CM2 Acted notes page 40 – q 10.5 (iii)

[bobby sanger]

CM2 Acted notes page 40 – q 10.5 (iii)


Answer states:

Mu = 0 and sigma = 0.

Using equation from part (i), we require

g'(t)Bt + ½*6Bt = 0


Which equation are we using from part (i) to form g'(t)Bt + ½*6Bt = 0 and what are the steps involved to get to this point?

best wishes

bobby

 

[Joe Hook]

Hi Bobby,

So we're using the final equation of part (i), the one that following "For a martingale we require zero drift and hence".

With g(t) a function of time our overall expression here: Bt^3 +g(t)Bt is itself a function of both time and standard Brownian motion. With mu=0 and sigma=1 we can take the first partial differential of Bt^3 + g(t)Bt with respect to t and the second partial derivative with respect to Bt and plug into our equation from part (i).

Hopefully this gets you there.
Let me know if any issues remain unsolved.
Joe