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I am having difficulty in recovering some result in smile dynamics of Bergomi https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1520443, the paper gives $(1-3\alpha x +(6\alpha^2 - \frac{5}{2}\beta)x^2)$ and $(x - (2\alpha-\sigma_0\alpha')x^2)$ part, while my own calculation gives $(1-\alpha x +(\alpha^2 - \frac{1}{2}\beta)x^2)$ and $(x - (\alpha-\sigma_0\alpha')x^2)$ respectively, which will give different SSR ratio for this model. So I think something wrong with my calculation.

Quickly sum up, in the paper it states given a model of smile as a function of moneyness, $$\hat{\sigma}(x) = \sigma_0(1+\alpha(\sigma_0)x+\frac{1}{2}\beta(\sigma_0)x^2)$$ one can calculate below greeks of option $Q$, whose BS price is $P^{BS}(\hat{\sigma}(x))$, as: $$\frac{1}{2}S^2\frac{d^2Q}{dS^2}=\frac{1}{2}\frac{SN'(d)}{\sigma_0\sqrt{T}}(1-3\alpha x +(6\alpha^2 - \frac{5}{2}\beta)x^2)$$ $$S\sigma_0\frac{d^2Q}{dSd\sigma_0}=\frac{SN'(d)}{\sigma_0\sqrt{T}}(x - (2\alpha-\sigma_0\alpha')x^2)$$

Here is how I approached it. By checking the context I suppose calculation of $\frac{d^2Q}{dSd\sigma_0}$ it should be using BS greeks at $\sigma=\hat{\sigma}(x)$ and link $\frac{d}{d\sigma_0}=\frac{d\hat{\sigma}}{d\sigma_0}\frac{d}{d\hat{\sigma}}$. Giving a go based on that, I get gamma theta as: $$\frac{1}{2}S^2\frac{d^2P^{BS}(\hat{\sigma})}{dS^2}=\frac{1}{2}\frac{SN'(d)}{\sigma_0\sqrt{T}}\frac{\sigma_0}{\hat{\sigma}(x)}$$ and doing taylor expansion at order 2 in x and order 0 in T, $$\frac{\sigma_0}{\hat{\sigma}(x)}:=f(x)=\frac{1}{1+\alpha(\sigma_0)x+\frac{1}{2}\beta(\sigma_0)x^2}\sim 1-\alpha x + (\alpha^2-\frac{1}{2}\beta)x^2$$

Similar to vanna theta, I get $$S\sigma_0\frac{d^2P^{BS}}{dSd\sigma_0}=S\sigma_0\times vanna^{BS}\times\frac{d\hat{\sigma}}{d\sigma_0}=\frac{SN'(d)}{\sigma_0\sqrt{T}}\sigma_0^2\sqrt{T}\frac{x+\frac{1}{2}\hat{\sigma}^2T}{\hat{\sigma}\sqrt{T}}\frac{1}{\hat{\sigma}}(\frac{\hat{\sigma}}{\sigma_0}+\sigma_0(\alpha'x+\frac{1}{2}\beta'x^2))$$, where $\alpha'=\frac{d\alpha}{\sigma_0}$. And finally still using $f(x)$ above I get $$S\sigma_0\frac{d^2P^{BS}}{dSd\sigma_0}=\frac{SN'(d)}{\sigma_0\sqrt{T}}(x - (\alpha-\sigma_0\alpha')x^2)$$

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