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Originally posted by: tibbar
hi,

if we are defining Ito integrals from first principles, and choose to use the Riemann-Stieltjes integral definition, we can calculate the integral as the limit as n->infty of:

or:

The 1st of these definitions turns out to be the "correct" one, since it produces martingales. The integral however, does not exist in a Riemann-Stieltjes sense, since the two limits do not coincide.

Now, can anyone give any further justification for choosing the 1st integral definition?

I can see that by setting f(B_t) = B_t taking expectations of each integral, we get for the first integral definition E(...) = 0, whereas the 2nd integral gives E(...) = t
(using E(B_s x B_ t) = min(s,t) )

and since when f(B_t) = B_t, the integrals should "intuitively" represent B_t, we will pick definition 1, to ensure that E(B_t) = 0.

But this all seems too convenient to me, is there some more technical reason we define the Ito integral this way, other than it just works?

Thx.

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X'_t (ω ) = I (m_t f )(ω ).
(8)
Sadly, this candidate has problems; for each 0 ≤ t ≤ T the integral
I (m_t f ) is only defined as an element of L_2 (dP ), so the value of I (m_t f ) can
be specified arbitrarily on any set A_t ∈ F_t with P (A_t ) = 0. The union of
the A_t over all t in [0, T ] can be as large as the full set Ω, so, in the end,
the process X_t suggested by (8) might not be continuous for any ω ∈ Ω.
This observation is troubling, but it is not devastating. With care (and
help from Doob?s maximal inequality) one can prove that there exists a
unique continuous martingale X_t which agrees with X_t with probability one
for each fixed t ∈ [0, T ]. The next theorem gives a more precise statement
of this crucial fact.

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