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I often see Vega in the Heston model specified as: \begin{align*} \nu & = \frac{\partial C}{\partial v} = \frac{\partial C}{\partial v_0} 2 \sqrt{v_0} \end{align*} where $v = \sqrt{v_0}$.

Why are we setting $v = \sqrt{v_0}$?

Where is the square-root and "$2$" coming from?

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    $\begingroup$ Let $V$ denote the variance and $v$ the volatility, i.e. $V=v^2$. The natural argument for stochastic volatility models is typically the variance. However, using the chain rule, we can compute vega in terms of the volatility: $$\frac{\partial C}{\partial v}=\frac{\partial C}{\partial V}\frac{\partial V}{\partial v}=\frac{\partial C}{\partial V}\frac{\partial v^2}{\partial v}=\frac{\partial C}{\partial V}2v=\frac{\partial C}{\partial V}2\sqrt{V}.$$ Writing $V_0$ and $v_0=\sqrt{V_0}$ emphasises that we talk about the spot variance and volatility. $\endgroup$ – Kevin Jan 11 at 22:33
  • $\begingroup$ @Kevin Thank you! I would market your question as the answer if I could:-). $\endgroup$ – Modvinden Jan 12 at 7:01
  • $\begingroup$ I added the comment as an answer:) $\endgroup$ – Kevin Jan 12 at 7:47
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Let $V$ denote the variance and $v$ the volatility, i.e. $V=v^2$. The natural argument for the option price under a stochastic volatility model is typically the variance, i.e. $C_\text{SV}=C_\text{SV}(S_0,V_0,...)$. However, using the chain rule, we can compute vega in terms of the volatility: $$\nu=\frac{\partial C_\text{SV}}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}\frac{\partial V}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}\frac{\partial v^2}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}2v=\frac{\partial C_\text{SV}}{\partial V}2\sqrt{V}.$$ We do this in order to resemble the Black-Scholes vega which is the partial derivative of the Black-Scholes option price, $C_\text{BS}=C_\text{BS}(S_0,\sigma,...)$, with respect to $\sigma$. Of course, when we have $\frac{\partial C_\text{BS}}{\partial \sigma}$, we can easily infer $\frac{\partial C_\text{BS}}{\partial \sigma^2}$ using again the chain rule. This would be the Black-Scholes vega with respect to the variance.

The model-free put-call parity implies that European-style put and call options have the same vega (with respect to volatility or variance).

Writing $V_0$ and $v_0=\sqrt{V_0}$ emphasises that we talk about the spot variance and volatility.

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    $\begingroup$ Nice and clean. Just to make sure: The Heston Vega is computed using the Heston call price function $C(S_0,v_0,\ldots)$ (obviously). $\endgroup$ – Kermittfrog Jan 12 at 8:02
  • $\begingroup$ @Kermittfrog Fair point. I added the point in case it caused any confusion for anyone. Thanks! $\endgroup$ – Kevin Jan 12 at 8:18

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