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As we know, if an asset S follows geometric Brownian motion, under risk neutral measure, it can be expressed as $\frac{dS}{S}=rdt+\sigma dW$, by applying Ito's lemma, $d(lnS)=(r-0.5*σ^2)dt+σdW(t)$, for me, the mathematical conversion from $\frac{dS}{S}$ to $d(lnS)$ makes sense, but I'm trying to make sense intuitively why the drift changes from $r$ to $(r-0.5*σ^2)$. Here is my understanding (but not 100% sure about it): $dlnS$ is continuously compounded rate of stock price, due to continuously compounded feature, it takes into account volatility (or standard deviation), so its real drift should be subtracted by this volatility component, I'm wondering if my understanding is correct?

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    $\begingroup$ This is a result from working with stochastic Ito integrals. It is a result from working with rough objects that is hard to reason if you only know regular smooth calculus. $\endgroup$
    – oliversm
    Commented Jan 8, 2020 at 21:50

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Because $\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac{1}{2}\sigma^2T} > 1$, you need that correction to ensure that your asset grows on average at rate $\mu$ (or $r$ in the risk-neutral measure).

This is pretty well explained in the chapter on BS model from Hull’s book Options, futures and other derivatives!

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    $\begingroup$ $$\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac12 \sigma^2T} > 1$$ $\endgroup$
    – Canardini
    Commented Jan 9, 2020 at 17:50
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    $\begingroup$ Amended - Thanks @Canardini $\endgroup$
    – siou0107
    Commented Jan 10, 2020 at 8:11
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    $\begingroup$ Note that the result is not necessarily a result of working with ito integrals. It's due to the difference between arithmetic and geometric returns. Google for "variance drain" and there should be some relevant literature that shows up. Here's one but there are probably many others. bogleheads.org/wiki/Variance_drain $\endgroup$
    – mark leeds
    Commented Jan 10, 2020 at 17:50

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