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Re: Measure and Integration

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  • Noel Vaillant
    ... Hi, you have f:R- [0,+oo[ measurable such that int f(u)du
    Message 1 of 8 , Oct 1, 2003
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      > Let F(x) = Integral [-infinite to x] f(u)du be an absolutely
      > continous distribution function. Prove that corresponding
      > probability measure m is given by:

      Hi,

      you have f:R->[0,+oo[ measurable such that int f(u)du<+oo. You define
      F(x)=int_oo^x f(u)du and you call m the measure on R such that:
      m(]-oo,x])=F(x) for all x in R (the fact that such measure exists is
      assumed to be granted) . You want to prove that for all B in B(R) you
      have m(B)=int_B f(u)du. Define n(B)=int_B f(u)du. It is not difficult
      to show that n(.) is a finite measure on R, and you want to prove
      that m=n. Define:

      D={B in B(R), n(B)=m(B)}

      and:

      C={]-oo,x]: x in R}

      Then C is a subset of D which is closed under finite intersection. It
      is not difficult to show that D is a dynkin sytem on R (see
      definition on Tutorial 1 of probability.net, it is important to use
      the fact that the mesures m and n are finite). From the dynkin system
      theorem (see Tutorial 1), the sigma algebra s(C) on R generated by C
      is a subset of D. However, s(C) is nothing but B(R) (exercise). So B
      (R) is a subset of D and finally D=B(R). This proves that m=n, and we
      are done.

      Regards. Noel.
    • azr124
      Hi.... I m lost in some parts of the proof: * Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?] * How can I prove that D is a Dynkin system ? I
      Message 2 of 8 , Oct 1, 2003
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        Hi.... I'm lost in some parts of the proof:

        * Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?]

        * How can I prove that D is a Dynkin system ?

        I understand the argument when you try to see if the two sets are
        equal.... but why is so important that the measures have to be
        finite ?

        Thanks !!!

        Alejandro

        --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
        wrote:
        > > Let F(x) = Integral [-infinite to x] f(u)du be an absolutely
        > > continous distribution function. Prove that corresponding
        > > probability measure m is given by:
        >
        > Hi,
        >
        > you have f:R->[0,+oo[ measurable such that int f(u)du<+oo. You
        define
        > F(x)=int_oo^x f(u)du and you call m the measure on R such that:
        > m(]-oo,x])=F(x) for all x in R (the fact that such measure exists
        is
        > assumed to be granted) . You want to prove that for all B in B(R)
        you
        > have m(B)=int_B f(u)du. Define n(B)=int_B f(u)du. It is not
        difficult
        > to show that n(.) is a finite measure on R, and you want to prove
        > that m=n. Define:
        >
        > D={B in B(R), n(B)=m(B)}
        >
        > and:
        >
        > C={]-oo,x]: x in R}
        >
        > Then C is a subset of D which is closed under finite intersection.
        It
        > is not difficult to show that D is a dynkin sytem on R (see
        > definition on Tutorial 1 of probability.net, it is important to use
        > the fact that the mesures m and n are finite). From the dynkin
        system
        > theorem (see Tutorial 1), the sigma algebra s(C) on R generated by
        C
        > is a subset of D. However, s(C) is nothing but B(R) (exercise). So B
        > (R) is a subset of D and finally D=B(R). This proves that m=n, and
        we
        > are done.
        >
        > Regards. Noel.
      • Noel Vaillant
        ... I would not say assuming but just defining . You then have to prove that n is indeed a measure (i.e. that it is countably additive). (see theorem 21 in
        Message 3 of 8 , Oct 2, 2003
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          --- In probability@yahoogroups.com, "azr124" <azr124@y...> wrote:
          > Hi.... I'm lost in some parts of the proof:
          >
          > * Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?]

          I would not say "assuming" but just "defining". You then have to
          prove that n is indeed a measure (i.e. that it is countably additive).
          (see theorem 21 in Tutorial 5 and exercises preceding it if you are
          stuck on the proof)

          >
          > * How can I prove that D is a Dynkin system ?

          If you read the definition in Tutorial 1, I don't think you ll find
          it too difficult: you need to show that:

          1. R lies in D
          2. if A,B are in D and A<B (inclusion) then B\A (set difference) lies
          in D
          3. if (An) is increasing sequence (i.e. An<A(n+1)) of elements of D,
          and A=\/_n An, then A is also an element of D.



          > I understand the argument when you try to see if the two sets are
          > equal.... but why is so important that the measures have to be
          > finite ?

          This is important when proving 2. of the dynkin system:
          suppose A,B are in D and A<B then:

          m(B)=m(B\A)+m(A) (A and B\A are disjoint)

          and since m is finite, you can write m(B\A)=m(B)-m(A) (if m wasnt
          finite you'd potentially be writing +oo-(+oo) which makes no sense)
          Likewise, since n is finite, n(B\A)=n(B)-n(A). Now since A,B are in D
          we have m(A)=n(A) and m(B)=n(B). So m(B\A)=n(B\A) and we have proved
          that B\A lies in D....

          Regards. Noel.
        • The Webmaster
          Hello: I m really stuck on the below as far as setting up the problem. Since I don t know how much is bet in total I don t see how can I work out the below
          Message 4 of 8 , Oct 5, 2003
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            Hello:

            I'm really stuck on the below as far as setting up the
            problem. Since I don't know how much is bet in total
            I don't see how can I work out the below problem? Is
            there an algebraic way to determine this?

            Can someone please help? Thanks.

            --------------------------------------------------
            I have the following 3 horse race with the following
            quoted odds by the bookmaker:

            horse #1: 6-1 for
            horse #2: 7-2 against
            horse #3: 50-1 against

            So:
            $6 bet on #1 would return $7 if he won, 0 otherwise.
            $2 bet on #2 would return $9 if he won, 0 otherwise.
            $1 bet on #3 would return $51 if he won, 0 otherwise.

            We can interpret this as a one period market with 3
            assets: $1 bet on #1, $1 bet on #2, and $1 bet on #3.

            What are the associated "risk-neutral" probabilities?
            What is the bookmaker taking on every bet?


            __________________________________
            Do you Yahoo!?
            The New Yahoo! Shopping - with improved product search
            http://shopping.yahoo.com
          • The Webmaster
            Hi Group, I sent this out before, but haven t heard back from anyone. Can anyone help me with this? Thanks, M.W. __________________________________ Do you
            Message 5 of 8 , Oct 9, 2003
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              Hi Group,

              I sent this out before, but haven't heard back from
              anyone. Can anyone help me with this?

              Thanks,
              M.W.

              __________________________________
              Do you Yahoo!?
              The New Yahoo! Shopping - with improved product search
              http://shopping.yahoo.com

              [Non-text portions of this message have been removed]
            • The Webmaster
              Can anyone help me regarding the below? Thanks. -M.W. ... __________________________________ Do you Yahoo!? The New Yahoo! Shopping - with improved product
              Message 6 of 8 , Oct 9, 2003
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                Can anyone help me regarding the below?
                Thanks.

                -M.W.

                >
                > Hello:
                >
                > I'm really stuck on the below as far as setting up
                > the
                > problem. Since I don't know how much is bet in
                > total
                > I don't see how can I work out the below problem?
                > Is
                > there an algebraic way to determine this?
                >
                > Can someone please help? Thanks.
                >
                > --------------------------------------------------
                > I have the following 3 horse race with the following
                > quoted odds by the bookmaker:
                >
                > horse #1: 6-1 for
                > horse #2: 7-2 against
                > horse #3: 50-1 against
                >
                > So:
                > $6 bet on #1 would return $7 if he won, 0 otherwise.
                > $2 bet on #2 would return $9 if he won, 0 otherwise.
                > $1 bet on #3 would return $51 if he won, 0
                > otherwise.
                >
                > We can interpret this as a one period market with 3
                > assets: $1 bet on #1, $1 bet on #2, and $1 bet on
                > #3.
                >
                > What are the associated "risk-neutral"
                > probabilities?
                > What is the bookmaker taking on every bet?
                >
                >
                > __________________________________
                > Do you Yahoo!?
                > The New Yahoo! Shopping - with improved product
                > search
                > http://shopping.yahoo.com
                >
                >


                __________________________________
                Do you Yahoo!?
                The New Yahoo! Shopping - with improved product search
                http://shopping.yahoo.com
              • Noel Vaillant
                ... Hi M.W. you can view the economy as having one time period and three possible states of the world in the future. State 1: horse 1 wins. State 2: horse 2
                Message 7 of 8 , Oct 9, 2003
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                  > > $6 bet on #1 would return $7 if he won, 0 otherwise.
                  > > $2 bet on #2 would return $9 if he won, 0 otherwise.
                  > > $1 bet on #3 would return $51 if he won, 0
                  > > otherwise.

                  Hi M.W.

                  you can view the economy as having one time period and three possible
                  states of the world in the future. State 1: horse 1 wins. State 2:
                  horse 2 wins etc... A triplet (a,b,c) can represent a financial asset:

                  you receive a in state 1
                  you receive b in state 2
                  you receive c in state 3

                  Betting on horse 1, is just like buying the asset (7,0,0), the price
                  of which is 6. The price of (0,9,0) is 2. The price of (0,0,51) is 1.
                  It follows that the price of (1,0,0) is 6/7, that of (0,1,0) is 2/9
                  and that of (0,0,1) is 1/51. These (I believe) is what we call the
                  risk neutral probabilities. Indeed the price of (a,b,c) is (by
                  linearity):

                  Price(a,b,c)=a*6/7+b*2/9+c*1/51

                  So formally, 6/7, 2/9 and 1/51 look like probabilities and Price
                  (a,b,c) is the expected payoff with respect to these probabilitities.
                  Now in practice, the bookmaker will sell (1,0,0), (0,1,0) and (0,0,1)
                  to different people, so as to be as unexposed as possible: so the
                  bookmaker is trying to sell (1,1,1). Their liability is 1 whatever
                  happens. However, they have received:

                  6/7+2/9+1/51=1.09897 ...

                  So the bookmaker is making a profit of about 10% on its turnover...

                  Hope this helps.

                  Regards. Noel
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