- 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.

--- 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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