Really struggling in this question:

Consider a market with two assets $(B,S)$ whose price dynamics satisfy \begin{equation} dB_t = B_t r dt \end{equation} \begin{equation} \quad \quad \quad \quad \, \, \, \, \, \, \, dS_t = S_t ( r dt + \sqrt{v_t} dW_t) \end{equation} \begin{equation} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \, \, \, \, dv_t = (a -b v_t ) dt + c \sqrt {v_t} ( \rho dW_t + \sqrt{ 1- \rho^2} dZ_t), \end{equation} where $r, a, b, c \text{ and } \rho$ are constants, with $a,b>0$ and $-1 \leq \rho \leq 1$, and $W$ and $Z$ are independent Brownian motions.

Let $F: [0,T] \times \mathbb{R}_{+} \times \mathbb{R}_{+} \rightarrow \mathbb{R}_{+} $ satisfy the PDE \begin{equation} \frac{\partial F}{\partial t} + Sr \frac{\partial F}{\partial S} + (a-b v_t) \frac{\partial F}{\partial v} + \frac{1}{2} S^2 v\frac{{\partial}^2 F}{\partial S^2} + c \rho Sv \frac{{\partial}^2 F}{\partial S \partial v} + \frac{1}{2} c^2 v \frac{ {\partial}^2 F}{\partial v^2} = rF, \end{equation} with boundary conditions $F(T,S,v) = \sqrt{S}$.

Introduce a contingent claim with payout $\xi_T = \sqrt{S_T}$.

The problem is to show that in the augmented market, there is a strictly positive Ito process $(Y_t)_{t \geq 0}$ such that $(Y_t ( B_t, S_t, \xi_t))_{t \geq 0}$ is a local martingale, if the time-$t$ price of the contingent claim is given by $\xi_t = F(t, S_t, v_t)$.

What I have done so far (applying Ito's formula and using the PDE): \begin{equation} d \xi_t = r F(t, S_t, v_t) dt + \bigg( \frac{\partial F}{\partial S} (t, S_t, v_t) \sqrt{v_t} + \frac{\partial F}{\partial v} (t, S_t, v_t) c \sqrt{v_t} \rho \bigg) dW_t + \frac{\partial F}{\partial v} (t, S_t, v_t) c \sqrt{v_t} \sqrt{1-\rho^2} dZ_t. \end{equation} I try out $Y$ with $dY_t = m_t dt + n_t dW_t + q_t dZ_t$, for processes $(m_t)$, $(n_t)$ and $(q_t)$.

The first condition that $(Y_t B_t)$ is a local martingale tells us that $m_t=0$.

The second condition that $(Y_t B_t)$ is a local martingale seems to tell us that $n_t= Y_t ( \frac{-r}{1+\sqrt{v_t}})$.

Unfortunately, the expression for $q_t$ is so complicated that I cannot conclude from there that $(Y_t)$ is strictly positive. Any ideas???


2 Answers 2


Show that the discounted expectation price of the new security is the same as the solution of the PDE. Once this is done all three assets have discounted price processes which are martingales so there can be no arbitrage.

  • $\begingroup$ My work so far is shown above. $\endgroup$
    – Richard
    Commented Nov 28, 2014 at 2:14
  • $\begingroup$ Please see my edit above. $\endgroup$
    – Richard
    Commented Nov 28, 2014 at 18:15

Mark Joshi has pretty much solved it. To add to it, you can see that from Feynamn Kac (see remarks in http://en.wikipedia.org/wiki/Feynman–Kac_formula ) it follows that $$ F(t,S,v) = B_t \mathbf{E}\left[ \frac{ \sqrt{ S_T } }{ B_T } \big \vert S_t = S, v_t = v \right], $$ where the expectation is taken with respect to a measure where $W$ and $Z$ are independent Brownian Motions. Note how this measure coincides with the risk neutral measure if the process $M_t = S_tB_t^{-1}$ is a martingale: risk neutral measure is the measure so that the underlying expressed in the savings account units is a martingale.

So because of the equality above, if you show that $M_t$ is a martingale, then the price given as a solution to the pde is indeed arbitrage free. Now, this is easy, since $$ dM_t = \sqrt{v_t} M_t dW_t, $$ which is a martingale in the Heston Model, by Novikov's condition (http://en.wikipedia.org/wiki/Novikov's_condition): $$ \mathbf{E} \left[ e^{ \int_0^T v_r dr} \right] < \infty. $$

  • $\begingroup$ How do you know that the Novikov's condition is satisfied: This is not given as a condition in the model of this question? $\endgroup$
    – Richard
    Commented Dec 2, 2014 at 18:30
  • $\begingroup$ is Heston's model, you can show that it satisfies novikov's condition $\endgroup$ Commented Dec 3, 2014 at 4:15
  • $\begingroup$ How can you show that? Is the argument complicated? $\endgroup$
    – Richard
    Commented Dec 3, 2014 at 10:07
  • $\begingroup$ @Richard: You can check this article, in particular, Appendix A for the Novikov condition. $\endgroup$
    – Gordon
    Commented Nov 4, 2016 at 17:31

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