Solving Ornstein-Uhlenbeck process

[maz1987]

In the derivation of the solution to the Ornstein-Uhlenbeck process, we use the factor exp(γt) as the integrating factor, or "guessing" the form of the solution as Ut.exp(-γt).

However I don't see why we cannot proceed with changing the dummy variable from t to s, and integrating both sides of the original equation dXt = -γXtdt + σdWt.

More specifically:

dXs = -γXsds + σdWs

integrate both sides between 0 and t to obtain:

Xt - X0 = -γtXt + σ(Wt - W0)

Xt (1 + γt) = X0 + σWt (since W0 = 0)

Xt = [X0 + σdWt] / (1 + γt)

Can someone please point out what is incorrect about my working.
Thanks

 

[Oxymoron]

In the derivation of the solution to the Ornstein-Uhlenbeck process, we use the factor exp(γt) as the integrating factor, or "guessing" the form of the solution as Ut.exp(-γt).

However I don't see why we cannot proceed with changing the dummy variable from t to s, and integrating both sides of the original equation dXt = -γXtdt + σdWt.

More specifically:

dXs = -γXsds + σdWs

integrate both sides between 0 and t to obtain:

Xt - X0 = -γtXt + σ(Wt - W0)

Xt (1 + γt) = X0 + σWt (since W0 = 0)

Xt = [X0 + σdWt] / (1 + γt)

Can someone please point out what is incorrect about my working.
Thanks

dXs integration would depend on Xs - which is in the other side of the equation

 

[Whippet1]

It is because integrating -γXsds doesn't give -γtXt because Xs is itself a variable (that changes with time) and not a constant.

 

[maz1987]

Thanks for the replies.

I'm still a bit confused here:

ASET solution to April 2011 Q10 integrates the following expression:

d(r_s.exp(as)) = a b exp(as) ds + sigma exp(as) dW_s

But here s is included in all terms of the equation, so why is it ok to integrate each of those separately but not what I wrote above? I'm not doubting your replies - I'm I'm just a bit confused that's all!

Thanks