Vega as a Change to the Initial Variance in a Stochastic Volatility Model

Summary

The document examines how to interpret variance vega in a stochastic volatility model. It considers changing the variance at the valuation time by a small amount and asks whether that change should be added to the variance path at every later time, or whether the variance dynamics themselves should also be altered. The accepted explanation clarifies that the stochastic differential equations continue to describe the same evolution rules; the shock changes the initial condition for variance.

This distinction matters when calculating sensitivity: the model is evaluated from a different starting variance, while its drift and diffusion functions remain as specified. The discussion is conceptual and does not provide a numerical example, a pricing formula, or a particular calibration procedure. Its answer addresses the setup in the question, so practical vega estimates still depend on the model, the size and convention of the variance shock, and how the option price is recalculated under the changed initial state.

Key ideas

• A variance shock for vega changes the starting variance used to value the claim.
• The drift and diffusion equations continue to govern the variance process after the initial condition changes.
• Changing the initial state does not itself redefine the model’s variance dynamics.
• The explanation is conceptual and does not specify a numerical sensitivity procedure.
• Vega conventions depend on the model and on how the variance perturbation is defined.

Tags

# What is vega, really?


# 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?

## Answer by Ivan (score 3, accepted)

https://quant.stackexchange.com/a/53066

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: $v_0 \rightarrow v_0 + \epsilon$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.