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I agree with Gordon's deduction if the stock price is distributed that way under risk neutral measure. With sufficiently large K it should be monotone, but for other cases, there could be different cases. I think it can be helpful if you create a list of options with different strikes, time to maturity and spot prices to observe the multivariate relationship ...


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The implied Black-Scholes skew will be downward sloping in the limit on both the left and the right. (I believe @Gordon's derivation claiming upward slope may have a sign error somewhere). Left Side For the left side it is sufficient to note that the lognormal model has no density below zero while the normal model has strictly positive density in that ...


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Since $S_T = S_0 + \sigma W_T$, \begin{align*} C &:= E\left((S_T-K)^+ \right)\\ &= E\left((S_0+\sigma W_T-K)^+ \right)\\ &=\int_{\frac{K-S_0}{\sigma \sqrt{T}}}^{\infty}(S_0+\sigma\sqrt{T} x-K) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=(S_0-K)\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right)+\frac{\sigma\sqrt{T}}{\sqrt{2\pi}}e^{-\frac{(S_0-K)...


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The only problem I see with this approach, which remains completely valid from a theoretical perspective, is the embedded (and probably not accounted for) calibration risk: what if your LV surface does not allow you to correctly reproduce the observed vanilla option prices in the first place? In that case, you'll have lost information in the process and ...


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It is valid to do that, but if your local volatility surface is calibrated to the same OTM options, then your price will converge to the same answer. A local volatility surface is mainly a way of treating path-dependent options consistently with the option volatility surface. Variance swaps are path dependent on the face of it, but as you note the math ...



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