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I'm trying to calculate the hedging quantities of the Heston model. I undestand that the replicating portfolio consist of one option, $V = V(S,v,t)$, $\Delta$ stocks and $\phi$ units of the option to hedge volatility, $U(S,v,t)$. The quantities are found by: \begin{align} \phi = - \frac{\partial V}{\partial v} / \frac{\partial U}{\partial v} = - \nu_V / \nu_U \quad \text{and} \quad \Delta = - \phi \frac{\partial U}{\partial S} - \frac{\partial V}{\partial S}. \end{align} Next, I need to calculate these quantities. As pointed out by Zhu(2010), the dynamics of the volatility in the Heston model is given by two parameters, the mean reversion level, $\theta$, and the initial level of the variance, $v_0$. He therefore suggest to base the calculation of vega on both parameters by defining vega as a gradient of two partial differentials: \begin{align*} \nu & = (\nu_1, \nu_2) = \left( \frac{\partial C}{\partial v}, \frac{\partial C}{\partial \omega} \right) = \left( \frac{\partial C}{\partial v_0} 2 \sqrt{v_0}, \frac{\partial C}{\partial \theta} 2 \sqrt{\theta} \right), \end{align*} where $\omega = \sqrt{\theta}$ and $v = \sqrt{v_0}$.

Zhu(2010) further states that "The cash amount of mean Vega labeled as mean cash Vega is the total differential: $$ \nu_{cash} = 2\frac{\partial C}{\partial V_0}v_0 \Delta v_0 + 2\frac{\partial C}{\partial V_0}\theta\Delta \theta$$"

My questions:

  1. as the we now has that vega is a gradients, how do I calculate $\phi$? I'm implementing this hedging procedure, so I need to return a number - not a gradient?
  2. I don't understand what Zhu means with $\nu_{cash}$? Is this the quantities that I to use for calculating $\phi$? If so, what is $\Delta$ here?

Thank you in advance!

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    $\begingroup$ I do not understand Zhu's definition of vega. Strictly speaking under SV models vega would mean sensitivity to the instantaneous volatility (or variance), and not some vector as Zhu is suggesting. So I would say vega is $\partial C / \partial v_0$ period. Cash vega, or dollar vega, is the dollar amount that would accompany a change of $d v_0$ in the instantaneous variance, which is $(\partial C / \partial v_0) dv_0$. $\endgroup$
    – user34971
    Oct 26, 2020 at 16:58
  • $\begingroup$ In "The Heston Model and Its Extensions, it is stated that: "since $v_0$ and $\theta$ are responsible for the initial and long-term level of the variance, Zhu (2010) recommends basing vega on those two parameters. Both parameters represent variance, so to create measures of sensitivity to volatility Zhu (2010) defines two vegas, one based on $v =\sqrt{v_0}$ and the other based on $\omega =\sqrt{\theta}$". Doesn't he have a point? $\endgroup$
    – Modvinden
    Oct 26, 2020 at 17:56
  • $\begingroup$ The Heston SV model is known to be incomplete, i.e. the market price of volatility risk is not specified. Options could be used as hedging instrument, but problems may arise when the underlying price moves far away from the strike and the option’s vega value decreases. Gatheral argues variance swaps or portfolios containing them can complete the market. To solve Heston's PDE one can make the assumption that the market price of risk is linear in the instantaneous variance, this way you can perfectly hedge the vega via variance swaps. $\endgroup$
    – FunnyBuzer
    Oct 26, 2020 at 18:14
  • $\begingroup$ @FunnyBuzer, thank you for your comment. I don't see how this answers the question I have? $\endgroup$
    – Modvinden
    Oct 26, 2020 at 18:22
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    $\begingroup$ A distinction should be made here between in model hedging and out model hedging. The former assumes the model is perfect and in that case vega is sensitivity to instantaneous variance only. Out of model hedging assumes that the model is not perfect and hence other parameters need to be hedged. In that case basically you can define vega as you like (ad long as it is somewhat logical). But then you also need to hedge mean reversion speed, vol of vol, and correlation then. $\endgroup$
    – user34971
    Oct 26, 2020 at 19:06

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Zhu makes sense to me.

The vega cash in Black-Scholes corresponds to a shift of the vol surface by 1%.

If you bump only $v_0$ in Heston, you bump only the short maturities, and if your structure is also dependent on the long maturities, the vega will be vastly underestimated. So you need to bump the $\theta$ as well. I think it implicitly assumes that the other Heston parameters have little relation with a parallel shift, and that a parallel shift can be approximated by the sum of the two independent shifts.

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