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


by operationsres
Tags: calculus, eusdbs, stochastic
operationsres
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#1
Nov9-12, 08:55 AM
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Why does:

[itex] \int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s)[/itex]

for stochastic process [itex]X(t)[/itex] and Wiener process [itex]B(t)[/itex]?

Also, why is the following true:

[itex] \int_0^t d(e^{-us} X(s)) = e^{-ut}X(t) - X(0)[/itex]
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SammyS
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Nov9-12, 09:36 AM
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Quote Quote by operationsres View Post
Why does:

[itex] \int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s)[/itex]

for stochastic process [itex]X(t)[/itex] and Wiener process [itex]B(t)[/itex]?
Also, why is the following true:

[itex] \int_0^t d(e^{-us} X(s)) = e^{-ut}X(t) - X(0)[/itex]
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))\ .[/itex]


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

So that [itex]\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\ .[/itex]
operationsres
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#3
Nov9-12, 09:43 AM
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Quote Quote by SammyS View Post
So that [itex]\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\ .[/itex]
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[/itex]

[itex]= -u \int_0^t e^{-us}X(s)ds + \int_0^t e^{-us}ds[/itex]

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

SammyS
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Nov9-12, 09:48 AM
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Stochastic calculus: [itex] \int_0^td(e^{-us}X(s))=\sigma\int_0^t e^{-us}dB(s)[/itex]


Quote Quote by operationsres View Post
Working with this, we have that:

[itex]\int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right) = \int_{0}^{t} \left( -ue^{-us}X(s) + e^{-us} \right)ds[/itex]

[itex]= -u \int_0^t e^{-us}X(s)ds + \int_0^t e^{-us}ds[/itex]

So I'm still not sure how I can get to the identity based on what you've provided?
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\ .[/itex]

What is the anti-derivative of [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ ?[/itex]
Mute
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Nov9-12, 09:55 AM
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Quote Quote by operationsres View Post
Why does:

[itex] \int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s)[/itex]

for stochastic process [itex]X(t)[/itex] and Wiener process [itex]B(t)[/itex]?
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.
operationsres
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Nov9-12, 10:05 AM
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Quote Quote by SammyS View Post
What is the anti-derivative of [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ ?[/itex]
[itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ = \int_0^t \frac{d}{ds}\left(e^{-us} X(s)\right)ds[/itex]

[itex]\displaystyle \ \ = d(e^{-ut}X(t))[/itex]

It seems I've gone in a circle (obviously I didn't do what you were asking for).
operationsres
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#7
Nov9-12, 10:06 AM
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Quote Quote by Mute View Post
Presumably because that is the solution to some stochastic differential equation, maybe something like

$$dX_t = -uX_t dt + \sigma dB_t.$$
(Ito interpretation)

A differential stochastic differential equation would yield a different solution.
In fact it is. Nice!
Mute
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#8
Nov9-12, 10:10 AM
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Quote Quote by operationsres View Post
In fact it is. Nice!
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.
operationsres
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Nov9-12, 10:12 AM
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Quote Quote by Mute View Post
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.
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)[/itex]

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.).
Ray Vickson
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Nov9-12, 10:41 AM
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Quote Quote by SammyS View Post
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\ .[/itex]

What is the anti-derivative of [itex]\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ ?[/itex]
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
SammyS
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Nov9-12, 11:00 AM
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Quote Quote by SammyS View Post
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\ .[/itex]

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

is [itex]\displaystyle \ \ e^{-us} X(s)\ .[/itex]

So that

[itex]\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)\ .[/itex]
operationsres
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Nov9-12, 11:10 AM
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nice, I understand, well done Sammy!
Ray Vickson
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#13
Nov9-12, 12:14 PM
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Quote Quote by operationsres View Post
Why does:

[itex] \int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s)[/itex]

for stochastic process [itex]X(t)[/itex] and Wiener process [itex]B(t)[/itex]?

Also, why is the following true:

[itex] \int_0^t d(e^{-us} X(s)) = e^{-ut}X(t) - X(0)[/itex]
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
[tex] f_x(x,t) = e^{-ut},\: f_{xx}(x,t) = 0, \; f_t(x,t) = -u e^{-ut} x,[/tex]
so Ito's Lemma gives
[tex] 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,[/tex]
so
[tex] \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.[/tex]

RGV


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