# 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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Consider the following analogy: you can hedge a derivative in a deterministic-volatility model using either futures, or spot underlying. The hedge ratio will change, but all the mathematics to effectively eliminate stochastic portfolio PL is the same, and must work out to be equivalent.

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 $V_2$.

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nice point. One way of looking at it, I think, is that you have just two Brownian motions, so in a sense your space is just 2 dimensional. Thus, as long as $V$ and $V_1$ are not "linearly dependent", you're spanning the space, and you're done, and it doesn't really matter what $V_1$ you're choosing.

Now, this is of course a very hand-waving argument, in particular as Brownian motion is so "weird" (think about it: you could take the 1st, 3rd, 5th etc. digit of a Brownian motion, and the 2nd, 4th, 6th digit etc of that same Brownian motion, and you'd have two processes that are, well, complicated, and might well be independent (not sure how whether they'd be Brownian though?)).

In your example, as @Brian B highlighted, it is conceivable that the price hedged with $V_1$ and the price hedged with $V_2$ are different, but then there's an arbitrage opportunity right there. If $V, V_1, V_2$ are driven by only two Brownian motions, you could then construct a portfolio that would have positive instantaneous excess return (try it).

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