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Can someone give me an intuition of what is the stochastique integral $\int_0^T X_s\mathrm d B_s$ where $(B_t)_t$ is the brownian motion or $\int_0^T X_s\mathrm d M_s$ when $(M_t)_t$ is a martingale ? I really don't see the intuition behind those stochastic integral. Be intuition, I mean, concretely what is this integral. For example, $\int_a^b f(x)dx$ in Riemann sense, is the area under the curve $y=f(x)$ between $a$ and $b$. Or $\int_\Gamma F\cdot d\gamma $, is the circulation of the vector field $F$ into the curve $\Gamma$, or $\iint_\Sigma F\cdot d\sigma $ is the flow of $F$ throuw the surface $\Sigma$...

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If you have an integral of the form $\int_0^T f(t) dt$ the intuition is: $$ \lim_{n\rightarrow \infty} \sum_{i = 0}^n f(t_i) \cdot (t_{i+1} - t_i), $$

where $0 = t_0<\dots< t_{n+1} = T$ and $\Delta(t_0,\dots,t_{n+1}) := \max(|t_{i+1} - t_i|, i =0,\dots,n) \rightarrow 0$.

Now just replace this "time" by Brownian time $B_t$ to obtain something like:

$$ \lim_{n\rightarrow \infty} \sum_{i=0}^n f(t_i) \cdot (B_{t_{i+1}} - B_{t_i}) $$ so we take the same as for Rieman integrals but with the time scale replaced by a Brownian motion.

For a formal definition you can for example look at this course in particular lecture 15.

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  • $\begingroup$ I don't need a definition ! Just to see concretely what it is. For example, with Riemann intégral, it's the area under the curve... $\endgroup$ – user380364 Dec 11 '16 at 10:43
  • $\begingroup$ Oke, but why is it the "area under the curve" because your $x-$axis is $[0,T]$ and $t_0,\dots,t_{n+1}$ are all just points on the $x-$axis and we take the above sum. Now replace your $x-$axis by the $B_{t}$, then you can still kind of see it as the "area under the curve" but with an adapted $x-$axis $\endgroup$ – HolyMonk Dec 11 '16 at 10:45
  • $\begingroup$ Also see mathoverflow.net/questions/29750/… $\endgroup$ – HolyMonk Dec 11 '16 at 10:47

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