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The formula works for total variance, not "strike specific" variance that you need to construct basket vol surface from components, because single historical correlation (or correlation matrix) just does not provide enough information to uniquely reconstruct expected distribution of basket returns (unless for a trivial case where all components are gaussian, ...


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Using a simple moving average is a trick to take the noise out of the estimated quantity. He calculates a measure of volatility that is the SMA of the estimates of historical volatilities for the past 21 days. This is a crude way to estimate volatility. If you read further you will see that the author uses a GARCH(1,1) model to estimate volatility.


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Yes, there is a unique time homogeneous local vol model. This is proven in http://www.sciencedirect.com/science/article/pii/S0304414912002487. There is a slight generalization required that if the option-implied density is zero somewhere, the corresponding local vol is infinite in that region, giving a "gap diffusion". No, there is no nice formula for the ...


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No. In practice the local volatility model has a finite number of slices, so a single slice works as well. Now the problem is : how to compute the time derivative ? Well without adding any information you know that $$ C(0,K) = (S_0-K)_+ $$ so you could try $$ C_\tau = \frac{C(\tau,K)-C(0,K)}{\tau} $$ but it is a very crude approximation. What you may want ...


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I think I have figured this out. The key to the understanding is to think of the options' vegas as "key-strike vegas" compared to the var swap/replication portfolio's vega, which is analogous to "key rate durations of a bond portfolio" to the total effective duration of the portfolio.


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The variance swap's Vega that is equal to the variance notional refers to the realized variance. The Black-Scholes vega refers to the market implied volatility. Now if you want, you can estimate the realized variance at expiry from the volatility of the options (for instance taking the atm variance arbitrarily), and that's often what people do. But that's ...


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Assuming that we are talking about volatility as the standard deviation of uncorrelated random variables (in this case this would mean no autocorrelation) the variance is additive, which means that we get $\sqrt{.15^2+.2^2}=.25=25\%$. You can illustrate this result by simulation in R: > sd(rnorm(1e7,sd=.15)+rnorm(1e7,sd=.2)) [1] 0.2500001 If you want ...



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