Deriving Variance Swap Payoffs with Itô’s Lemma and Carr–Madan
Summary
The discussion explains two identities used in variance swap analysis. Applying Itô’s lemma to the log of an asset price shows that its quadratic variation can be written as twice the integral of the asset’s proportional price change minus twice the change in log price. This resolves the questioner’s missing factor of two in the stochastic integral term.
The answer then uses the Carr–Madan representation to express a smooth convex payoff through a linear term and integrals of calls and puts across strikes. Applying the representation to the logarithm produces the option-payoff decomposition relevant to the log-moment formula. The post gives the derivation’s setup and identifies the applicable result, but does not spell out every integration step. The stochastic differential equation shown assumes a diffusion with a volatility process; the explanation does not address jumps or other model extensions.
Key ideas
- Itô’s lemma links log-price changes to proportional price changes and quadratic variation.
- The variance identity contains a factor of two on the stochastic integral term.
- Carr–Madan decomposes a smooth convex payoff into a linear position and a strip of options.
- Applying the payoff decomposition to a logarithm connects option prices with variance-related quantities.
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Full text
# Variance swaps and the Log-Moment formula
# Variance swaps and the Log-Moment formula
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
## Answer by ir7 (score 2, accepted)
https://quant.stackexchange.com/a/63602
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 $$ f(x) = \log (x), \; \; \; x = S_T, \; \; \; x_0 = S_0. $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.