# smile dynamics IV appendix 4

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 $$S\sigma_0\frac{d^2P^{BS}}{dSd\sigma_0}=\frac{SN'(d)}{\sigma_0\sqrt{T}}(x - (\alpha-\sigma_0\alpha')x^2)$$