Can anyone solve the following SDE please?

dX = a*dt + bX*dW,

where a, b are constant and W is BM.

Any thought is appreciated,

Thanks

7 posts • Page **1** of **1**

Can anyone solve the following SDE please?

dX = a*dt + bX*dW,

where a, b are constant and W is BM.

Any thought is appreciated,

Thanks

dX = a*dt + bX*dW,

where a, b are constant and W is BM.

Any thought is appreciated,

Thanks

- elviszhang
**Posts:**4**Joined:**Wed Jul 02, 2008 9:44 pm

First try pretending W(t) is deterministic and solve the differential equation. Then try and see how to fiddle your answer so that when using Ito's lemma on it you get the original SDE.

If you still can't do it try Oksendal's way from his book which is applying the Ito product rule on d[X(t)f(t)] where f(t)=exp(-b*W(t)+1/2*b^2*t) or something like that and then the answer drops out nicely.

But the first way is what I guess the interviewer wants you to do and then you will see where Oksendal gets his f(t) from.

If you still can't do it try Oksendal's way from his book which is applying the Ito product rule on d[X(t)f(t)] where f(t)=exp(-b*W(t)+1/2*b^2*t) or something like that and then the answer drops out nicely.

But the first way is what I guess the interviewer wants you to do and then you will see where Oksendal gets his f(t) from.

- sruggiero
**Posts:**13**Joined:**Sat Jul 11, 2009 7:36 pm

Thanks sruggiero,

I followed the methods from Oksendal and got the answer. The key thing is the "integrating factor".

I followed the methods from Oksendal and got the answer. The key thing is the "integrating factor".

- elviszhang
**Posts:**4**Joined:**Wed Jul 02, 2008 9:44 pm

Finding integration factors may be cumbersome. Here is a "general" trick.

Suppose, the function X_{t} takes the following form :

X_{t} = f(W_{t} , t )

We can make the total differential of this, using Ito's Formula :

dX_{t} = \partial_{W} f(W_{t} , t ) dW + \left( \partial_{t} + \frac{1}{2} \partial^{2}_{WW} \right) f(W_{t} , t ) dt

Comparing this with the original equation

dX_{t} = rdt + \alpha X_{t} dW

we obtain the following set of equations :

(1) \partial_{W} f(W_{t} , t ) = \alpha f(W_{t} , t )

(2) \left( \partial_{t} + \frac{1}{2} \partial^{2}_{WW} \right) f(W_{t} , t ) = r

From (1) one obtains, that :

(3) f(W_{t} , t ) = C(t) exp \left[ \alpha W \right]

We substitute this into (2) to obtain the following inhomogenous first order, linear ODE :

C' + \frac{1}{2} \alpha^{2} C = r exp \left[ - \alpha W_{t} \right]

with C' = \partial{d}{dt} C

We shall handle the RHS of this ODE as an expression which is time dependant ( yet this dependence is pure stochastic ). Plausibly we can do this, since we want to INTEGRATE the RHS and the Ito's correction must be taken into account when we differentiate with respect to W_{t}. This integration is a standard stuff, one can do it with varying the coefficients.

After doing so we obtain, that

C(t) = exp \left[ - \alpha^{2} \frac{t}{2} \right] \int_{0}^{t} r exp \left[ - \alpha W_{s} \right + \alpha^{2} \frac{t}{2} ] ds + C_{0} exp \left[ \alpha^{2} \frac{t}{2} ]

YOu can substitute this into (3) to obtain the solution.

Note, that during the solution procedure we obtained automatically the integrating factor. The integrating factor was obtained quite automatically.

Suppose, the function X_{t} takes the following form :

X_{t} = f(W_{t} , t )

We can make the total differential of this, using Ito's Formula :

dX_{t} = \partial_{W} f(W_{t} , t ) dW + \left( \partial_{t} + \frac{1}{2} \partial^{2}_{WW} \right) f(W_{t} , t ) dt

Comparing this with the original equation

dX_{t} = rdt + \alpha X_{t} dW

we obtain the following set of equations :

(1) \partial_{W} f(W_{t} , t ) = \alpha f(W_{t} , t )

(2) \left( \partial_{t} + \frac{1}{2} \partial^{2}_{WW} \right) f(W_{t} , t ) = r

From (1) one obtains, that :

(3) f(W_{t} , t ) = C(t) exp \left[ \alpha W \right]

We substitute this into (2) to obtain the following inhomogenous first order, linear ODE :

C' + \frac{1}{2} \alpha^{2} C = r exp \left[ - \alpha W_{t} \right]

with C' = \partial{d}{dt} C

We shall handle the RHS of this ODE as an expression which is time dependant ( yet this dependence is pure stochastic ). Plausibly we can do this, since we want to INTEGRATE the RHS and the Ito's correction must be taken into account when we differentiate with respect to W_{t}. This integration is a standard stuff, one can do it with varying the coefficients.

After doing so we obtain, that

C(t) = exp \left[ - \alpha^{2} \frac{t}{2} \right] \int_{0}^{t} r exp \left[ - \alpha W_{s} \right + \alpha^{2} \frac{t}{2} ] ds + C_{0} exp \left[ \alpha^{2} \frac{t}{2} ]

YOu can substitute this into (3) to obtain the solution.

Note, that during the solution procedure we obtained automatically the integrating factor. The integrating factor was obtained quite automatically.

- nablaQuadrat
**Posts:**18**Joined:**Thu Sep 09, 2010 5:46 pm

PS : Mark, it has been very ugly.

Would it be a big effort for you to export the LaTeX engine from Wilmott ?

Would it be a big effort for you to export the LaTeX engine from Wilmott ?

- nablaQuadrat
**Posts:**18**Joined:**Thu Sep 09, 2010 5:46 pm

working out how to do the latex thing does not appeal...

A couple of solutions are posting pdfs, and developing an image using wilmott.com and then posting it here!

A couple of solutions are posting pdfs, and developing an image using wilmott.com and then posting it here!

- mj
- Site Admin
**Posts:**1380**Joined:**Fri Jul 27, 2007 7:21 am

Linear SDEs can be solved by separation of variables etc. If I remember, a general trick is perhaps to assume a soln X=g1(t)g2(w)

dX=g1_dash*g2*dt+g1*g2_dash*dW+0.5*g1*g2_double_dash*dt=a*dt + b*g1*g2*dW

therefore, b*g1*g2=g1*g2_dash (collect dW terms)

g1_dash*g2+0.5*g1*g2_double_dash=a (collect dt terms)

dX=g1_dash*g2*dt+g1*g2_dash*dW+0.5*g1*g2_double_dash*dt=a*dt + b*g1*g2*dW

therefore, b*g1*g2=g1*g2_dash (collect dW terms)

g1_dash*g2+0.5*g1*g2_double_dash=a (collect dt terms)

- sraks
**Posts:**30**Joined:**Tue Feb 16, 2010 3:31 am

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