Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)$

1. Nov 9, 2012

operationsres

Why does:

$\int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s)$

for stochastic process $X(t)$ and Wiener process $B(t)$?

Also, why is the following true:

$\int_0^t d(e^{-us} X(s)) = e^{-ut}X(t) - X(0)$

2. Nov 9, 2012

SammyS

Staff Emeritus
Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i I'll answer the second part. I assume the limits of integration are for the variable, s, and not for the differential quantity, [itex]\displaystyle d(e^{-us} X(s))\ .$

We can write that differential as $\displaystyle \ \ d(e^{-us} X(s)) = \left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds\ .$

So that $\displaystyle \ \ \int_{s=0}^{s=t}d(e^{-us} X(s)) = \int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds\ .$

3. Nov 9, 2012

operationsres

Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i Working with this, we have that: [itex]\int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds = \int_{0}^{t} \left( -ue^{-us}X(s) + e^{-us} \right)ds$

$= -u \int_0^t e^{-us}X(s)ds + \int_0^t e^{-us}ds$

So I'm still not sure how I can get to the identity based on what you've provided?

Last edited: Nov 9, 2012
4. Nov 9, 2012

SammyS

Staff Emeritus
Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i Use the Fundamental Theorem of Calculus to evaluate the definite integral [itex]\displaystyle \ \ \int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds\ .$

What is the anti-derivative of $\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ ?$

5. Nov 9, 2012

Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i Presumably because that is the solution to some stochastic differential equation, maybe something like dX_t = -uX_t dt + \sigma dB_t. (Ito interpretation) (I believe that is not actually the correct stochastic differential equation for that integral; it was just a first guess. Look up Ornstein-Uhlenbeck process for more information and the actually stochastic differential equation. ) The important point is that a different stochastic differential equation would yield a different integral-form solution. Last edited: Nov 9, 2012 6. Nov 9, 2012 operationsres Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ = \int_0^t \frac{d}{ds}\left(e^{-us} X(s)\right)ds$

$\displaystyle \ \ = d(e^{-ut}X(t))$

It seems I've gone in a circle (obviously I didn't do what you were asking for).

7. Nov 9, 2012

Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i In fact it is. Nice! 8. Nov 9, 2012 Mute Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i Actually, I think the SDE I guessed in that post was not quite correct, as it may ignore a drift term. The full SDE is likely the one corresponding to the Ornstein-Uhlenbeck process, as I just mentioned in an edit to my previous post. 9. Nov 9, 2012 operationsres Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i Dear Mute, Since you know about the OU process, can I ask if this is an accurate representation of a mean-reverting OU? [itex] dX(t) = (m-X(t))dt + \sigma X(t) dB(t)$

where m is the mean-reversion term, B(t) is standard Brownian Motion.

I ask because (i) This is what's in my tutorial question list, (ii) Wikipedia and all other external sources I've seen state this process without X(t) in the latter part of the expression (e.g. Wikipedia et al.).

10. Nov 9, 2012

Re: Stochastic calculus: $\int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i You need to avoid things like (d/dt)[exp(-ut) X(t)], because, typically, X(t) is a seriously non-differentiable function, and the usual calculus rules do not apply. Instead, we need to deal with stochastic differentials (essentially, the opposite of stochastic integrals), and we need to use Ito's Lemma or similar results to evaluate things. For example, if B is a standard Brownian motion, we have d(B(t)^2) = 2 B(t) dB(t) + dt, and d(exp(B)) = exp(B) dB + (1/2)*exp(B) dt, etc. RGV Last edited: Nov 9, 2012 11. Nov 9, 2012 SammyS Staff Emeritus Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i The anti-derivative of [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\$

is $\displaystyle \ \ e^{-us} X(s)\ .$

So that

$\displaystyle \ \ \int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds=\left(e^{-ut} X(t)\right)-\left(e^{-u0} X(0)\right)\ .$

12. Nov 9, 2012

operationsres

Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i

nice, I understand, well done Sammy!

13. Nov 9, 2012

Ray Vickson

Re: Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/i

I don't think the expression in this thread's title is correct; however, the second one you wrote is OK (and is, basically, equivalent to the notion of a stochastic differential as an anti-stochastic integral---almost the opposite of how it is done in ordinary calculus). If we have dX = σ dB, and we let f(x,t) = exp(-ut)*x, then
$$f_x(x,t) = e^{-ut},\: f_{xx}(x,t) = 0, \; f_t(x,t) = -u e^{-ut} x,$$
so Ito's Lemma gives
$$d[f(X(t),t)] = \left( 0 f_x + \frac{1}{2} \sigma^2 f_{xx} + f_t \right) dt + \sigma f_x dB = -u e^{-ut} X(t) dt + \sigma e^{-ut} dB,$$
so
$$\int_0^t d\left(e^{-us} X(s) \right) = \sigma \int_0^t e^{-us} dB(s) - \int_0^t u e^{-us} X(s) ds.$$

RGV