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You want to work directly with $\overline{X}$, and not some other r.v. with the same distribution, since equivalence in distribution doesn't imply that correlation remains the same. For ease of notation, I'll assume that $\mu = 0$ and $\sigma = 1$. I claim that $$ \text{cov}\left(\overline{X},X \right) = \frac{1}{t} \int_0^t s \ ds. $$ Note that this is ...


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Your analysis is correct. From the risk neutral process $$dS_t = rS_tdt + \sigma S_tdW_t$$ we get $$\mathbb{E}(S_\tau|\mathbb{F}_t) = S_te^{r(\tau-t)}$$ and $$\mathbb{E}(\int_{t}^{T}S_\tau d\tau|\mathbb{F}_t) = \frac{S_t}{r}[e^{r(T-t)}-1]$$ Hence, as $S \rightarrow \infty$ $$C(S,A,t) \sim \frac{S_te^{-r(T-t)}}{r(T-T_0)}[e^{r(T-t)}-1] = ...


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There is no difference in information, though the fitting algorithm may increase in complexity. First note that in practice you never have an entire curve or surface of prices $C(K,T)$ of any kind of option. You only have a finite number of observations and even those typically have a bid and an offer. I would therefore argue that the correct picture of ...


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The easiest way is to use single-expiry volatility that you would get from your volatility surface. It is usually good enough for government work (e.g. to get a sense if you are getting raped by a dealer or to understand your vega risk). A better way is to use local volatility model and the whole volatility surface up to the date of expiry. There is also a ...



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