Suppose I have $$dS_t = \mu(S_t,t) dt + \sigma(S_t,t)dW_t$$ What would be the process satisfying the following process of $y_t$? $$y_t = \int_0^t S_u du + \int_0^t S_u dW_u$$

I'm not quite sure about differentiating $y_t$. The following is what I did $$\frac{\partial y_t}{\partial S_t} =dt +dW_t $$ and $$\frac{\partial^2 dy_t}{\partial dS_t^2} = 0$$

Are these right? Then Ito's Formula gives $$dy_t = (dt+dW_t)dS_t = \sigma(S_t,t)dt $$

But this feels wrong.


A stochastic differential equation is nothing more than a short-hand notation for a corresponding integral equation. So the initial SDE you provided actually means

$$ \int_0^t d S_u = \int_0^t \mu(S_u, u) du + \int_0^t\sigma(S_u, u) dW_u$$

This is how the SDE is defined (see e.g. here). The reason is that you cannot differentiate a Brownian motion. It does not have a derivative according to the usual definition of calculus (taking limits etc).

Things like $\frac{\partial y}{\partial S_t}$ just don't make sense in the world of stochastic calculus.

OK, so, back to your equation. Note that it can be written as:

$$ \int_0^t dy_u = \int_0^t S_u du + \int_0^t S_u dW_u$$

with $y_0 = 0$. Then the corresponding SDE is simply obtained by removing the integration signs:

$$dy_t = S_t dt + S_t dW_t$$

That's it! And why? Well, again, because this SDE is actually defined as the corresponding integral equation. There is no corresponding differential equation which involves actual derivatives.

  • $\begingroup$ There is something missing: You write "should be really be interpreted as a" ...? $\endgroup$ – vonjd Feb 8 '16 at 14:21
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    $\begingroup$ Woops. Removed the half-sentence. It was just more of the same: only integrals exists; there are no derivatives. $\endgroup$ – Olaf Feb 8 '16 at 15:14
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    $\begingroup$ Thank you very much for your explanation! It confused me but now it's clear. $\endgroup$ – Kenneth Chen Feb 8 '16 at 16:41

@Olaf gave a clear answer. Another way to see this is as system of 2 SDEs:

\begin{cases} dS_u &= \mu(S_u,u) du + \sigma(S_u,u) dW_u \\ dy_u &= S_u du + S_u dW_u. \end{cases}

E.g. if we want to simulate this system using Euler discretuzation then we perform \begin{cases} S_u + \Delta S_u &= S_u + \mu(S_u,u) \Delta t + \sigma(S_u,u) \epsilon \sqrt{\Delta t} \\ y_u + \Delta y_u &=y_u + S_u \Delta t + S_u \epsilon \sqrt{\Delta t} \end{cases} for a chosen time step $\Delta t$ and where $\epsilon$ is standard normal and sampled in each iteration.

  • $\begingroup$ @Richard R.H.S. is just increment in $S_u$, so, don't you think RHS must include $S_u$ in the equation ? $\endgroup$ – Neeraj Feb 9 '16 at 15:07

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