# What is vega, really?

Assume for now we are working in a stohastic volatility (SV) setting, $$dS_r = \sqrt{v_r} S_r dW$$ and $$dv_r = a(v_r,r)dr + b(v_r,r) dZ$$ with $$dWdZ = \rho dr$$

Let $$C(S_t,v_t,t)$$ denote the SV price of a claim today. Let's define (variance) vega as the change in the option value if time $$t$$ variance is shocked/displaced by some amount $$\varepsilon$$: $$v_t \rightarrow v_t' = v_t + \varepsilon$$ Now let's look at what happens to the instantaneous variance for all $$u>t$$ after this shock: \begin{align} v_u' &= v_t + \varepsilon + \int_t^u d(v_r + \varepsilon) \\ &= v_t + \varepsilon + \int_t^u dv_r \\ &= v_u + \varepsilon \end{align}

My question is, isn't then $$C(S_t,v_t + \varepsilon,t) = E_t [ F(S_T)]$$ where now $$dS_r = \sqrt{v_r + \varepsilon}\, S_r dW$$ and $$dv_r = a(v_r,r)dr + b(v_r,r) dZ$$ or is \begin{align} d(v_r + \varepsilon) &= a(v_r + \varepsilon,r)dr + b(v_r + \varepsilon,r) dZ \\ &\neq dv_r \end{align} an the argument above is incorrect?

Both equations for $$S, v$$ should remain the same as they govern the evolution of these quantities over time regardless of initial conditions. It is the initial condition (unstated here) that must change: $$v_0 \rightarrow v_0 + \epsilon$$.
• Thanks! Yes, as you say the equations should stay the same, only the initial condition changes. Hence writing $dS_u = \sqrt{v_u + \varepsilon} S_u dW$ is incorrect. Agree? That said, I am free to define another type of Vega, let's call it parallel path vega, where now $dS_u = \sqrt{v_u + \varepsilon} S_u dW$, basically this is a vega where the entire variance path is shifted up or down. The down-side of this is that variance can turn negative, but taking the shift size small enough it should be OK. The upside, which I can prove, is that then parallel path vega can be computed model-free. – ilovevolatility Apr 5 at 16:27