Recentered Expansion for Equivalent Black Volatility

# Change of expansion point for singular perturbation solution in Equivalent Black Volatilities


# Change of expansion point for singular perturbation solution in Equivalent Black Volatilities












In the paper Equivalent Black Volatilities, an peturbative solution is derived for the equivalent Black volatility of a vanilla call option under the dynamics $dF_t = a(t) A(F_t) dW_t$ by Taylor expanding $A$ about $K$. We get that the appropriate integrated volatility is $$\sqrt{\tau^*} = A(K) \sqrt{\tau} [1 + \frac{\nu_1}{2}\cdot (f-K) + \frac{2\nu_2 -\nu_1^2}{12}(f-K)^2+\frac{2\nu_2-\nu_1^2}{24}A^2(K)\tau+\cdots]$$ for $f$ the current value of the forward, $\tau=\int_t^T a^2(s) ds$, and $\nu_i$ the ratio of the $i$th derivative of $A$ to $A$ itself: $A^{(i)}(K)/A(K)$.

The authors then point out that the first two terms of this expansion are $\tau[A(K) + A'(K) (f-K)]$ and state that this suggests expanding $A$ instead about the point $f_{av}:=(f+K)/2$. They then present another equation for $\tau^*$ as

$$A(f_{av}) \sqrt{\tau}[1+\frac{\gamma_2-2\gamma_1^2}{24}(f-K)^2 + \frac{2\gamma_2-\gamma_1^2}{24}A^2(f_{av})\tau+\cdots]$$ where $\gamma_i = A^{(i)}(f_{av})/A(f_{av})$

I'm struggling to understand why those first terms should suggest this new expansion (though I imagine one can somehow see in advance that it will eliminate the linear term in $f-K$) or how the calculation of the new expression is performed. Is there a simple way to get to the second formula from the first, or does one need to repeat the full perturbation analysis?

## Answer by Zach Effman (score 0, accepted)

https://quant.stackexchange.com/a/77293

I hadn't struggled long enough -- all that's happening is we write $K = f_{av} + \frac{K-f}{2}$ and replace $A(K)$ and its derivatives by the Taylor expansion $A(f_{av}) + \frac{K-f}{2}A'(f_{av})+\cdots$ everywhere in the first expression for the vol. This has the nice effect of removing the linear term $\frac{f-K}{2}A'(K)$. $f_{av}$ is chosen for the expansion point specifically because $K - f_{av} = -\frac{f-K}{2}$. The full second expression is just a matter of algebra on the Taylor expansion.