# Why does the price of a derivative not depend on the derivative with which you hedge volatility risk?

I'm trying to derive the valuation equation under a general stochastic volatility model. What one can read in the literature is the following reasoning:

One considers a replicating self-financing portfolio $V$ with $\delta$ underlying and $\delta_1$ units of another derivative $V_1$. One writes Ito on one hand, and the self-financing equation on the other hand, and then one identifies the terms in front of the two Brownian motions and in front of $dt$.

The first two identifications give $\delta$ and $\delta_1$, and the last identification gives us a PDE in $V$ and $V_1$. Then what is commonly done is to write it with a left hand side depending on $V$ only, and a right hand side depending on $V_1$ only. So you get $f(V) = f(V_1)$. We could have chosen $V_2$ instead of $V_1$ so one gets $f(V) = f(V_1) = f(V_2)$. Thus $f(W)$ does not depend on the derivative $W$ one chooses, and is called the market price of the volatility risk.

What I cannot understand in this reasoning is why $V$ does not depend on the derivative $V_1$ you choose to hedge the volatility risk in your portfolio with. As far as I see it, one should write $V(V_1)$ instead of $V$. Then one has $f(V(V_1)) = f(V_1)$ and $f(V(V_2)) = f(V_2)$ so one gets no unique market price of the volatility risk.

Does anyone know why the price of a derivative does not depend on the derivative you choose to hedge against the volatility risk?

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A similar situation applies here: any triangle of (nontrivial) derivative securities can be shown to have an equivalent set of hedge ratios from any two of them (assumed observable, liquid etc) to form a price of the third. Basically the hedge ratio for $V_2$ in terms of $V_1$ is perfectly symmetric to the one for $V_1$ in terms of $V_2$.