2005 April Q5

[Darragh Kelly]

Hi,

For Part (a) of the question, I'm able to derieve all parts of the SDE except for the df/d(T-t) term. When doing this I let T-t = tau so I have f(x,tau) = exp(-tau*x + sigma^3*tau^3/6). When I differenciate wrt to tau I get -x+sigma^2*tau*f(x,tau) = -rt+sigma^2*(T-t)*B(t,T). In examiners report I think they have rt-igma^2*(T-t)*B(t,T) when they differenicate, so my signs are opposite to the solution. The rest of my SDE lines up with report.

Many thanks,

Darragh

 

[Steve Hales]

As you've switched to using tau, your Taylor's expansion will need to have a term that looks like @f/@tau*d(tau). Notice that d(tau) = - dt, which is maybe where the confusion has arisen. It's difficult to say without your full working.
Hope that helps.

 

[Darragh Kelly]

Thanks I see what you mean, all sorted now.
Cheers for your help.
Darragh