In the formula, the stock return is modelled as a brownian motion that is a drift + a stochastic term, ok I get that. But the drift term is then modelled as r - volatility ^ 2 / 2. I am not sure how they derive this "volatility ^ 2 / 2". Is this derived out of the Ito Lemma??


1 Answer 1


This drift comes from making the discounted stock a martingale in the risk-neutral measure $\mathbb Q$

You start with a stock in $\mathbb P$ having this form: $$ dS_t = \mu S_t dt + \sigma S_t dW_t $$ You also have a discount factor $e^{rt}$.

The idea is to remove the drift of the discounted process in $\mathbb Q$ so you get (after applying Girsanov's theorem) a martingale:

$$ d\hat S_t = \sigma \hat S_t d \tilde W_t $$ where $\hat S_t$ is the discounted stock and $\tilde W_t$ is a $\mathbb Q$-brownian motion.

If you solve this last SDE you get

$$ \hat S_t = \hat S_0\exp(\sigma W_t - \frac{1}{2}\sigma^2t) $$ Multiplying with $e^{rt}$ on both sides you get the un-discounted process and the drift you were asking about.

But the gist of why you get the correction term $\frac{1}{2}\sigma^2t$ is when solving the SDE $$ dX_t = \sigma X_t dW_t $$ you get $X_t = X_0 \exp(\sigma W_t - \frac{1}{2}\sigma^2t)$

  • $\begingroup$ In solving the SDE, I assume it is the Ito Lemma we apply to arrive at the solution right? $\endgroup$
    – Liam
    May 23, 2015 at 10:26
  • $\begingroup$ @Liam yes, exactly. $\endgroup$
    – Slug Pue
    May 23, 2015 at 10:30

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