Can anyone explain to me the logic behind the choice of (Ito) function to apply to Ito processes to solve SDE's, or is it just a case of learning the standard ones? For example, in April 2014 ASET Q5 part ii), why do we let f(t) = exp{at}r(t)?
Hi Pete, Easy answer... Yes, just learn them. If you see geo BM, eg dFt = 27Ft dt +4 Ft dZt, always start with d(ln Ft). If you see OU process, eg dQt = 27Qt dt+ 4 dZt, always start with d(Qt e^-27t). If you see OU process with an extra deterministic bit floating about, eg dQt = 5 dt + 27Qt dt+ 4 dZt, still always start with d(Qt e^-27t). THE END, go and pass your exam. However, more interesting answer, in terms of the logic... Divide dFt = 27Ft dt +4 Ft dZt by Ft... dFt / Ft = 27dt +4 dZt Oh look, dFt / Ft, differential of something over itself, that's differentiating log of a function, isn't it? Nearly, it doesn't quite work out like that when Ft is a stochastic quantity but it sure gives us a good idea of what the answer might be close to. I know, let's work out d(ln Ft) and see what happens. The logic on dQt = 27Qt dt+ 4 dZt would be similar. Oh look, I can use an integrating factor here, e^-27t, which is, of course, product rule backwards. I know, let's work out d(Qt e^-27t) and see what happens. Good luck! John
