Re: expectation of e^{Int Y(s) ds}

On 06.09.2007 14:49, Jess wrote:
On Sep 6, 10:42 pm, Jannick Asmus <jannick.n...@xxxxxx> wrote:
On 06.09.2007 14:24, Jess wrote:

On Sep 6, 10:04 pm, Jannick Asmus <jannick.n...@xxxxxx> wrote:
Alternatively you could derive the stochastic differential equation
(SDE) fulfilled by the stochastic process exp[int Y](t). Then taking
expectations of the SDE reveals the expectation at once. But ... this
approach uses the Itô calculus as well. Do you have it in stock?
I also tried to derive SDE of Z(t) = exp{Integral from 0 to of Y(s)
ds} and got
dZ(t) = Z(t)*Y(t)*dt
Oups, where did the dY(t)-term disappear? Did you really apply Itô's
rule? ;) ... maybe you try one more time

BTW: This is the easier way to get to the aim.

Then how can I read off E(Z(t)) at once? I thought I had to do E(Z(t))
= E[dZ(t) / (Y(t) * dt)]
which is rather hard to find. I guess there must be some clever
method to do E(Z(t))?
Thanks,
Jess
--
J.


Many thanks, please also ignore my request to look at this problem in
my previous email: I posted the message before seeing this reply from
you. :)

I derived the answer this way:

Z(t) = exp{X(t)}

where

X(t) = Integral from 0 to t of Y(s) ds

Then

dZ(t) = Z(t)*{dX(t) + 1/2 * dX(t) * dX(t)} -- equation (1)

dX(t) = Y(t) * dt -- equation (2)

Hence I got the result as dZ(t) = Z(t) * {Y(t) * dt}

That's fine. :-)

I think I must have made a mistake at either (1) or (2). (1) looks
correct to me, but (2) also seems correct... Can you please point out
my problem? :)

Everything seems to be correct. I made a mistake on my little piece of paper which resulted in a SDE which could be solved like this. :-( The tricky thing I applied was that I took to expectation of the SDE yielding an ordinary differential equation for the expectation. In many situations it is a real help.

So we need to stick to the first approach you have already gone through.

--
Best wishes,
J.
.

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