# Calculating vega in Heston?

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?

• 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. Jan 11, 2021 at 22:33
• @Kevin Thank you! I would market your question as the answer if I could:-). Jan 12, 2021 at 7:01
• I added the comment as an answer:) Jan 12, 2021 at 7:47

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.
Writing $$V_0$$ and $$v_0=\sqrt{V_0}$$ emphasises that we talk about the spot variance and volatility.
• 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). Jan 12, 2021 at 8:02