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My instructor has mostly self contained notes, where our textbook is mostly a reference. She has it written that $$ S_t = S_0e^{(\mu - \frac{\sigma^2}{2})t + \sigma W_t} \iff dS_t = S_t(\mu\, dt + \sigma\, dW_t). $$ I feel that basic differentiation of the exponential e implies that on the right hand side we should have $dS_t = S_t((\mu - \frac{\sigma^2}{2})\,dt + \sigma\, dW_t)$.

I'd appreciate understanding why the $\frac{\sigma^2}{2}$ disappears from the differentiation when this is a basic rule about differentiating the exponential $e$.

user7348
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1 Answers1

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The answer comes from Ito Lemma.

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This differs from the regular calculus that there is an additional term on the right hand side. Stochastic calculus is different to regular calculus in that a Brownian Motion can't be differentiated with respect to t.

The additional term explains your observation.

Furthermore, this has nothing to do with Black-Scholes. Black-Scholes is a hedging strategy that replicates an option. Your formula is a GBM, and is lognormal distributed. It just happens that in the BS world, we assume the stock price exhibit log-normal returns.