus      ∫                            R                +              (      e          θ      x        −  1  )  Q  (  d  x  )  =      ∫          (      0      ,      ∞      )        θ      e          θ      t                                    Q          (          {          x          :          x          ≥          t          }          )                ⏟                    =      Q      (      [      t      ,      ∞      )      )        d  t(*) The gist of the proof is                              ∫          X                ϕ        (        u        (        x        )        )        μ        (        d        x        )                            =                  ∫                      (            0            ,            ∞            )                          μ        (        ϕ        ∘        u        ≥        t        )        d        t        =                                          =                  ∫                      (            0            ,            ∞            )                          μ        (        ϕ        ∘        u        ≥        ϕ        (        y        )        )                  ϕ          ′                (        y        )        d        y        =                                          =                  ∫                      (            0            ,            ∞            )                          μ        (        u        ≥        y        )                  ϕ          ′                (        y        )        d        y            but some technicalities are to be considered.

How does the equation come up: <msubsup> ∫ <mrow class="MJX-TeXAtom

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