I'am wondering if there is a standard definition to the Vega of an exotic product when the underlying model is not Black-Scholes.

Let me give some examples :

  • What is the Vega if the price is obtained by a local volatility model. Is it obtained by a parallel shift of local vol ?

  • What is the Vega when the model is a stochastic vol ? Is it a sensitivity to spot vol ?

  • Or the Vega is obtained by parallel bumping implied volatility surface ?

In the first place I thought Vega was obtained with the following steps :

  • Price the exotic with the relevant (well calibrated) model.

  • Find the constant implied volatility of Black-Scholes model that gives the same price (let us call it the $\text{ExoIV}$)

  • Find the sensitivity of BS price to $\text{ExoIV}$.

But this might be quite complicated either numerically or merely because no close formula exists for the exotic derivative in Black-Scholes framework.

  • 3
    $\begingroup$ It depends what you define as the Vega. If you are looking for the "Black-Scholes" Vega of an exotic product priced under another specific model then it would indeed be the sensitivity of the latter price to a full parallel shift of the IV surface used to calibrate your exotic model in the first place. $\endgroup$
    – Quantuple
    Commented Jul 30, 2018 at 8:39
  • $\begingroup$ If I understand you well, you believe the steps I mentioned about ExoIV are equivalent to compute Vega by bumping IV surface ? $\endgroup$
    – Jiem
    Commented Jul 31, 2018 at 18:18

2 Answers 2


In the interest rate world, the vega of an exotic is usually defined by bumping all the relevant volatilities by a multiplicative factor , typically 1.01, 1.05 or 1.10. This could be done in at least two ways (1) bump all the input volatilities or (2) bump directly all the volatilities in the model. To illustrate the difference: some models take as inputs a set of standard swaptions. These are then used by some interpolation scheme to calculate all other required volatilities. In (1), we bump the inputs. In (2), we bump the outputs of the interpolation. There shouldn't be much difference.

  • $\begingroup$ So from your answer and @Quantuple one, I conclude no standard exists. May be just market practices. Thank you both $\endgroup$
    – Jiem
    Commented Jul 31, 2018 at 18:20

In this book : Zhu, J. (2010). Applications of Fourier Transform to Smile Modeling: Theory and Implementation. Zhu defines two vegas for the Heston model :

$ \nu_1 = \frac{ \partial C}{ \partial v} = \frac{ \partial C}{ \partial v_0} 2 \sqrt{v_0}$ and $ \nu_2 = \frac{ \partial C}{ \partial w} = \frac{ \partial C}{ \partial \theta } 2 \sqrt{\theta} $

With $v = \sqrt{v_0}$ and $w = \sqrt{\theta}$ , $v_0$ and $\theta $ being respectively the mean reversion level and the initial level of variance in the Heston model.

You can see more about the greeks in the Heston model in this book : Fabrice D. Rouah The Heston Model and Its Extensions in Matlab and C#


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