Reading an Integral Process as a Stochastic Differential Equation
Summary
The document clarifies how to obtain the stochastic differential for a process defined as the sum of an ordinary time integral and an Itô integral. Rather than differentiating with respect to the stochastic state or Brownian motion, it treats the integral equation as the definition of the corresponding SDE. The process therefore has a time-integral term and a Brownian-integral term, each with the original state as its integrand. Ordinary derivatives such as those proposed in the question are not meaningful tools for this step because Brownian paths are not differentiable in the usual calculus sense.
A second answer illustrates the relationship by writing the state and accumulated process as a coupled SDE system, then gives an Euler-style discretization using a normal random draw scaled by the square root of the time step. This offers a simulation interpretation, but the example does not discuss numerical error, convergence, or assumptions on the coefficients. The key lesson is to distinguish a stochastic integral identity from ordinary differentiation and to translate between integral and differential notation accordingly.
Key ideas
- An SDE is shorthand for its associated integral equation.
- Brownian motion is not differentiable in the usual calculus sense, so ordinary partial derivatives with respect to it are not appropriate.
- The process defined by the two integrals has drift and diffusion terms inherited directly from their integrands.
- An Euler discretization approximates the Brownian increment with a normal draw scaled by the square root of the time step.
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# Ito Formula for Stochastic Integral
# Ito Formula for Stochastic Integral
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.
## Answer by Olaf (score 8, accepted)
https://quant.stackexchange.com/a/23163
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.
## Answer by Richi Wa (score 1)
https://quant.stackexchange.com/a/23184
@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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.