- #1
mathfied
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Hi Everyone,
A problem I have here. I am trying to solve a problem involving Ito Integrals and Riemann interals.
Prove
[tex]\int^{T}_{0} tdW(t) = TW(T) -\int^{T}_{0} W(t)dt [/tex]
I want to solve this question WITHOUT using Ito's Lemma directly.
OK I know that in general, the Ito integral and the Riemann integral are going to be slightly different. The Ito Integral will have an extra term.
So in my question, I know if I was to generally just integrate the Riemann integral, I would get:
[tex]\int^{T}_{0} tdW(t) = TW(T) [/tex]
But I know the Ito integral adds the extra:
[tex] -\int^{T}_{0} W(t)dt [/tex]
But how does this arise? Is it possible to show how using the mean square error ? I keep reading the mean square error when reading about this but don't really see the connection. Maybe someone could kindly help out please?
Thanks
A problem I have here. I am trying to solve a problem involving Ito Integrals and Riemann interals.
Homework Statement
Prove
[tex]\int^{T}_{0} tdW(t) = TW(T) -\int^{T}_{0} W(t)dt [/tex]
Homework Equations
I want to solve this question WITHOUT using Ito's Lemma directly.
The Attempt at a Solution
OK I know that in general, the Ito integral and the Riemann integral are going to be slightly different. The Ito Integral will have an extra term.
So in my question, I know if I was to generally just integrate the Riemann integral, I would get:
[tex]\int^{T}_{0} tdW(t) = TW(T) [/tex]
But I know the Ito integral adds the extra:
[tex] -\int^{T}_{0} W(t)dt [/tex]
But how does this arise? Is it possible to show how using the mean square error ? I keep reading the mean square error when reading about this but don't really see the connection. Maybe someone could kindly help out please?
Thanks