Using the Ito formula for the differential of a (function of) Brownian motion

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I am getting troubled about how to use the Ito formula for a function f=f(x,s): $$ df(B_t,t)=f_t(B_t,t)dt+f_x(B_t,t)dB_t+(1/2)f_{xx}(B_t,t)dt $$

Say for the function $f(x,t)=\cos(xt)$. We have the function $$ f(B_t,t)=\cos(B_tt) $$

By applying the formula we get: $$ df(B_t,t)=f_t(B_t,t)dt+f_x(B_t,t)dB_t+(1/2)f_{xx}(B_t,t)dt =-\sin(tB_t)B_tdt-\\ -\sin(B_tt)tdB_t-(1/2)\cos(B_tt)t^2dt $$

Am I getting this right? Is this the right result? If not, where have I messed up?

Thanks in advance for any replies :)