## Re: Simple Ito Calculus Help Request

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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)
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.
• 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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• 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 "