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  • shuva gupta
    Hi Noel, Please help me with this one : Suppose E( |X| ) d] ,set ~mu=E(~X)
    Message 1 of 18 , Nov 4, 2004
      Hi Noel,
      Please help me with this one :

      Suppose E( |X| )< infinity, mu=E( |X| ) Fix c<d .
      Let ~X= c , x, d accordingly as [X<c],[c<=X<=d],[X>d] ,set ~mu=E(~X)
      Prove: E( |~X - ~mu|^r )<=E(|X -mu|^r) for all r>= 1

      Regards,
      Shuva

      PS: I solved the problem for mu=~mu but didn't need the condition r>=1, ie true for any r
      Not very sure whether the problem is correct .If wrong can you get an counter example.
      You many have to use Cr Inequality , Minkowski's , or Jensen's or Whatever .


      Tony Wong <tw813@...> wrote:

      hello all,

      I have the following question which I hope someone can
      help with:

      Given X_t is a brownian motion with drift rate u and
      volatility sigma, and X_0 = a >0, define T(b) with b>a
      to be the first time X_t hits the level b. Now define
      Y_t = X_{min(t, T(b)}. (i.e. Y_t=X_t if T(b) is
      greater than t and Y_t=b for all t>= T(b) ). For each
      fixed t, Y_t has a mixed distribution with a point
      mass at b and a continuous density on (-infinity, b).

      My question is: how can one derive the density part??
      (For the case where X_t is the standard brownian
      motion, this question is not too hard.) Also, can
      someone tell me from which book, I could get an answer
      for this question??

      Many thanks,

      Tony


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    • shuva gupta
      Hi Noel, Can you please help me with this Suppose E|X| d] Set ~mu=E(~X) Show
      Message 2 of 18 , Nov 4, 2004
        Hi Noel,

        Can you please help me with this

        Suppose E|X|< infinity, mu=E(X). Fix c<d ,

        Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]

        Set ~mu=E(~X)

        Show that E ( |~X - ~mu|^r ) <=E ( |X - mu|^r ) for all r>=1

        Regards,

        Shuva

        PS: I have been able to do it for mu = ~mu ( did not need the condn r>=1here )but couldn't for the case mu not equals ~mu.I am noy sure whether the problem is correct so can you please provide me with a counter example if you think it is not true.

        You many need to use Cr Inequality ,Jensen or Whatever .....












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      • jason1990
        Here s a heuristic sketch of a very special case. Let s suppose that c=- infty, that X has a nice density ( nice here means whatever it needs to mean so
        Message 3 of 18 , Nov 5, 2004
          Here's a heuristic sketch of a very special case. Let's suppose that
          c=-\infty, that X has a "nice" density ("nice" here means whatever
          it needs to mean so that all my dubious differentiations are
          justified), and that r=2k for some positive integer k.

          Then ~X=min(X,d). Define g(t)=E[min(X,t)] and observe that

          g'(t) = E[1_{X > t}] = P(X > t).

          Also note that ~mu=g(d).

          Now define h(t) = E[(min(X,t) - g(t))^{2k}], so that E|~X - ~mu|^r =
          h(d). Note that

          h'(t) = 2k E[(min(X,t) - g(t))^{2k-1} (1_{X > t} - g'(t))].

          If we let Y be an independent copy of X, then

          1_{X > t} - g'(t)
          1_{X > t} P(Y < t) - 1_{X < t} P(Y > t)
          = E[1_{X > t > Y}|X] - E[1_{X < t < Y}|X].

          Hence,

          h'(t) = 2k E[(min(X,t) - g(t))^{2k-1} 1_{X > t > Y}]
          - 2k E[(min(X,t) - g(t))^{2k-1} 1_{X < t < Y}]
          = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
          - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].

          Since 2k-1 is odd,

          (X - g(t))^{2k-1} 1_{X < t < Y}
          <= (t - g(t))^{2k-1} 1_{X < t < Y}.

          Therefore,

          h'(t) >= 2k(t - g(t))^{2k-1} E[1_{X > t > Y} - 1_{X < t < Y}].

          But since X and Y are iid, this expectation is 0, so h is an
          increasing function in t. Thus,

          E|~X - ~mu|^r = h(d)
          <= lim_{t->\infty} h(t)
          = E|X - mu|^r. QED

          This argument is so convoluted, I don't have much confidence in it.
          (Where's my mistake?) Even if it is correct, hopefully someone else
          (Noel?) can provide something more transparent.

          --- In probability@yahoogroups.com, shuva gupta <shuvagupta@y...>
          wrote:
          >
          >
          > Hi Noel,
          >
          > Can you please help me with this
          >
          > Suppose E|X|< infinity, mu=E(X). Fix c<d ,
          >
          > Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]
          >
          > Set ~mu=E(~X)
          >
          > Show that E ( |~X - ~mu|^r ) <=E ( |X - mu|^r ) for all r>=1
          >
          > Regards,
          >
          > Shuva
          >
          > PS: I have been able to do it for mu = ~mu ( did not need the
          condn r>=1here )but couldn't for the case mu not equals ~mu.I am
          noy sure whether the problem is correct so can you please provide me
          with a counter example if you think it is not true.
          >
          > You many need to use Cr Inequality ,Jensen or Whatever .....
          >
          >
          >
          >
          >
          >
          >
          >
          >
          >
          >
          >
          > ---------------------------------
          > Do you Yahoo!?
          > Check out the new Yahoo! Front Page. www.yahoo.com/a
          >
          > [Non-text portions of this message have been removed]
        • jason1990
          Well, here s one mistake: I wrote ... What I wanted to say was ... But that s clearly false. I then used this false fact to show that h (t) =0, which
          Message 4 of 18 , Nov 5, 2004
            Well, here's one mistake:

            I wrote

            > Since 2k-1 is odd,
            >
            > (X - g(t))^{2k-1} 1_{X < t < Y}
            > <= (t - g(t))^{2k-1} 1_{X < t < Y}.

            What I wanted to say was

            > Since 2k-1 is odd,
            >
            > (X - g(t))^{2k-1} 1_{X < t < Y}
            > <= (t - g(t))^{2k-1} 1_{X > t > Y}.

            But that's clearly false. I then used this false "fact" to show that
            h'(t)>=0, which finished the "proof".

            But I think it's salvagable. Here's another way to show h'(t)>=0.
            Start with what we know:

            h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
            - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].

            Since X and Y are iid, I can interchange them in the first
            expectation, giving

            h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
            - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].

            Now we use the fact that (X - g(t))^{2k-1} <= (t - g(t))^{2k-1}
            whenever X < t to conclude that h'(t)>=0.

            Any other glaring mistakes? (Besides the lack of rigor?) :)

            --- In probability@yahoogroups.com, "jason1990" <jason1990@y...>
            wrote:
            >
            >
            > Here's a heuristic sketch of a very special case. Let's suppose
            that
            > c=-\infty, that X has a "nice" density ("nice" here means whatever
            > it needs to mean so that all my dubious differentiations are
            > justified), and that r=2k for some positive integer k.
            >
            > Then ~X=min(X,d). Define g(t)=E[min(X,t)] and observe that
            >
            > g'(t) = E[1_{X > t}] = P(X > t).
            >
            > Also note that ~mu=g(d).
            >
            > Now define h(t) = E[(min(X,t) - g(t))^{2k}], so that E|~X - ~mu|^r
            =
            > h(d). Note that
            >
            > h'(t) = 2k E[(min(X,t) - g(t))^{2k-1} (1_{X > t} - g'(t))].
            >
            > If we let Y be an independent copy of X, then
            >
            > 1_{X > t} - g'(t)
            > 1_{X > t} P(Y < t) - 1_{X < t} P(Y > t)
            > = E[1_{X > t > Y}|X] - E[1_{X < t < Y}|X].
            >
            > Hence,
            >
            > h'(t) = 2k E[(min(X,t) - g(t))^{2k-1} 1_{X > t > Y}]
            > - 2k E[(min(X,t) - g(t))^{2k-1} 1_{X < t < Y}]
            > = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
            > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
            >
            > Since 2k-1 is odd,
            >
            > (X - g(t))^{2k-1} 1_{X < t < Y}
            > <= (t - g(t))^{2k-1} 1_{X < t < Y}.
            >
            > Therefore,
            >
            > h'(t) >= 2k(t - g(t))^{2k-1} E[1_{X > t > Y} - 1_{X < t < Y}].
            >
            > But since X and Y are iid, this expectation is 0, so h is an
            > increasing function in t. Thus,
            >
            > E|~X - ~mu|^r = h(d)
            > <= lim_{t->\infty} h(t)
            > = E|X - mu|^r. QED
            >
            > This argument is so convoluted, I don't have much confidence in
            it.
            > (Where's my mistake?) Even if it is correct, hopefully someone
            else
            > (Noel?) can provide something more transparent.
            >
            > --- In probability@yahoogroups.com, shuva gupta <shuvagupta@y...>
            > wrote:
            > >
            > >
            > > Hi Noel,
            > >
            > > Can you please help me with this
            > >
            > > Suppose E|X|< infinity, mu=E(X). Fix c<d ,
            > >
            > > Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]
            > >
            > > Set ~mu=E(~X)
            > >
            > > Show that E ( |~X - ~mu|^r ) <=E ( |X - mu|^r ) for all r>=1
          • Noel Vaillant
            ... Jason, I find this argument very impressive (there is a lot of ideas in it). I am a bit uneasy about differentiation under E[...] but I am confident that
            Message 5 of 18 , Nov 5, 2004
              > This argument is so convoluted, I don't have much confidence in it.
              > (Where's my mistake?) Even if it is correct, hopefully someone else
              > (Noel?) can provide something more transparent.

              Jason,

              I find this argument very impressive (there is a lot of ideas in it).
              I am a bit uneasy about differentiation under E[...] but I am
              confident that this is not a flaw in your proof (i.e. I am confident
              every equality could be rigorously justified). I know you have
              mentioned a mistake (cf your next post), but I haven't seen any on
              first reading :-) So I am gonna go through your next post now.

              Noel.
            • Noel Vaillant
              ... Since (2k-1) is odd, x- x^(2k-1)is non-decreasing on R and since X - g(t)
              Message 6 of 18 , Nov 5, 2004
                > > Since 2k-1 is odd,
                > >
                > > (X - g(t))^{2k-1} 1_{X < t < Y}
                > > <= (t - g(t))^{2k-1} 1_{X < t < Y}.

                Since (2k-1) is odd, x->x^(2k-1)is non-decreasing on R
                and since X - g(t) <= t - g(t) on {X < t}, the inequality
                you wrote seems fine to me.

                I also think this inequality allows you to conclude
                that h'(t) >= 0.

                h'(t)
                = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}]
                >= 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                - 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                = 2k(t-g(t))^{2k-1}E[1_{X > t > Y} - 1_{X < t < Y}]
                =0

                which is pretty much what you wrote in your first post.
                What am I missing?

                > What I wanted to say was
                >
                > > Since 2k-1 is odd,
                > >
                > > (X - g(t))^{2k-1} 1_{X < t < Y}
                > > <= (t - g(t))^{2k-1} 1_{X > t > Y}.
                >
                > But that's clearly false.

                I wonder why you wanted to write this in the first place.



                > But I think it's salvagable. Here's another way to show
                >h'(t)>=0. Start with what we know:
                >
                > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                >
                > Since X and Y are iid, I can interchange them in the first
                > expectation, giving
                >
                > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                >
                > Now we use the fact that (X - g(t))^{2k-1} <= (t - g(t))^{2k-1}
                > whenever X < t to conclude that h'(t)>=0.
                >
                > Any other glaring mistakes? (Besides the lack of rigor?) :)

                This seems to me like the same as what you wrote in the first place,
                except you use iid property, before taking inequalities.

                Any way, it seems to me your proof is good. I certainly haven't
                seen a flaw.

                Noel.
              • jason1990
                Yes, you re right. I thought about this in the morning on the bus and again this evening before going to a movie. I had two lines of reasoning in my head at
                Message 7 of 18 , Nov 5, 2004
                  Yes, you're right. I thought about this in the morning on the bus
                  and again this evening before going to a movie. I had two lines of
                  reasoning in my head at once and when I reread my post, I confused
                  myself.

                  About differentiating under the expectation, something occurred to
                  me at the theater. For any nonnegative random variable Y and any
                  r>0, E[Y^r]=\int_0^\infty{rz^{r-1}P(Y>z)dz}. So suppose we're given
                  a real t and we want to compute E[min(X,t)]. Let Y=t-min(X,t). Then
                  Y is nonnegative and we can apply the above to get

                  E[min(X,t)] = t - \int_{-\infty}^t{P(X < z)dz}.

                  We can differentiate this with no problem. Something similar should
                  be possible with the other expectation. I don't think this is
                  necessary to justify the differentiation, but it makes me wonder
                  whether the whole differentiation approach is unnecessary.

                  Another thing that is curious: the only property of the map x->x^
                  {2k} that was used is the fact that it has a nondecreasing
                  derivative. In other words, it is convex. So perhaps the original
                  poster's claim is true not only for functions of the form x->|x|^r
                  where r>=1, but for all convex functions. If so, then maybe Jensen's
                  inequality would be useful in creating a simpler proof.

                  I just feel that it shouldn't be this hard.

                  --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
                  wrote:
                  >
                  >
                  > > > Since 2k-1 is odd,
                  > > >
                  > > > (X - g(t))^{2k-1} 1_{X < t < Y}
                  > > > <= (t - g(t))^{2k-1} 1_{X < t < Y}.
                  >
                  > Since (2k-1) is odd, x->x^(2k-1)is non-decreasing on R
                  > and since X - g(t) <= t - g(t) on {X < t}, the inequality
                  > you wrote seems fine to me.
                  >
                  > I also think this inequality allows you to conclude
                  > that h'(t) >= 0.
                  >
                  > h'(t)
                  > = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                  > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}]
                  > >= 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                  > - 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                  > = 2k(t-g(t))^{2k-1}E[1_{X > t > Y} - 1_{X < t < Y}]
                  > =0
                  >
                  > which is pretty much what you wrote in your first post.
                  > What am I missing?
                  >
                  > > What I wanted to say was
                  > >
                  > > > Since 2k-1 is odd,
                  > > >
                  > > > (X - g(t))^{2k-1} 1_{X < t < Y}
                  > > > <= (t - g(t))^{2k-1} 1_{X > t > Y}.
                  > >
                  > > But that's clearly false.
                  >
                  > I wonder why you wanted to write this in the first place.
                  >
                  >
                  >
                  > > But I think it's salvagable. Here's another way to show
                  > >h'(t)>=0. Start with what we know:
                  > >
                  > > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                  > > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                  > >
                  > > Since X and Y are iid, I can interchange them in the first
                  > > expectation, giving
                  > >
                  > > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                  > > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                  > >
                  > > Now we use the fact that (X - g(t))^{2k-1} <= (t - g(t))^{2k-1}
                  > > whenever X < t to conclude that h'(t)>=0.
                  > >
                  > > Any other glaring mistakes? (Besides the lack of rigor?) :)
                  >
                  > This seems to me like the same as what you wrote in the first
                  place,
                  > except you use iid property, before taking inequalities.
                  >
                  > Any way, it seems to me your proof is good. I certainly haven't
                  > seen a flaw.
                  >
                  > Noel.
                • Noel Vaillant
                  ... Very good. I agree. ... Yes. ... Yes. ... Well, I d be happy already to crack this for r in [0,+oo[ ... Yes, I am guessing there should be a simpler proof,
                  Message 8 of 18 , Nov 5, 2004
                    > About differentiating under the expectation, something occurred to
                    > me at the theater. For any nonnegative random variable Y and any
                    > r>0, E[Y^r]=\int_0^\infty{rz^{r-1}P(Y>z)dz}. So suppose we're given
                    > a real t and we want to compute E[min(X,t)]. Let Y=t-min(X,t). Then
                    > Y is nonnegative and we can apply the above to get
                    >
                    > E[min(X,t)] = t - \int_{-\infty}^t{P(X < z)dz}.
                    >

                    Very good. I agree.

                    > We can differentiate this with no problem.

                    Yes.

                    > Something similar should be possible with the other
                    > expectation.

                    Yes.

                    > Another thing that is curious: the only property of the map x->x^
                    > {2k} that was used is the fact that it has a nondecreasing
                    > derivative. In other words, it is convex. So perhaps the original
                    > poster's claim is true not only for functions of the form x->|x|^r
                    > where r>=1, but for all convex functions.

                    Well, I'd be happy already to crack this for r in [0,+oo[

                    > If so, then maybe Jensen's inequality would be useful in
                    >creating a simpler proof.

                    Yes, I am guessing there should be a simpler proof, but I
                    can't find it. I have been looking for a while now.
                    Even looked for counter-example for r=2.
                    I am going to give up soon :-)


                    Noel.
                  • Noel Vaillant
                    I am very sorry Shuva, I have been stuck for 2 hours on this. I need to move on, otherwise I ll go crazy. I think Jason has a good chance to find a complete
                    Message 9 of 18 , Nov 5, 2004
                      I am very sorry Shuva,
                      I have been stuck for 2 hours on this.
                      I need to move on, otherwise I ll go crazy.
                      I think Jason has a good chance to find a complete proof.

                      Noel.
                    • shuva gupta
                      Thanks Jason Noel and Myriam, Actually the problem has a very simple solution if we exploit the fact that g(x)=|x|^r, r =1 is a convex function. ie we use the
                      Message 10 of 18 , Nov 8, 2004
                        Thanks Jason Noel and Myriam,
                        Actually the problem has a very simple solution if we exploit the fact that
                        g(x)=|x|^r, r>=1 is a convex function.
                        ie
                        we use the property g(x)-g(y)>=(x-y)*g'(y) when g(.)is convex.
                        Put x=X-E(X)
                        y= ~X - E(~X) Then take expectation on both sides .
                        Regards,
                        Shuva
                        PS A friend of mine found this solution in the book Probability by Chow and Teicher 1st Edition ( pg 102-103)

                        jason1990 <jason1990@...> wrote:


                        Yes, you're right. I thought about this in the morning on the bus
                        and again this evening before going to a movie. I had two lines of
                        reasoning in my head at once and when I reread my post, I confused
                        myself.

                        About differentiating under the expectation, something occurred to
                        me at the theater. For any nonnegative random variable Y and any
                        r>0, E[Y^r]=\int_0^\infty{rz^{r-1}P(Y>z)dz}. So suppose we're given
                        a real t and we want to compute E[min(X,t)]. Let Y=t-min(X,t). Then
                        Y is nonnegative and we can apply the above to get

                        E[min(X,t)] = t - \int_{-\infty}^t{P(X < z)dz}.

                        We can differentiate this with no problem. Something similar should
                        be possible with the other expectation. I don't think this is
                        necessary to justify the differentiation, but it makes me wonder
                        whether the whole differentiation approach is unnecessary.

                        Another thing that is curious: the only property of the map x->x^
                        {2k} that was used is the fact that it has a nondecreasing
                        derivative. In other words, it is convex. So perhaps the original
                        poster's claim is true not only for functions of the form x->|x|^r
                        where r>=1, but for all convex functions. If so, then maybe Jensen's
                        inequality would be useful in creating a simpler proof.

                        I just feel that it shouldn't be this hard.

                        --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
                        wrote:
                        >
                        >
                        > > > Since 2k-1 is odd,
                        > > >
                        > > > (X - g(t))^{2k-1} 1_{X < t < Y}
                        > > > <= (t - g(t))^{2k-1} 1_{X < t < Y}.
                        >
                        > Since (2k-1) is odd, x->x^(2k-1)is non-decreasing on R
                        > and since X - g(t) <= t - g(t) on {X < t}, the inequality
                        > you wrote seems fine to me.
                        >
                        > I also think this inequality allows you to conclude
                        > that h'(t) >= 0.
                        >
                        > h'(t)
                        > = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                        > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}]
                        > >= 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                        > - 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                        > = 2k(t-g(t))^{2k-1}E[1_{X > t > Y} - 1_{X < t < Y}]
                        > =0
                        >
                        > which is pretty much what you wrote in your first post.
                        > What am I missing?
                        >
                        > > What I wanted to say was
                        > >
                        > > > Since 2k-1 is odd,
                        > > >
                        > > > (X - g(t))^{2k-1} 1_{X < t < Y}
                        > > > <= (t - g(t))^{2k-1} 1_{X > t > Y}.
                        > >
                        > > But that's clearly false.
                        >
                        > I wonder why you wanted to write this in the first place.
                        >
                        >
                        >
                        > > But I think it's salvagable. Here's another way to show
                        > >h'(t)>=0. Start with what we know:
                        > >
                        > > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X > t > Y}]
                        > > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                        > >
                        > > Since X and Y are iid, I can interchange them in the first
                        > > expectation, giving
                        > >
                        > > h'(t) = 2k E[(t - g(t))^{2k-1} 1_{X < t < Y}]
                        > > - 2k E[(X - g(t))^{2k-1} 1_{X < t < Y}].
                        > >
                        > > Now we use the fact that (X - g(t))^{2k-1} <= (t - g(t))^{2k-1}
                        > > whenever X < t to conclude that h'(t)>=0.
                        > >
                        > > Any other glaring mistakes? (Besides the lack of rigor?) :)
                        >
                        > This seems to me like the same as what you wrote in the first
                        place,
                        > except you use iid property, before taking inequalities.
                        >
                        > Any way, it seems to me your proof is good. I certainly haven't
                        > seen a flaw.
                        >
                        > Noel.







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                      • Noel Vaillant
                        Is it obvious that E[(x-y)g (y)] =0 for this choice of x and y? I am not sure I understand. Noel.
                        Message 11 of 18 , Nov 8, 2004
                          Is it obvious that E[(x-y)g'(y)]>=0 for this choice of x and y?
                          I am not sure I understand.

                          Noel.


                          > g(x)=|x|^r, r>=1 is a convex function.
                          > ie
                          > we use the property g(x)-g(y)>=(x-y)*g'(y) when g(.)is convex.
                          > Put x=X-E(X)
                          > y= ~X - E(~X) Then take expectation on both sides .
                        • jason1990
                          For what it s worth, it s not obvious at all to me. But maybe we re both missing something. The book is apparently Probability theory : independence,
                          Message 12 of 18 , Nov 10, 2004
                            For what it's worth, it's not obvious at all to me. But maybe we're
                            both missing something. The book is apparently "Probability theory :
                            independence, interchangeability, martingales" by Yuan Shih Chow,
                            Henry Teicher. I have a request on that book just to see what's
                            going on. If someone can enlighten me while I wait, that would be
                            great.

                            --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
                            wrote:
                            >
                            >
                            >
                            > Is it obvious that E[(x-y)g'(y)]>=0 for this choice of x and y?
                            > I am not sure I understand.
                            >
                            > Noel.
                            >
                            >
                            > > g(x)=|x|^r, r>=1 is a convex function.
                            > > ie
                            > > we use the property g(x)-g(y)>=(x-y)*g'(y) when g(.)is convex.
                            > > Put x=X-E(X)
                            > > y= ~X - E(~X) Then take expectation on both sides .
                          • shuva gupta
                            Sorry folks, for the confusion, Sorry Jason you are right the book is Probability theory : independence, interchangeability, martingales by Yuan Shih Chow,
                            Message 13 of 18 , Nov 10, 2004
                              Sorry folks, for the confusion,
                              Sorry Jason you are right the book is
                              "Probability theory : independence, interchangeability, martingales" by Yuan Shih Chow,
                              Henry Teicher.

                              If G(.) is a convex fxn then this propert hold :

                              G(x)-G(y) >=(x-y)G'r(y) { G'r(y):= rt hand derivative of G(y) at the pt y moreover G'(.) is an increasing function.}

                              Now take x=X-E(X) y=~X-E(~X)

                              therefore
                              G(X-E(X))-G(~X-E(~X))>=(X-E(X)-(~X-E(~X))G'r(~X-E(~X))

                              Now it can be shown that ((X-E(X)-(~X-E(~X))G'r(~X-E(~X)) is >=
                              (X-E(X)-(~X-E(~X))*K ( where K is some constant, the detailed argument is in the book , using the fact that G'r(.) is increasing and also the fact the function a(x)=x-~x is monotonically incraesing (where ~x=a if x<=a, =x if a=<x <=b and =b if x>=b; a<b)

                              Thus we have
                              G(X-E(X))-G(~X-E(~X))>=(X-E(X)-(~X-E(~X))*K
                              Now taking expectation on both sides we have

                              E(G(X-E(X)))- E(G(~X-E(~X)))>=E((X-E(X)-(~X-E(~X))*K)

                              Since E((X-E(X)-(~X-E(~X))*K)=0 thus we have

                              E(G(X-E(X)))- E(G(~X-E(~X)))>=0

                              Take G(x)=|x|^r (r>=1) and we get the desired result ie
                              E(|X-E(X)|^r)>=E(|~X-E(~X)|^r)


                              If it is still not clear please let me know and I will be glad to put in more details ( or maybe I can also scan a couple of relevant pages from the book and send it across only if you ask ....I dont want to unnecessary overload the inboxes:-))))))))))

                              Best Regards,
                              Shuva



                              jason1990 <jason1990@...> wrote:


                              For what it's worth, it's not obvious at all to me. But maybe we're
                              both missing something. The book is apparently "Probability theory :
                              independence, interchangeability, martingales" by Yuan Shih Chow,
                              Henry Teicher. I have a request on that book just to see what's
                              going on. If someone can enlighten me while I wait, that would be
                              great.

                              --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
                              wrote:
                              >
                              >
                              >
                              > Is it obvious that E[(x-y)g'(y)]>=0 for this choice of x and y?
                              > I am not sure I understand.
                              >
                              > Noel.
                              >
                              >
                              > > g(x)=|x|^r, r>=1 is a convex function.
                              > > ie
                              > > we use the property g(x)-g(y)>=(x-y)*g'(y) when g(.)is convex.
                              > > Put x=X-E(X)
                              > > y= ~X - E(~X) Then take expectation on both sides .







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                              [Non-text portions of this message have been removed]
                            • Noel Vaillant
                              ... ((X-E(X)-(~X-E(~X))G r(~X-E(~X)) = (X-E(X)-(~X-E(~X))*K ... I feel very dumb here. Tried but failed. Will someone take me out of my misery? I know
                              Message 14 of 18 , Nov 11, 2004
                                > Now it can be shown that:

                                ((X-E(X)-(~X-E(~X))G'r(~X-E(~X)) >= (X-E(X)-(~X-E(~X))*K

                                > where K is some constant, [...] using the fact that G'r(.) is
                                >increasing and [...] a(x)=x-~x is monotonically increasing

                                I feel very dumb here. Tried but failed. Will someone take me
                                out of my misery? I know (~X-E(~X)) is bounded, and since G'
                                is non-decreasing, |G'r(~X-E(~X))| is also bounded. But somehow
                                I can't manage to conclude (and use the hint about a(x))

                                I have been spending so much time on this. I may as well
                                go to the end, so it won't have been for nothing :-)

                                Noel.
                              • jason1990
                                I think I see it now. Define f(t) = t - ~t - E(X) + E(~X) and g(t) = G r(~t - E(~X)). Both functions are nondecreasing, and we can find t_0 such that f(t)
                                Message 15 of 18 , Nov 11, 2004
                                  I think I see it now. Define

                                  f(t) = t - ~t - E(X) + E(~X) and
                                  g(t) = G'r(~t - E(~X)).

                                  Both functions are nondecreasing, and we can find t_0 such that

                                  f(t) <= 0 for t <= t_0 and
                                  f(t) >= 0 for t >= t_0.

                                  Hence, if X >= t_0, then since f(X) >= 0 and g(X) >= g(t_0), we have

                                  f(X)g(X) >= f(X)g(t_0).

                                  Also, if X <= t_0, then f(X) <= 0 and g(X) <= g(t_0), so

                                  f(X)g(X) >= f(X)g(t_0).

                                  So I guess we take K = g(t_0). I think this works. I wouldn't call
                                  it obvious, though.

                                  --- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
                                  wrote:
                                  >
                                  >
                                  > > Now it can be shown that:
                                  >
                                  > ((X-E(X)-(~X-E(~X))G'r(~X-E(~X)) >= (X-E(X)-(~X-E(~X))*K
                                  >
                                  > > where K is some constant, [...] using the fact that G'r(.) is
                                  > >increasing and [...] a(x)=x-~x is monotonically increasing
                                  >
                                  > I feel very dumb here. Tried but failed. Will someone take me
                                  > out of my misery? I know (~X-E(~X)) is bounded, and since G'
                                  > is non-decreasing, |G'r(~X-E(~X))| is also bounded. But somehow
                                  > I can't manage to conclude (and use the hint about a(x))
                                  >
                                  > I have been spending so much time on this. I may as well
                                  > go to the end, so it won't have been for nothing :-)
                                  >
                                  > Noel.
                                • Noel Vaillant
                                  Thank you very much Jason. This looks very good to me, and is a huge relief :-) Noel.
                                  Message 16 of 18 , Nov 11, 2004
                                    Thank you very much Jason. This looks very good to me,
                                    and is a huge relief :-)


                                    Noel.



                                    > I think I see it now. Define
                                    >
                                    > f(t) = t - ~t - E(X) + E(~X) and
                                    > g(t) = G'r(~t - E(~X)).
                                    >
                                    > Both functions are nondecreasing, and we can find t_0 such that
                                    >
                                    > f(t) <= 0 for t <= t_0 and
                                    > f(t) >= 0 for t >= t_0.
                                    >
                                    > Hence, if X >= t_0, then since f(X) >= 0 and g(X) >= g(t_0), we have
                                    >
                                    > f(X)g(X) >= f(X)g(t_0).
                                    >
                                    > Also, if X <= t_0, then f(X) <= 0 and g(X) <= g(t_0), so
                                    >
                                    > f(X)g(X) >= f(X)g(t_0).
                                    >
                                    > So I guess we take K = g(t_0).
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