Hi all, for Apr 2007 Q7, we have to solve the SDE: dSt=St(mu*dt+sig*dZt) and for Apr 2010 Q1, we have to solve the SDE: dXt=-lam*Xt*dt + sig*dWt. Both Zt and Wt are standard BM. The approach to the first one was to apply Ito lemma on f=ln(St) whereas for the second one, we have to start with f=exp(lam*t) Xt. The choice of f obviously affects the answer but I'm not sure why we have to use these particular choices of f. If we use f=ln(Xt) for the second question, I don't seems to be able to get the answer. Could someone please explain to me why should we start off the derivation with those particular choices of f? Much thanks!!
(Let Zt be a STANDARD Brownian motion) Geometric Brownian motion Let's call it Jt, so we can see that the letters really aren't important. This process has a drift and a volatility that are BOTH proportional to the process itself: eg dJt = Jt[37dt + 96dZt] We ALWAYS solve this by starting with f(Jt) = log Jt. LEARN THIS starting point Ornstein-Uhlenbeck process Let's call it Kt This process has a drift that is proportional to the process itself but a constant volatility: eg dKt = 564Ktdt + 26dZt We ALWAYS solve this by starting with f(Kt,t) = Kt exp(-564t). LEARN THIS starting point Vasicek model drt = a(mu - rt) dt + bdZt = - a rt dt + bdZt + a.mu.dt Ignore the a.mu.dt term - this will just sit there behaving itself. This is essentially the Ornstein-Uhlenbeck process again so we ALWAYS start with: f(rt,t) = rt exp(at) Test for whether you've got it... Write down the starting points for solving the following SDEs: 1) dFt = 5dZt - Ftdt 2) dQt = 2Qt(dZt - dt) 3) dYt = 12(2Yt - 6) dt + 3dZt Good luck! John
Thanks for the reply John! Just wondering will there be another kind of process that we will get ask to solve in the exam? Like: dFt = 2dt + 5FtdZt or dYt = 2dt + (3-Yt)dZt If so, how can we decide the form of f? Thanks!!
Good question! I've never thought about this one, drift constant, volatility proportional to the process itself. I've never seen this one solved analytically - that's not to say that it can't be done. I just tried a few things and found myself going round in circles! I might get out of this question by saying "beyond CT8". Famous last words?! :-| You could try posting on the ST6 forum, I will also check with the ST6 tutors. If there is a function that leads to a neat solution, they are the most likely to have seen it. For now though, the only 2 that I would be able to "solve" in the CT8 exam under time pressure are the 2 that we have looked at John
"beyond CT8" is good enough for me at the moment. Maybe I will hack this question once again in a few years time when I do the ST6. Thanks John !!
My colleague, Mike Lewry (who teaches ST6) has told me that it's not in ST6 either. We've come up with one possible idea for "solving" dXt = a dt + b Xt dZt Apply Taylor's formula in 2 variables (X and Z) to the function f(Xt, Zt) = Xt exp(-bZt). You get Xt = X0 exp(-bZt) + integral from 0 to t of (a+b^2) Xs exp(-bZs) ds Though I'm not sure if this is any more solved than where we started! Let's stick with "beyond CT8" ;-) John
Really useful. Great to see numbers involved! As a check are the answers: f(Ft,t) = Ft e^t f(Qt) = ln Qt F(Yt,t) = Yt e^-24t
