I have a question that might be trivial to most of you, but somehow I'm not able to solve it by myself. I have a disagreement with my colleague on the distributional properties of a Geometric Brownian Motion: my point of view is, that if you estimate the parameters $\mu$ and $\sigma$ on log return (and assume that they are normal), the GBM at point $t$ has indeed an expected Value of $X_0\exp{((\mu+\sigma^2/2)t)}$ (properties of log-normal) and not $X_0 \exp{(\mu t)}$ as found in the literature, since there $\mu$ is the drift term of the differential equation of the stock price process itself, not of its log-returns. I tried to confirm my view via several derivations and numerical examples. My colleague though is still not convinced, cause I use $\text{d }{\ln{\!X}}$ in Ito's Lemma for the log returns, but he argues that they are $\ln{(X_t/X_{t-1})}$. Apparently, my knowledge of differential equations is too limited to find the step from the log-returns to their corresponding SDE. In the literature I only find derivations where they start with the sde of $X$. I'm looking foward to your hints.


If you consider $X_1$ a random variable which is normally distributed with mean $\mu$ and variance $\sigma^2$ them $S_1 = \exp(X_1)$ is log-normally distributed with mean $\exp(\mu + \sigma^2/2)$ and variance $(\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)$. This follows from the definitions of the normal distribution and the log-normal distribution and deriving the expectation and variance from these definitions and you can find it here.

If you define a process $X_t$ by $$ dX_t = \mu dt + \sigma dB_t, $$ then the process has expectaion $\mu t$ and variance $\sigma^2 t$ by basic SDE theory. If you consider the process $S_t = \exp(X_t)$ then you get by Ito's lemma $$ \frac{dS_t}{S_t} = \mu dt + \sigma dB_t + \sigma^2/2 dt = (\mu + \sigma^2/2) dt + \sigma dB_t. $$ and because of considerations above we have $E[S_t/S_0] = \exp(\mu t + \sigma^2/2 t)$ and variance $(\exp(\sigma^2 t)-1)\exp(2\mu t+\sigma^2 t)$. Thus a drift of $\mu$ in the process of the log-returns gives you a drift of $\mu + \sigma^2/2$ in the process of geometric returns. Now you can define $\tilde{\mu} = \mu + \sigma^2/2$ and rewrite the processes with $\tilde{\mu}$ then you have drift $\tilde{\mu}$ for the geometric returns and $\tilde{\mu} - \sigma^2/2$ for the log-returns.

EDIT: After having written this, I found nice lecture notes by Karl Sigman about this in the web.

  • $\begingroup$ I know how to apply Ito's Lemma here, my question was more on how to show that dlnX is the corresponding SDE for ln(X+dt/X), thus the change of the log of the ratio of X, and not of the log of X itself. $\endgroup$ – Arne Sep 25 '13 at 7:49
  • $\begingroup$ I'm aware, of course, that $ln(\frac{X_{t+dt}}{X_{t}}) $ equals $ln(X_{t+dt})-ln(X_{t})$, but is that equal to $dln(X)$? $\endgroup$ – Arne Sep 25 '13 at 10:11
  • $\begingroup$ $ln(X_t)$ can be written in integral form as $ln(X_0) + \int_0^t d ln(X_u)$. So if you are interested in $ln(X_u) - ln(X_v)$ for $u>v$ then this is $\int_u^v d ln(X_u)$. $\endgroup$ – Ric Sep 25 '13 at 14:25
  • $\begingroup$ Thank you very much! I knew I couldn't get around the integral notation... $\endgroup$ – Arne Sep 25 '13 at 18:42

To complete the perfect answer of Richard, I would add that pretending that the expected value of the GBM at $t$ is $X_0\exp(\mu t)$ amounts to claim that $E(exp X) = exp(EX)$ which is wrong “because the exponential is not linear.” This is why there is this $\sigma^2/2$ term popping up, it is sometimes known as the “convexity correction”—the exponential being convex you have $E(exp X) \ge exp(EX)$ and you may find a more precise statement in Jensen's inequality.

This could provide you with an argument which is easier to grasp as a mathematical derivation. This topic is adressed in Hull's classical book on derivatives (12.3 The expected return in the 5th. edition).

  • $\begingroup$ Nice addition, "Jensen" is an important key word I was missing. $\endgroup$ – Ric Oct 4 '13 at 13:42

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