market completion in stochastic volatility model Hi all,
Consider a stochastic volatility model. As there are two sources of risk and one asset only, this is an imcomplete market. One can complete the market by considering a derivative V1 used to hedge the volatility risk. My question is: Do derivatives prices depends on the derivative V1 one chooses to complete the market? And if no, why?
Many thanks in advance.
 A: The question is incomplete itself! What kind of derivative prices are you interested in ? Second, do assume the presence of a numeraire to compute the prices after completion of the market ? Otherwise, the answer is trivially yes since the unit of account chnages the prices (consider for instance the case you introduce some $V_1$ and $V_2=2\times V_1$. Details of the question need to be fixed...
A: Assuming there are no arbitrage opportunities, the price of a derivative depends on the prices of other derivatives available in the market. If you introduce a derivative without giving it a price, or by giving it a price that is already theoretically determined by your initial setup, then you do not affect the prices of other derivatives.
To take a simpler example, suppose we have two sources of risk, two (biased) coin tosses $\omega_1$ and $\omega_2$ and the only tradable asset is $S=\$1_{H(\omega_1)} 1_{H(\omega_2)}$ i.e. we get 1 dollar if both coin tosses are heads and 0 otherwise. 
Now we can complete the market by introducing two options having payoffs
$$
V_1=h S +\$1_{H(\omega_2)}
$$
$$
V_2=\$1_{H(\omega_1)}
$$
where $h$ is a huge number,
Assume there are no arbitrage opportunities and the price of the option $V_1$ is $\$1$. Then  $S=\$1$ must be very unlikely and so the original asset $S$ is now essentially worthless. But if we don't assume the price of any introduced option to be given then introducing the option has no effect.
