Thanks for visiting my question.
Im am currently working on this paper (https://arxiv.org/abs/2305.02523) and I am stuck at page 21 (Theorem 14 proof).
First these SDE's were defined:
\begin{align*}
dZ_t&=Y_tdt\\
dY_t&=X_tdt\\
dX_t&=A(t,X_t)dt+B(t,X_t)dW_t
\end{align*}
And we define "the value of the contract" by:
$P=de^{-rt} v(t,x,y,z)$
Then the authors use the general Ito formula to derive:
($v_x$ is the derivative with respect to x)
$dv(t,x,y,z)=v_tdt+v_xdx+\frac{1}{2}v_{xx}d[x,x]+v_ydy+v_zdz$
up to this point I understand the proof.
Then they want to calculate $dP$ which is given without further explanation by:
$dP=e^{-rt}(-rv+v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+ v_xB(t,X_t)e^{-rt}dW_t$
I dont unterstand how to get to this formula
My attempt:
starting from
$dv(t,x,y,z)=v_tdt+v_xdx+\frac{1}{2}v_{xx}d[x,x]+v_ydy+v_zdz$
i have plugged in the SDE's:
e.g. $dz=ydt$, $dy=xdt$, $dx=A(t,X_t)dt+B(t,X_t)dW_t$ and $d[x,x]=B(t,X_t)^2dt$
$\Rightarrow dv(t,x,y,z)=(v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+v_xB(t,X_t)dW_t$
Now i used:
$P=de^{-rt} v(t,x,y,z)=(de^{-rt})v(t,x,y,z)+e^{-rt} (dv(t,x,y,z))$ Is this correct?
assuming this is correct and with our results from bevor we get:
$P=e^{-rt}(-rv+v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+ v_xB(t,X_t)e^{-rt}dW_t$
But that is the same formula that the authors get for $dP$
So there must be a mistake
I dont realy understand how to handle $P=de^{-rt} v(t,x,y,z)$ and $dP=d(de^{-rt} v(t,x,y,z))$