2

I'm currently picking up some introductory questions on stochastic calculus, and ran into some issues with one of them.

Let $Z_t = t^2B_t - 2\int^t_0sB_sds$, where $B$ is a standard Brownian Motion. Now to apply Ito's Lemma:
If we let $Z_t = f(t,B_t)$, where $f(t,x) = t^2x-2\int^t_0sxds$, then $$f_t(t,x) = 2tx - 2tx = 0,\quad f_x(t,x) = t^2 - 2\int^t_0sds = t^2-t^2 = 0, \quad f_{xx} = 0$$ But obviously I made a mistake somewhere because the answer is supposed to be $dZ_t = 2tB_tdt + t^2dB_t − 2tB_t$, whereas I am getting $0$ everywhere. Would appreciate some help. Thanks

user619755
  • 620
  • Did you mean $dZ_t = 2tB_tdt + t^2dB_t − 2tB_t {\color{red}{dt}}$? – Jose Avilez Jan 20 '24 at 16:02
  • @JoseAvilez according to the solution there is no dt, but there may be a mistake – user619755 Jan 20 '24 at 16:07
  • 4
    Obviously your choice of $f$ is poor since it is 0 identically... – LNT Jan 20 '24 at 18:04
  • 2
    You cannot write $Z_t$ as a function of $t$ and $B_t$ because it depends through $\int_0^tsB_s,ds$ on the entire path of $B$ until $t,.$ This is where your mistake starts. Thank god, that $ds$-integral is differentiable. What is $d$ of it? To $t^2B_t$ you can apply the Ito formula. – Kurt G. Jan 20 '24 at 18:04

1 Answers1

2

Notice that $Y_t = \int_0^t sB_s ds $ is differentiable, so $dY_t = tB_tdt$. We may apply Itô's product rule to $X_t = t^2B_t$ to get $dX_t = 2tB_tdt + t^2 dB_t$. Putting these together, you get $$dZ_t = 2tB_t dt + t^2 dB_t - 2tB_tdt$$

Jose Avilez
  • 12,710