2
$\begingroup$

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??

$\endgroup$
4
$\begingroup$

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 $X_t = X_0 \exp(\sigma W_t - \frac{1}{2}\sigma^2t)$

$\endgroup$
  • $\begingroup$ In solving the SDE, I assume it is the Ito Lemma we apply to arrive at the solution right? $\endgroup$ – Liam May 23 '15 at 10:26
  • $\begingroup$ @Liam yes, exactly. $\endgroup$ – Slug Pue May 23 '15 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.