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I was looking at the paper of Raval and Jaquier The Log Moment Formula For Implied Volatility available here : https://arxiv.org/pdf/2101.08145.pdf

On the page 4 they wrote(with $<logS>_T$ and $<S>_t$ quadratic terms ) :

$<logS>_T$ = $\int_{0}^{T}\frac{1}{S_t^2}d<S>_t = -2 log(\frac{S_T}{S_0}) + 2\int_{0}^{T}\frac{1}{S_t}dS_t$

I don't understand well the last step of the derivation as I find:

$-2\frac{S_T - S_0}{S_0} + 2\int_{0}^{T}\frac{1}{S_t}dS_t$

Moreover, the authors define :

$-log(\frac{S_T}{S_0}) = \frac{S_T - S_0}{S_0} + \int_{S_0}^{\infty}(\frac{S_t-K}{K^2})^+dK + \int_{0}^{S_0}(\frac{K-S_t}{K^2})^+dK$

Which I couldn't demonstrate. Could someone help me please.

Thank you

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Starting with $$dS_t = rS_tdt +\sigma_t S_tdW_t,$$

Ito Lemma in two steps gives:

$$d\log S_t = S_t^{-1} dS_t - 2^{-1}S_t^{-2} (dS_t)^2 \; \; \; (*)$$

$$d\log S_t = (r-2^{-1}\sigma^2_t) dt + \sigma_t dW_t \; \; \; (**)$$

From (**) (and starting SDE) we get

$$ d[\log S]_t = (d\log S_t)^2 = \sigma_t^2 dt = S_t^{-2} (dS_t)^2 $$

From (*) we then get:

$$ d\log S_t = S_t^{-1} dS_t - 2^{-1} d[\log S]_t $$

So:

$$ d[\log S]_t = - 2 d\log S_t + 2 S_t^{-1} dS_t $$

(you are missing the factor $2$ in the last term).

The second equation focuses on the $\log$ contract payoff and it is an application of Lemma 3.6 in the paper, resulting in

$$f(S_T)=f(S_0) + f'(S_0) (S_T - S_0) + \int_0^{S_0} f''(K) (K-S_T)^+ d K $$ $$+ \int_{S_0}^{\infty} f''(K) (S_T-K)^+ d K, $$

known as the Carr-Madan formula (for any convex and smooth $f$). See a proof here on SE Quant. We can take $$ f(x) = \log (x), \; \; \; x = S_T, \; \; \; x_0 = S_0. $$

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  • $\begingroup$ Thank you very much but why is there a minus sign in front of the log such :$-log(\frac{S_T}{S_0}) = \frac{S_T - S_0}{S_0} + \int_{S_0}^{\infty}(\frac{S_t-K}{K^2})^+dK + \int_{0}^{S_0}(\frac{K-S_t}{K^2})^+dK$. $\endgroup$
    – lays
    Apr 27 at 14:22
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    $\begingroup$ The literal application to $f=\log $ gives: $\log(\frac{S_T}{S_0}) = \frac{S_T - S_0}{S_0} - \int_{S_0}^{\infty}{K^{-2}}({S_T-K})^+dK - \int_{0}^{S_0}{K^{-2}}({K-S_T})^+dK$. So you have a sign error in your formula. $\endgroup$
    – ir7
    Apr 27 at 15:01

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