This question applies Ito's lemma to the following function: f(Bt) = Bt^4 i.e. find SDE for d Bt^4 The solution (i'm using the Acted revision notes) differentiates f to get: f ' = 4 Bt^3 and f '' = 12Bt^2 I don't see how this is allowed \ possible because the core reading tells us that Bt is at no point differentiable ! If anyone can help out I would be most grateful. Chris.
Chris, This is technically not how Itos lemma is applied. Ito's lemma stipulates that you define a deterministic function, and then evaluate it at a random point. So here your function is f(x) = x^4 (deterministic) so you have f'(x) = 4x^3 and f''(x) = 12x Now evaluate them at the random point Bt; f'(Bt) = 4*Bt^3 f''(Bt) = 12Bt If the notes have done what you say then they are incorrect.
Devon, Thank you for your reply - I think I understand it a little better. They write f(x) = x^4 first i.e it's deterministic function and they differentiate to get f'(x) =4x^3 and f''(x) = 12x^2 These f'(x) and f''(x) terms are then plugged into Ito's formula with x = Bt to get d f(Bt). So I guess it is being evaluated at a random point Bt, as you say. However my brain is finding it hard to accept. The fact that the solution to the question is d f(Bt) = d Bt ^4 = (4Bt^3) dBt + (6Bt^2) dt seems to be a stochastic differential that defines IN GENERAL the process Bt^4 and not a process at any ONE PARTICULAR POINT. The problem probably lies with me not having a good enough foundation in calculus though. Thanks for your help. Chris.
