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A Geometric Brownian motion satisfying the SDE $dS_t = rS_t dt+\sigma S_t dW_t$ has the analytic solution $$S_t = S_0\exp\left\{\left(r-\frac{\sigma^2}{2}\right)t\right\}\exp\{\sigma W_t\}$$

Recently in an interview I was asked the following (I am paraphrasing):

The magnitude of uncertainty of the movement of $S_t$ is represented by $\sigma$ and is clearly captured in the term $\exp\{\sigma W_t\}$. But intuitively, why does $\sigma$ appear again in the term $r-\frac{\sigma^2}{2}$? That is, why are we deducting $\frac{\sigma^2}{2}$ from our drift $r$? What is the interpretation?

Does anybody know how to interpret it?

(I originally asked my question on MSE http://math.stackexchange.com/questions/722368/geometric-brownian-motion-volatility-interpretation, but it was suggested I seek proper help here)

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3 Answers 3

up vote 3 down vote accepted

I will try to answer this a bit differently.

The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$ Then with some abuse of notation: $$ E[d(S_t/S_0)] = \mu t + \frac{\sigma^2}{2} t, $$ and we get the converxity term. So it is again the crucial story of Ito calculus: second order terms don't vanish (as in usual calculus) - they just stay. If you want to see this from the SDE then you have to use the Stratonivich forumulation (see e.g. here).

The intuitive answer: Just look at $E[\exp(\sigma W_t)]$. You can think of this as $$ \exp(\sigma W_t) \approx \exp(Z \sqrt{t} \sigma) $$ where $Z$ takes the values $\pm 1$ with probability $1/2$ (note that the noise gets $\sqrt t $ whereas a drift term would get a $t$). Then the expectation is $$ E[\exp(\sigma W_t)] \approx \frac12 (\exp(\sqrt{t} \sigma)+\exp(-\sqrt{t} \sigma)), $$ using the Taylor series expansion this is $$ \frac12 ((1+ \sqrt{t} \sigma + \frac{t \sigma^2}{2} + \text{terms of higher order}) + (1- \sqrt{t} \sigma + \frac{t \sigma^2}{2} + \text{terms of higher order} )), $$ and you see that the terms of order $\sqrt{t}$ cancel out. You get something like $$ E[\exp(\sigma W_t)] \approx 1 + \frac{t \sigma^2}{2} + \text{terms of higher order} \approx \exp(\frac{t \sigma^2}{2}). $$

As a last comment if you have $\exp(\sigma W_t)$ and $W_t$ is symmetric then the positive outcomes draw the expectation up, $\exp(\sqrt{t} \sigma)$ is further away from $\exp(0)=1$ than $\exp(-\sqrt{t} \sigma)$ e.g. for $\sqrt{t} = 0.1$ and $\sigma=0.2$ you have $ 1.020201$ versus $0.9801987$ - thus if it goes up it goes further up from $1$ then if it goes down.


The very short answer: because $W_t$ is symmetric around $0$ but $\exp(x)$ is not symmetric around $1$.

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+1 nice :) very detlaid and nicely technical –  Probilitator Mar 24 at 15:04

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor expanding the $e^{\sigma W}$ around $W=0$ up to the quadratic term. The convexity thus produces a drift increasing with respect to $\sigma$. We know the drift should be $e^{rt}$. Therefore the factor in front should scale down the drift from the convexity measured by $\sigma$.

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you could mention Jensen Inequality here as the theoretical explanation for why $E[W_t] \leq E[e^{W_t}]$ –  Probilitator Mar 24 at 8:28
You meab $\exp(E[W_t]) \le E[\exp(W_t)]$ - right? –  Richard Mar 24 at 13:31
sorry - you are correct of cours –  Probilitator Mar 24 at 15:04
@Probilitator: I have added explicitly Jensen's inequality as you have suggested. –  Hans Mar 24 at 16:34

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$

is not yet a martingale for it is not dirftless.

From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol $\sigma$ equals $e^{(\mu+0.5\sigma^2)t}$(see the very detailed Wikipedia article) Thus by setting $\mu= (r-0.5\sigma^2)t$ we arrive at $E[S_t]=e^{rt}$

Now in most cases $r$ will denote the market risk free rate. Thus on average our stock will earn only that rate.

You can interpret the $-0.5\sigma^2$ to be the volatility-dependent drift adjustment which insures the risk neutrality of the process. Thus if judging by average returns the ivnestor won't care whether he will be invested in the risk-free portfolio or into the market portfolio.

To pick up the comment on MSE - the discounted expected payoff will then be $S_0$ and the discounted process $e^{-rt}S_t$ will be a martingale. This further supports that the thus created market setting is fair.

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