I found this process in a scientific paper:

$M_t = \int_{0}^t e^{-(t-u)} \frac{dS_u}{S_u}$


$dS_t = S_t (\phi M_t + (1-\phi)\mu_t) dt + \sigma S_t dW_t$

and I want to compute the differential $dM_t$. In my opinion, I don't even need Ito since $f(t,s)$ in the Ito function is really only a function of $t$. So from my basic calculus skills:

$\frac{dM_t}{dt} = e^{-(t-t)} \frac{dS_t}{S_t} - \int_0^t e^{-(t-u)} \frac{dS_u}{S_u} \\ \iff dM_t = \frac{dS_t}{S_t} dt - M_t dt$

In the paper though they derive:

$dM_t = \frac{dS_t}{S_t} - M_t dt$

missing one dt term. I am pretty sure I am wrong (it's not a crap paper) but I don't understand where exactly. For example, if I want to calculate the derivative of

$F(x,y) = \int_0^x f(x,y) dy$

w.r.t. x, then this should be

$\frac{dF}{dx}(x,y) = f(x,x) + \int_0^x \frac{\partial f}{\partial x}(x,y) dy$

which is what I applied above.. ?!

  • $\begingroup$ You don't have a "classical" integral of the form $$F(t) = \int_0^t f(t,y) \, dy$$ but a stochastic integral $$F(t) = \int_0^t f(t,s) \, dW_s.$$ This means that the calculations rule from classical analysis do not apply. You really have to use Itô's formula in order to calculate the differential $dM_t$. $\endgroup$ – saz May 20 '15 at 17:26
  • $\begingroup$ Use $M_t=e^{-t}N_t$ with $dN_t=e^{t}dS_t/S_t$ (by definition) and $dM_t=e^{-t}dN_t-e^{-t}N_tdt$ (Itô with $f(x,t)=e^{-t}x$) hence, indeed, $dM_t=dS_t/S_t-M_tdt$. $\endgroup$ – Did May 20 '15 at 21:10

even if would apply the classical calculus rules, then I think you did not use correctly the last formula your write. Namely, in the first term of the last equation, there is no $dy$. Hence, following your logic, you need to divide the first term in your equation by $dt$. This then gives the correct equation. Hope this helps.

  • $\begingroup$ Hm, are you sure about that? If I have for example $F(x) = \int_0^x f(y) dy$, then clearly $F'(x) = f(x)$ and there is no $dy$ either or? $\endgroup$ – marky2k May 20 '15 at 20:12

I agree, what I mean is that if you want to use the formula $\frac{dF}{dx}(x,y)= f(x,x) $+ 2nd term, then the $f(x,x)$ is the full expression after the integral symbol, but divided by the $dy$

Therefore, applying this method formally, you should write $\frac{dM}{dt} = e^{-(t-u) } \frac{dS}{S dt}$ + 2nd term.


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