# SDE for $e^\int$ expressions

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.

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

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

i.e.

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

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

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

i.e.

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

• 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. Commented Dec 16, 2017 at 12:26
• @jdods Yeah, sure $f$ and $g$ can be random ... you can plug in any of the examples you mentioned in your comment.
– saz
Commented Dec 16, 2017 at 12:30
• 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? Commented Dec 16, 2017 at 13:44
• I now see that I was making this way more complicated than it actually is. Thanks for taking the time to clear it up! Commented Dec 16, 2017 at 13:50
• @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.
– 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:

$$\int_0^td(log(X_t))=log(X_t)-log(X_0)$$

• 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 $B$? Commented Dec 16, 2017 at 11:47
• 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. Commented Dec 16, 2017 at 13:53
• 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?) Commented Dec 16, 2017 at 16:04
• @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. Commented Dec 16, 2017 at 22:44