Why Integrated Simple Returns Differ from Endpoint Returns

Summary

The note clarifies a common confusion in variance swap derivations: integrating the instantaneous simple return dS/S does not produce the endpoint simple return (S_t−S_0)/S_0. By contrast, integrating the differential of log price gives the total log return, ln(S_t/S_0). The distinction explains why the integrated simple return need not equal exp(total log return)−1.

The key is that dS/S uses the changing price S as its denominator at each instant, while the endpoint simple return uses the initial price S_0. The document gives the integral identities as its explanation, but does not develop the variance swap payoff derivation or discuss expectations, stochastic calculus, or practical pricing. It is a concise conceptual correction, so readers seeking a complete derivation will need additional material.

Key ideas

• The integral of dS/S is not the endpoint simple return because its denominator changes with price.
• The integral of d(ln S) equals the log of the price ratio between the endpoint and the start.
• Exponentiating total log return and subtracting one gives the endpoint simple return, not the integral of dS/S.
• The distinction helps explain return terms used in variance swap derivations.

Tags

# Confused by derivation of variance swap payoff


# Confused by derivation of variance swap payoff












I'm trying to follow https://en.wikipedia.org/wiki/Variance_swap#Pricing_and_valuation

where it seems to me that they're just subtracting a simple return: $$ R_t = \frac{\mathrm{d}S_t}{S_t} = \mu \mathrm{d}t + \sigma \mathrm{d}Z_t $$

from the log return: $$ r_t = \mathrm{d}(\mathrm{log} S_t) = \left(\mu - \frac{\sigma^2}{2} \right) \mathrm{d}t + \sigma \mathrm{d}Z_t $$

to get

$$ R_t - r_t = \frac{\mathrm{d}S_t}{S_t} - \mathrm{d}(\mathrm{log}S_t) = \frac{\sigma^2}{2}\mathrm{d}t $$

I'm obviously missing something obvious, but how can that difference depend on the volatility and/or elapsed time? Isn't there a direct 1-to-1 mapping between simple and log returns ($ R_t = e^{r_t} - 1 $)? Is there some implicit expected value that I'm missing?

## Answer by ir7 (score 2, accepted)

https://quant.stackexchange.com/a/66074

With integral definitions of $r_t$ and $R_t$, we do have:

$$ r_t := \int_0^t d(\ln S_u) = \ln S_t - \ln S_0 \color{green}= \ln \left( \frac{S_t}{S_0}\right),$$

but:

$$ R_t := \int_0^t \frac{dS_u}{S_u} \color{red}{\not=} \frac{S_t - S_0}{S_0} \color{green}= {\rm e}^{r_t} -1. $$

In symbolic differentials language:

$$ \frac{dS_u}{S_u} \color{red}{\not=} \frac{dS_u}{S_0}. $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.