I have some problems with understanding delta hedging with the Heston model.

So far I've understood that you need both Vega and Delta to hedge with the Heston model. However, I don't know how to practically do it. I have used the Heston model to obtain both the delta and vega.

How do I practically use them to delta-hedge? Or are there other hedging strategies that could be good to use?

My intention is to compare the hedging performance of the Heston model with Black Scholes on S&P 500 options

  • $\begingroup$ Yes but I found a hard time understanding it completely. Do you know any literature or anything that provide the delta strategy for Heston step by step? $\endgroup$
    – HestonHelp
    Commented May 9, 2022 at 19:16
  • 1
    $\begingroup$ Do you understand hedging under the Black Scholes model? $\endgroup$
    – user34971
    Commented May 9, 2022 at 19:20
  • $\begingroup$ To a certain extent yes but maybe not 100% $\endgroup$
    – HestonHelp
    Commented May 9, 2022 at 19:28
  • $\begingroup$ Even though delta can be computed with a different model (or set of assumptions), it will still just be just a single number which tells you how much of the underlying (or ETF for SPX) one needs to buy / sell. Since you have computed the Greeks already, I am not sure what you think will be different? $\endgroup$
    – AKdemy
    Commented May 10, 2022 at 14:24

1 Answer 1


It depends on what you are hedging. One thing to consider is the relationship between the naked position and the hedge, i.e., if the naked position is a European option, then you could argue that the hedge should not be computed using a stochastic volatility model. If the naked position is an exotic option, and the hedge is European, then it may be of interest in some cases to use the Heston model for hedge ratios.

Let's ignore such practicalities, and motor ahead. I will show you how to construct a minimum variance delta hedge for the Heston model.

Let's write out the SDE for the asset $S$ and variance $v$:

$ \frac {d S_t }{S_t} = r dt + \sqrt{v_t} dW_{t;S} $

$ d v_t = \kappa ( v_\infty - v_t ) dt + \eta \sqrt{v_t} dW_{t;v} $

with the usual linear correlation structure

$ dW_{t;S} dW_{t;v} = \rho dt $

The asset-only hedged portfolio is long the option $V$ and short $\Delta$ amount of the asset:

$ \pi = V - \Delta S $

Over the next instance in time, the change in the portfolio value owing to changes in $S$ and $v$ are given by

$ d \pi = \left[ \frac{\partial V}{\partial S} -\Delta \right] dS + \frac{\partial V}{\partial v} d v $

The instantaneous variance of the portfolio is given by $ \mathbf{V} [ \pi ] dt = d \pi d \pi $, therefore

$ \mathbf{V} = \left[ \frac{\partial V}{\partial S} -\Delta \right]^2 S^2 v + 2 \left[ \frac{\partial V}{\partial S} -\Delta \right] \frac{\partial V}{\partial v} S n v \rho + \left[ \frac{\partial V}{\partial v} \right]^2 \eta^2 v $

The value of $\Delta$ that minimises the PnL variance of this portfolio is given by solving this equation:

$ \frac{ \partial \mathbf{V} }{\partial \Delta} = 0 $

Do so and you will see that the minimum variance delta hedge is given by

$ \Delta = \frac{\partial V}{\partial S} + \frac{\partial V}{\partial v} \frac{ \eta \rho}{S} $

The first term is just the vanilla model delta, i.e., the Black-Scholes delta, the second term is a product of the vanilla model variance vega and a skew term, where by variance vega I mean the change in the Black-Scholes price with respect to a change in the implied variance.

Further-more, note how the terms $r, \kappa, v_\infty, v$ do not appear in the minimum variance delta. That is to be expected, as we are minimising the variance, which will only be impacted by properties that appear in the quadratic variation, i.e., the diffusion coefficients of the SDEs, not the drifts.

  • 1
    $\begingroup$ Good answer but I think this might confuse the OP even more. Now there is the Heston delta hedge, the minimum-variance Heston delta hedge, the Heston delta hedge including the Vega hedge option, the model free Heston delta hedge (just read it off the slope, which can be done for homogenous models) which is the smile adjusted BS delta hedge, and the usual BS delta hedge without smile adjustment. $\endgroup$
    – user34971
    Commented May 11, 2022 at 8:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.