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Re: Simple Ito Calculus Help Request

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  • Noel Vaillant
    Hi M.W. with slightly different notations, if: s(t)=s(0)exp(mt + vw(t)) then: E[s(t)]=s(0)exp(mt)E[exp(vw(t))] Since w is assumed to be a brownian motion, w(t)
    Message 1 of 13 , Nov 1, 2003
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      Hi M.W.

      with slightly different notations, if:

      s(t)=s(0)exp(mt + vw(t))

      then:

      E[s(t)]=s(0)exp(mt)E[exp(vw(t))]

      Since w is assumed to be a brownian motion, w(t) is normally
      distributed with mean 0 and variance t. So vw(t) is normally
      distributed with mean 0 and variance (v^2)t. So X=vw(t)/(v.sqr(t))
      is normally distributed with mean 0 and variance 1, i.e. X~N(0,1).
      Hence, we have to compute the expectation:

      E[exp(aX)] where a=v.sqr(t) and X~N(0,1). This expectation is equal
      to the integral:

      E[exp(aX)]=K\int_R exp(ax-x^2/2)dx
      =K\int_R exp(-(x-a)^2/2 + a^2/2)dx
      =exp(a^2/2)K\int_R exp(-(x-a)^2/2)dx
      =exp(a^2/2)

      where K=1/sqr(2pi). Finally we obtain:

      E[s(t)]=s(0)exp(mt)exp(v^2t/2)

      as claimed. This solution did not use at all the SDE (stochastic
      differential equation):

      ds(t)=(m + v^2/2)s(t)dt+vs(t)dw(t)

      If you want to use this fact, write the SDE in integral form:

      s(t)-s(0)=int_0^t (m + v^2/2)s(u)du +int_0^t vs(u)dw(u)

      and take expectations on both sides:

      E[s(t)]-s(0)=E[int_0^t (m + v^2/2)s(u)du]

      where we have used the fact that E[int_0^t vs(u)dw(u)]=0
      (since w is a martingale, x(t)=int_0^t y(s)dw(s) is also
      a martingale for any y (almost :-) and consequently
      E[x(t)]=E[x(0)]=E[0]=0)

      Now, we can use Fubini theorem, and change the order of integration:

      E[int_0^t (m + v^2/2)s(u)du]=int_0^t (m + v^2/2)E[s(u)]du

      and if we define f(t)=E[s(t)], we obtain the equation:

      f(t)=s(0)+\int_0^t (m + v^2/2)f(u)du

      Note that f(0)=s(0), and if we differentiate on both side (a little
      bit of work should be nice here to show that you are allowed to do
      that) we obtain:

      f'(t)= (m + v^2/2)f(t)

      the solution of which (given that f(0)=s(0)) is:

      f(t)=s(0)exp[(m+v^2/2)t]



      > We showed in class using Ito's formula that
      > s(t) = s(0)e^[µt+ów(t)] solves the stochastic
      > differential equation ds = (µ + 1/2ó^2)sdt + ósdw
      > with initial condition s(0) = s0.
      >
      >
      > Conclude that E[s(t)] = s(0)e^(µ + 1/2ó^2)t.
    • The Webmaster
      Noel: Thanks for you very thoughtful exposition. Best Regards, M.W. ... __________________________________ Do you Yahoo!? Exclusive Video Premiere - Britney
      Message 2 of 13 , Nov 1, 2003
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        Noel:

        Thanks for you very thoughtful exposition.

        Best Regards,
        M.W.


        --- Noel Vaillant <vaillant@...> wrote:
        > Hi M.W.
        >
        > with slightly different notations, if:
        >
        > s(t)=s(0)exp(mt + vw(t))
        >
        > then:
        >
        > E[s(t)]=s(0)exp(mt)E[exp(vw(t))]
        >
        > Since w is assumed to be a brownian motion, w(t) is
        > normally
        > distributed with mean 0 and variance t. So vw(t) is
        > normally
        > distributed with mean 0 and variance (v^2)t. So
        > X=vw(t)/(v.sqr(t))
        > is normally distributed with mean 0 and variance 1,
        > i.e. X~N(0,1).
        > Hence, we have to compute the expectation:
        >
        > E[exp(aX)] where a=v.sqr(t) and X~N(0,1). This
        > expectation is equal
        > to the integral:
        >
        > E[exp(aX)]=K\int_R exp(ax-x^2/2)dx
        > =K\int_R exp(-(x-a)^2/2 + a^2/2)dx
        > =exp(a^2/2)K\int_R exp(-(x-a)^2/2)dx
        > =exp(a^2/2)
        >
        > where K=1/sqr(2pi). Finally we obtain:
        >
        > E[s(t)]=s(0)exp(mt)exp(v^2t/2)
        >
        > as claimed. This solution did not use at all the SDE
        > (stochastic
        > differential equation):
        >
        > ds(t)=(m + v^2/2)s(t)dt+vs(t)dw(t)
        >
        > If you want to use this fact, write the SDE in
        > integral form:
        >
        > s(t)-s(0)=int_0^t (m + v^2/2)s(u)du +int_0^t
        > vs(u)dw(u)
        >
        > and take expectations on both sides:
        >
        > E[s(t)]-s(0)=E[int_0^t (m + v^2/2)s(u)du]
        >
        > where we have used the fact that E[int_0^t
        > vs(u)dw(u)]=0
        > (since w is a martingale, x(t)=int_0^t y(s)dw(s) is
        > also
        > a martingale for any y (almost :-) and consequently
        > E[x(t)]=E[x(0)]=E[0]=0)
        >
        > Now, we can use Fubini theorem, and change the order
        > of integration:
        >
        > E[int_0^t (m + v^2/2)s(u)du]=int_0^t (m +
        > v^2/2)E[s(u)]du
        >
        > and if we define f(t)=E[s(t)], we obtain the
        > equation:
        >
        > f(t)=s(0)+\int_0^t (m + v^2/2)f(u)du
        >
        > Note that f(0)=s(0), and if we differentiate on both
        > side (a little
        > bit of work should be nice here to show that you are
        > allowed to do
        > that) we obtain:
        >
        > f'(t)= (m + v^2/2)f(t)
        >
        > the solution of which (given that f(0)=s(0)) is:
        >
        > f(t)=s(0)exp[(m+v^2/2)t]
        >
        >
        >
        > > We showed in class using Ito's formula that
        > > s(t) = s(0)e^[�t+�w(t)] solves the stochastic
        > > differential equation ds = (� + 1/2�^2)sdt + �sdw
        > > with initial condition s(0) = s0.
        > >
        > >
        > > Conclude that E[s(t)] = s(0)e^(� + 1/2�^2)t.
        >
        >
        >
        >


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      • arash nademi
        hi group I need a solution problems for probability and measure :patrik billingsley please send to me. by. ... Yahoo! Music Unlimited - Access over 1
        Message 3 of 13 , Oct 19, 2005
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          hi group
          I need a solution problems for "probability and measure :patrik billingsley "
          please send to me.
          by.



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