I want to correctly simulate a $\mathcal{Q}$ - martingale $S$, which is a geometric Brownian motion and an exponential of a process $X$, \begin{equation} X_t = X_0 + \mu t + \sigma B_t = X_{t-\Delta t} + \mu \Delta t + \sigma B_{\Delta t}, \end{equation} where $X_0 = 0$ and $B$ is a Brownian motion under $\mathcal{Q}$, such that \begin{equation} S_t = S_0 \exp(X_t) = S_0 \exp(\mu t + \sigma B_t) = S_{t-\Delta t} \exp(\mu \Delta t + \sigma B_{\Delta t}), \end{equation}
with $\mu = -\sigma^2/2$ from the martingale condition (no interest rates, or $r=0$).

But when I run many (eg. N=1000) simulations of $(X_t)_{t=\Delta t}^T$ over a one-year time horizon ($T=1$, using the first equation above for simulation) with $\Delta t = 1/250$, the average of $X_T$ is significantly lower than $X_0 = 0$, which implies that also $S_T$ is on average significantly lower than $S_0$.
This seems understandable to me since I learnt that the above equation for $S_t$ is the solution of the dynamics $dS/S = \mu dt + \sigma dB_t$, and that, from Ito's lemma applied to the latter, in order for $S$ to be a martingale, the drift $\mu$ of the process $X$ needs to equal $-\sigma^2/2$; thus $X$ should go down on average.
However, from the martingale property of $S$, I would expect $S_T$ to be on average on the level of $S_0$. What is wrong? Can anybody write a concise illustration of the concept?


1 Answer 1


The average of the exponentials is not the exponential of the average. It is always higher due to convexity (Jensen inequality). So there is no contradiction between the average of $X_T$ being negative and the average of $S_T$ being $S_0$.

So the question is: are your results really significantly different from what you would expect? Have you tried increasing your simulation number? Do you know how to compute a confidence interval as a function of N?

  • $\begingroup$ Indeed, here, $X$ is a super-martingale, while $E^{X}$ is a martingale. The mean of $X$ drop significantly does not imply that the mean of $e^X$ will also drop significantly. $\endgroup$
    – Gordon
    Commented Apr 1, 2015 at 17:03
  • $\begingroup$ Ok thanks! I guess the wrong part in my exposee was: "which implies that also $S_T$ is on average significantly lower than $S_0$." I computed the mean of $S_T$ and it gets closer to $S_0$ as I increase the simulation number $N (1000, 10000,...)$. Re AFK: How would you report a confidence interval $\mathcal{I}$ - denoting the mean of $S_T$ by $\bar{S}_T$ and and $sd =$ RMS $= \sqrt{ \frac{1}{N} \sum_{j=1}^N (S_T^{(j)} - \bar{S}_T)^2 }$, just by $\mathcal{I} = [\bar{S}_T - sd, \bar{S}_T + sd] \,$ ? $\endgroup$
    – Futurist
    Commented Apr 3, 2015 at 8:19

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