# Should price impact be the same for positive/negative implied volatility shocks?

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?

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.
• 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. – Quantuple Feb 27 '17 at 12:57