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Using standard notation: $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dW_t, \:\:X_0=x \tag{1}$$ Now in my script it is said that if we integrate both sides, we get: $$X_t=x+\int_0^t b(s,X_s)ds+\int_0^t\sigma(s,X_s)dW_s \tag{2}$$ It seems quite unintuitive for me because which integral do we use on both sides? The first integral is Lebesgue and the second one is Ito. And we differentiate under integral sign? I know that Fundamental Theorem of Calculus is not true in Ito calculus so how $$\int_0^t\frac{\partial \sigma(s,X_s)}{\partial X_s}dW_s=\sigma(s,X_s) \tag{3}$$ And why we ad $dt$ and $dW_t$ on right hand side in (1). I'm confused here.

I know it's a beginner stuff, but almost all authors just give this information and acts like as it was obvious. Thank you for any explanation of that convention.

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It is exactly the other way round: We define $(1)$ to be the same thing as $(2)$; it's just a more convenient way to write $(2)$. That this notation makes sense can be seen by integrating both sides of $(1)$ from $0$ to $T$:

$$\underbrace{\int_0^T dX_t}_{X_T-X_0 = X_T-x} = \int_0^T b(t,X_t) \, dt + \int_0^T \sigma(t,X_t) \,dW_t.$$

A similar situation pops up in real ("deterministic") analysis: If we write

$$dx_t = f(x_t) \, dt, \tag{4}$$

then we might interpret this equation in the following ways:

  • We read it as a differential equation, i.e. as $$\frac{dx_t}{dt} = f(x_t). \tag{5}$$ In order to make sense of this, $(x_t)_t$ has to be differentiable.
  • We read it as an integral equation, i.e. we give it a meaning by integrating both sides: $$x_t-x_0 = \int_0^t \, dx_s = \int_0^t f(x_s) \, ds. \tag{6}$$ More generally, we find $$\int_0^t g(s) \, dx_s = \int_0^t g(s) f(s) \, ds.$$

In both cases, $(4)$ is by definition (!) a shortcut for $(5)$ and $(6)$, respectively. In stochastic calculus, we cannot read the equation as a differential equation since the Brownian motion is nowhere differentiable. Consequently, we cannot expect the process to be differentiable. In contrast, the interpretation as an integral equation works fine and turns out quite useful if we have to calculate stochastic integrals.

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  • $\begingroup$ So it is just a definiton and we can't derive it, yes? $\endgroup$ – luka5z Sep 14 '14 at 7:13
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    $\begingroup$ @luka5z Yes, it's simply a notation (and its our job to give notations a meaning). Actually, it's quite similar for ODEs: We write e.g. $$dx_t = f(x_t) \, dt$$ - and without defining what we mean this notation doesn't make sense at all. $\endgroup$ – saz Sep 14 '14 at 8:35
  • $\begingroup$ Can you write a bit more about this similarity, maybe edit your answer please. I would be very grateful. $\endgroup$ – luka5z Sep 14 '14 at 11:04
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    $\begingroup$ @luka5z Done... $\endgroup$ – saz Sep 14 '14 at 12:55
  • $\begingroup$ I have a question open with a similar issue and you seem to justify the equlity by integrating. But then $(1)$ is not equal to $(2)$ but rather looks like some kind of measure which we take the integral of e.g $\mu=\int d\mu$ $\endgroup$ – user1 Oct 13 '18 at 17:52

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