Consider the following processes: $$X_t=e^{\int_0^t f(s,\omega)ds}$$ $$Y_t=e^{\int_0^t g(s,\omega)dB_s}$$ Assume $f$ and $g$ have whatever properties necessary to make this tractable, e.g. square integrable, etc. $B_t$ is standard 1D Brownian motion starting at the origin. I want to understand how to directly take the stochastic Ito differential of these.

QUESTION: What are the stochastic differential equations for $dX_t$ and $dY_t$?

Here is one attempt:

$$d X_t=X_tf(t,\omega)dt +X_t\left(\int_0^t f_{B_s}(s,\omega)ds\right)dB_t +\frac12X_t\left(\int_0^t f_{B_sB_s}(s,\omega)ds\right)dt$$

$$d Y_t=Y_tg(t,\omega)dt +Y_t g(t,\omega)dB_t +\frac12Y_t g_{B_s}(t,\omega)dt$$

Of course, I am passing a derivative w.r.t. $B_t$ into the integral, and I am unsure about if that is ok here, or generally.

Also, I thought about taking logs: $$\log X_t=\int_0^t f(s,\omega)ds$$ So that $$d(\log X_t)=f(t,\omega)dt$$ but by the Ito formula it also is $$d(\log X_t)=\frac{1}{X_t}dX_t-\frac12\frac{1}{X_t^2}(dX_t)^2.$$

The similar calculation with $Y_t$ gets me stuck in a similar situation. I'm not sure how to deal with the $(dX_t)^2$ here. I also tried integration by parts to no avail. I'm guessing there is a standard result or trick that can be applied or that I have some basic mistake here. Any help is appreciated.


2 Answers 2


Let me first recall Itô's formula (for Itô processes):

Let $(Z_t)_{t \geq 0}$ be an Itô process, i.e. a stochastic process of the form $$Z_t-Z_0 = \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds \tag{1}$$ for suitable (random) mappings $b$ and $\sigma$. Then it holds for any twice continuously differentiable function $F$ that $$F(Z_t) -F(Z_0) = \int_0^t F'(Z_s) \sigma(s) \, dB_s + \int_0^t \left( b(s) F'(Z_s) + \frac{1}{2} F''(Z_s) \sigma^2(s) \right) \, ds. \tag{2}$$

Itô's formula can be written more compactly in the following way:

$$F(Z_t)-F(Z_0) = \int_0^t F'(Z_s) \, dZ_s + \frac{1}{2} \int_0^t F''(Z_s) \, d \langle Z \rangle_s \tag{3}$$

where $$d\langle Z \rangle_s := \sigma(s)^2 \, ds$$ and (according to $(1)$) $$dZ_s = \sigma(s) \, dB_s + b(s) \, ds.$$

Let's come back to your examples. If you want to find the stochastic differential of a given process using Itô's formula, then you have to identify a) the function $F$ (twice differentiable, deterministic) and b) a suitable Itô process $(Z_t)_t$ as in $(1)$. For

$$X_t = \exp \left( \int_0^t f(s) \, ds \right)$$

we can write

$$X_t = F(Z_t) \quad \text{for} \quad F(x) := e^x \quad Z_t := \int_0^t f(s) \, ds.$$

Clearly, $f$ is twice differentiable and $Z_t$ is of the form $(1)$ (with $\sigma := 0$ and $b:=f$). Applying Itô's formula $(2)$, we find

$$X_t-X_0 = \int_0^t f(s) \underbrace{e^{Z_s}}_{X_s} \, ds$$


$$dX_t = f(s) X_s \, ds.$$

The reasoning for $(Y_t)_{t \geq 0}$ is very similar: We have $Y_t = F(Z_t)$ for $$F(x) := e^x \quad \text{and} \quad Z_t := \int_0^t g(s) \, dB_s.$$ This means that $(1)$ holds with $b=0$ and $\sigma := g$. Using $(2)$ we find

$$Y_t-Y_0 = \int_0^T \underbrace{e^{Z_s}}_{Y_s} g(s) \, dB_s + \frac{1}{2} \int_0^t \underbrace{e^{Z_s}}_{Y_s} g(s)^2 \, ds,$$


$$dY_s = Y_s g(s) \, dB_s + \frac{1}{2} Y_s g(s)^2 \, ds.$$

  • $\begingroup$ But I do mean for the integrand to be stochastic as well. E.g. $g=B_t$ (a case I can handle) or $g=tB_t^2$, etc. The way your have written the Ito formula with $F(Z)$ does help though and should still be applicable. I guess I'm mostly confused about taking derivatives of integrals, e.g. $\partial/\partial t$ of a $dB_t$ integral, and vice versa. $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 12:26
  • $\begingroup$ @jdods Yeah, sure $f$ and $g$ can be random ... you can plug in any of the examples you mentioned in your comment. $\endgroup$
    – saz
    Commented Dec 16, 2017 at 12:30
  • $\begingroup$ Ok, I think I understand. This would mean that the other answer below is incorrect, and that my initial thoughts about taking partial derivatives of the integrals is incorrect, yes? $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 13:44
  • $\begingroup$ I now see that I was making this way more complicated than it actually is. Thanks for taking the time to clear it up! $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 13:50
  • 1
    $\begingroup$ @jdods Hmh, yes. Note that Itô's formula involves only derivatives of the deterministic function $F$. (Taking a derivative with respect to $\omega$ is, in general, a very bad idea. For instance if $g(s,\omega) = B_s(\omega)$, then $g$ is not differentiable with respect to $\omega$ (because Brownian motion has rough paths).) You are welcome; I'm glad I could help you. $\endgroup$
    – saz
    Commented Dec 16, 2017 at 13:50

Using Itô–Doeblin Theorem, you should get:

$$ \begin{align*} d X_t&=X_tf(t,\omega)dt +X_t\left(\int_0^t f_{\omega}(s,\omega)ds\right)dB_t +\frac12\left[X_t\int_0^t f_{\omega\omega}(s,\omega)ds+X_t\left(\int_0^t f_{\omega}(s,\omega)ds\right)^2\right]dt\\ d Y_t&=Y_tg(t,\omega)dt +Y_t g(t,\omega)dB_t +\frac12\left(Y_t g_{\omega}(t,\omega)+Y_tg^2(t,\omega)\right)dt \end{align*} $$

And concerning the other part of your post, note:


  • $\begingroup$ Ah, I see my derivative product rule mistake now. So does this work (with the derivative w.r.t. $\omega$ and tacking on $dB$) generally, or only if the integrand is actually a function of $B$? $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 11:47
  • $\begingroup$ please see the answer above as I think it is the correct solution. I'm still a bit confused on why this method of taking partial derivatives of the integrals doesn't work, but can now see the easier solution of directly applying the Ito formula as above. $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 13:53
  • $\begingroup$ The problem is that $ω$ is the index into the underlying probability space, not the Wiener process. For the purposes of time differentiation, $ω$ is a constant, $t\mapsto B_t(ω)$ the path that belongs to $ω$ as value of the stochastic variable/process $B$. If you set all the $ω$-derivatives to zero, you get the result of the other answer. (Where does $Y_tg(t,ω)\,dt$ come from?) $\endgroup$ Commented Dec 16, 2017 at 16:04
  • $\begingroup$ @LutzL, yes, this answer is wrong, however I'm hoping the author leaves it up since I think it is still instructive and follows my initial erroneous line of reasoning. I will edit it to note that it is wrong of the author doesn't do so soon. $\endgroup$
    – jdods
    Commented Dec 16, 2017 at 22:44

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