I just have a question for the beginning of a proof:
Suppose $\frac{dS_{t}}{S_{t}}=(r_{t}-q_{t})dt+\sigma(t,S_{t})dW_{t}$ with $r,q,S$ stochastic.
In the book I read, it is written:
We define the Arrow-Debreu price $\psi(x',y',z',t)$ as the present value of a derivative that pays off $\delta([S_{t},r_{t},q_{t}]-[x',y',z'])$ at time $t$. This is related to the $t$-forward measure probability density of $(x,y,z)$, $\phi(x,y,z,t)$ by: $$\psi(x,y,z,t)=B(0,t)\ \phi(x,y,z,t)$$ as can be seen from the defining equation for $\psi$ and $\phi$: $V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\ \psi(x,y,z,t)\ dx\ dy\ dz$ and $V(S_{0},r_{0},q_{0},t=0)=B(0,t)\ \iiint V(x,y,z,t)\ \phi(x,y,z,t)\ dx\ dy\ dz$
With these 2 last equations I understand why: $\psi(x,y,z,t)=B(0,t)\ \phi(x,y,z,t)$. But I don't understand why $V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\ \psi(x,y,z,t)\ dx\ dy\ dz$.
Because for me we have $V(S_{0},r_{0},q_{0},t=0)=\mathbb{E}^{Q}[e^{-\int_{0}^{t}r_{s}ds}\ V(S_{t},r_{t},q_{t},t)]$ so where is the discount term $e^{-\int_{0}^{t}r_{s}ds}$ gone? Thanks