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?


1 Answer 1


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: $v_0 \rightarrow v_0 + \epsilon$.

  • $\begingroup$ 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. $\endgroup$
    – user34971
    Apr 5, 2020 at 16:27
  • $\begingroup$ Agree and yes I suppose you can define that custom vega concept. $\endgroup$
    – Ivan
    Apr 5, 2020 at 19:20

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