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


I think you are confused by the definitions and interpretations of $\psi(x,y,z,t)$ and $\phi(x,y,z,t)$.

  • The quantity $\phi(x,y,z,t)$ is a probability density function. Infinitesimally, it represents the probability of transitioning from an initial state $[S_t,r_t,q_t]=[S_0,r_0,q_0]$ at $t=0$ to a state $[S_t,r_t,q_t]=[x,y,z]$ at $t>0$. As such, $\phi(x,y,z,t)$ is the solution of the Fokker-Planck equation (or Kolmogorov forward equation) associated to the SDEs of $S_t, r_t$ and $q_t$ with initial condition $\phi(x,y,z,t=0)=\delta([x,y,z]-[S_0,r_0,q_0])$. Since you seem to be familiar with risk-neutral pricing, if $$\phi(x,y,z) = \frac{d\mathbb{Q}\left([S_t, r_t, q_t] \leq [x,y,z]\right)}{d[x,y,z]}$$ and $\mathbb{Q}$ represents the $t$-forward measure one can write: \begin{align} V(S_{0},r_{0},q_{0},t=0) &:= \mathbb{E}^\mathbb{Q}_0 \left[ B(0,t) V(S_t, r_t, q_t, t) \right] \\ &= \iiint B(0,t) V(S_t,r_t,q_t,t)\phi(S_t,r_t,q_t,t)dS_t dr_t dq_t \end{align}

  • The quantity $\psi(x,y,z,t)$ represents a state price, i.e. the price of a so-called Arrow-Debreu security which pays off $1$ unit of currency at time $t$ if and only if the world ends up in the specific state $[S_t,r_t,q_t]=[x,y,z]$, \begin{align} \psi(x,y,z,t) &= \mathbb{E}^\mathbb{Q}\left[ B(0,t) \delta([S_t,r_t,q_t]-[x,y,z]) \right] \\ &= \iiint B(0,t) \delta([S_t,r_t,q_t]-[x,y,z]) \phi(S_t, r_t, q_t, t) dS_t dr_t dq_t \\ &= B(0,t) \phi(x, y, z) \end{align} This means that you can in turn write $$ V(S_{0},r_{0},q_{0},t=0):=\iiint V(S_t,r_t,q_t,t)\psi(S_t,r_t,q_t,t)dS_t dr_t dq_t $$ which corresponds to pricing your contingent claim as a weighted sum of elementary securities of known prices $\psi(S_t,r_t,q_t,t)$ covering all future possible states of the world, better known as the state price density.

  • $\begingroup$ Suppose only q, S are stochastic, then for me : $V(t=0,S_{0},q_{0})=\mathbb{E}^{Q}[e^{-\int_{0}^{t}r_{s}ds}V(S_{t},q_{t},t)]$ ie $V(S_{0},q_{0},t=0)=\iiint e^{-\int_{0}^{t}r_{s}ds}V(x,y,t)\psi(x,y,z,t)dxdy$. and I undestand $\psi$ as the density probability function of $S_{t},q_{t}$ $\endgroup$ – glork Aug 5 '16 at 16:05
  • $\begingroup$ @glork no in your question you've defined $\psi(x, y, z, t)$ as an Arrow-Debreu price, that is the price of a contingent claim paying 1 at $t$ if the world ends up in a specific state $S_t=x$, $r_t=y$ and $q_t=z$. Being a price, it already contains some sort of discounting. $\endgroup$ – Quantuple Aug 7 '16 at 9:36
  • $\begingroup$ Ok let's say $\psi$ is a price that paysoff 1 or 0, so : $\psi(S_{t},r_{t},q_{t},t)=\iiint e^{-\int_{0}^{t}r_{s}ds}\delta(x-S_{t},y-r_{t},z-q_{t})dxdydz$. Now, how can we have : $V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\psi(x,y,z,t)dxdydz$. It seems that there are 2 definitions for $\psi$ ... $\endgroup$ – glork Aug 8 '16 at 7:57
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    $\begingroup$ Yes, and you should agree $\iiint \psi(x,y,z,t) dx dy dz = B(0,t)$ right? Since it represents something that pays 1 in all states $x$,$y$,$z$ of the world (= risk-free zero coupon bond). And this is verified by using $\psi(x,y,z,t) = B(0,t) \phi(x,y,z,t)$ and using the fact that $\phi(x,y,z,t)$ is a pdf. Indeed, $$\iiint \psi(x,y,z,t) dx dy dz = \iiint B(0,t) \phi(x,y,z,t) dx dy dz = B(0,t) \underbrace{\iiint \phi(x,y,z,t) dx dy dz}_{=1} $$ $\endgroup$ – Quantuple Aug 8 '16 at 8:03
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    $\begingroup$ I think you should investigate the concept of "state prices" or "Arrow-Debreu prices", basically, when you are writing $$V(S_{0},r_{0},q_{0},t=0)=\iiint V(x,y,z,t)\psi(x,y,z,t)dxdydz$$, in words you are expressing value of your contingent claim as a *weighted sum of elementary contingent claims delivering a payoff 1 in specific states of the world*. The weights are of course $V(x,y,z,t)$ since your claim pays $V(x,y,z,t)$ in state $x,y,z$ and not $1$. $\endgroup$ – Quantuple Aug 8 '16 at 8:08

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