I am using a vendor system to stress a portfolio which contains (among others) derivatives with implied volatility exposure.

The issue is that when using a 1000 bps implied volatility stress upwards and downwards the result is really close in both cases (with an opposite sign obviously)

Is this expected?


1 Answer 1


It may help you to notice that, for a bump in implied volatility $\delta \sigma$, the impact on the price of the derivative $V$ is given by: $$ \delta V = \underbrace{\frac{\partial V}{\partial \sigma}}_{\text{Vega}} \delta \sigma + \frac{1}{2} \underbrace{\frac{\partial^2 V}{\partial \sigma^2}}_{\text{Volga, Vomma}} (\delta \sigma)^2 + o((\delta \sigma)^2) $$

Hence, the positive ($\delta V^P$) and negative ($\delta V^N$) price impacts for respective bumps $\delta \sigma^P= \vert\delta\sigma\vert$ and $\delta\sigma^N = - \vert\delta\sigma\vert$:

$$ \delta V^P = \frac{\partial V}{\partial \sigma} \mid \delta \sigma \mid + \frac{1}{2} \frac{\partial^2 V}{\partial \sigma^2} (\delta \sigma)^2 $$ $$\delta V^N = -\frac{\partial V}{\partial \sigma} \mid \delta \sigma \mid + \frac{1}{2} \frac{\partial^2 V}{\partial \sigma^2} (\delta \sigma)^2 $$ hence $$ \delta V^P = -\delta V^N + \frac{\partial^2 V}{\partial \sigma^2} (\delta \sigma)^2 + o((\delta \sigma)^3) $$ and when no Volga (also called Vomma): $$ \delta V^P = - \delta V^N $$

For illustration purpose here is the Volga curve of a vanilla option of time to maturity $\tau$ as a function of forward moneyness $m=K/F(0,\tau)$. Observe how an ATM option has no Volga and how this changes as you move away from the money.

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  • $\begingroup$ Hi, not the original poster but one question. Why is the curve for volga not symmetric about ATM? $\endgroup$
    – nimbus3000
    Commented Feb 27, 2017 at 11:03
  • 1
    $\begingroup$ Short answer: Because $\sigma$ is the return volatility not the price volatility. Long(er) answer: The price of a European option writes $V(T,\theta) = \int_{0}^{+\infty} h(S,\theta) q(T,S) dS $ where $T$ is the maturity, $\theta$ some contract parameters (e.g. strike for call/put), and $q(T,S) = d\Bbb{Q}(S_T \leq s)/ds$ the distribution of $S_T$ under the risk-neutral measure. Under BS, $q(T,S)$ is fully characterised by its first 2 moments, the mean $F(0,T)$ (forward price) and the variance $F^2(0,T)(e^{\sigma^2 T}-1)$. Thus you see that the dependence on $\sigma$ is non-symmetric. $\endgroup$
    – Quantuple
    Commented Feb 27, 2017 at 12:57

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