I want to use the following stochastic model

$$\frac{\mathrm{d}S_{t}}{ S_{t}} = k(\theta - \ln S_{t}) \mathrm{d}t + \sigma\mathrm{d}W_{t}\quad (1)$$

using the change in variable $Z_t=ln(S_t)$

we obtain the following SDE

$$\mathrm{d}Z_{t} = k(\theta - \frac{\sigma^2}{2k} - Z_{t}) \mathrm{d}t + \sigma\mathrm{d}W_{t}\quad (2)$$

again we change variable $X_t=e^{kt}Z_t$

we obtain the following SDE

$$\mathrm{d}X_t = k(\theta -\frac{\sigma^2}{2k})e^{kt}\mathrm{d}t + \sigma e^{kt}\mathrm{d}W_{t} \quad (3)$$

This last equation can be integrated easily and we see that

$$(X_{t+1}-X_t)\sim N(\mu, \sigma)$$

My question is: How can I get the ditribution of $\frac{S_{t+1}-S_{t}}{ S_{t}}$ ? (at least its expectation and standard deviation under filtration $F_t$)

If we come back to the first equation it seams that if $\mathrm{d}t$ is small enough then it is a normal distribution with mean $k(\theta - \ln S_{t}) \mathrm{d}t$ and sd $\sigma\mathrm{d}t$, but I want to find mathematically if it is true, and if yes, under what assumptions.

  • $\begingroup$ It seems to me that you found the incorrect SDE for $X_t$, could you please provide your calculations? $\endgroup$
    – SBF
    Jul 15, 2011 at 8:11
  • $\begingroup$ @Gortaur: indeed, there was a $e^{kt}$ missing in the stochastic part $\endgroup$ Jul 15, 2011 at 8:24
  • $\begingroup$ I think you missed the more important term $\frac{1}{S^2}$, see my answer. Could you refer me to Ito formula you are using? $\endgroup$
    – SBF
    Jul 15, 2011 at 8:33
  • $\begingroup$ can you put some details on your filtration $F_t$? $\endgroup$
    – Beer4All
    Jul 15, 2011 at 9:45
  • $\begingroup$ $F_t$ is the natural filtration $F_t=\{S_i\}_{i=0..t}$ $\endgroup$ Jul 18, 2011 at 4:54

2 Answers 2


I've edited my answer since Berr4All showed that your equation is right.

What you still can do - is to use Fokker-Planck equation to derive a density.

  • $\begingroup$ Why not considering the empirical way (MC) to get the pdf? $\endgroup$
    – Beer4All
    Jul 15, 2011 at 9:48
  • $\begingroup$ @Beer4All: what kind of empirical way do you advise? What is pdf by the way? $\endgroup$
    – SBF
    Jul 15, 2011 at 9:56
  • $\begingroup$ @Beer4All: I see. I am not so happy with randomized methods like Monte-Carlo, so I better use numerical solutions of PDEs if it is possible. $\endgroup$
    – SBF
    Jul 15, 2011 at 10:18
  • $\begingroup$ pdf = probability density function. To get numerically an empiricall estimate of your density function: 1 -- draft sample of $X_t$, 2 -- trace the empirical cdf (cumulative distribution function) of this sample 3 -- then compute (numerically) the differentiated function which is nothing but the empirical density ;) $\endgroup$
    – Beer4All
    Jul 15, 2011 at 10:18
  • $\begingroup$ For me MC methods are often the fastest way to get rapid & painless solutions, useful to test your results. $\endgroup$
    – Beer4All
    Jul 15, 2011 at 10:25
  • 1 ) A first-order Taylor expansion gives $\ln \left(\frac{S_{t+\Delta_t}}{S_t}\right)\approx \frac{S_{t+\Delta_t}-S_{t}}{S_t}+o(\Delta_t)$ , thus unless $\Delta_t$ is not small you can drop the residual term and consider $Z_t\overset{law}{=}\frac{S_{t+\Delta_t}-S_{t}}{S_t}$.

  • 2 ) Calculation of the moments: we can proceed by using the classical Dynkin way

    we remind that $dZ_t=k(\theta-\frac{\sigma^2}{2k}-Z_t)dt+\sigma dW_t$, so

    $d(Z_t^p)\overset{Itô}{=}p\cdot(Z_t)^{p-1}dZ_t+\frac{\sigma^2}{2}dt = \left(\frac{\sigma^2}{2}+ Z_t^{p-1}\cdot{pk(\theta-\frac{\sigma^2}{2k})-Z_t^p\cdot pk} \right)dt+p\sigma Z_t^{p-1}dW_t$

    taking the expectation,

    $E\left( Z_t^p \right)=E\left( Z_0^p \right)+\int_0^t\left(\frac{\sigma^2}{2}+ E(Z_s^{p-1})\cdot{pk(\theta-\frac{\sigma^2}{2k})-E(Z_s^p)\cdot pk} \right)ds$


    $E\left( Z_t^p \right)=E\left( Z_0^p \right)+\frac{\sigma^2}{2}t+pk(\theta-\frac{\sigma^2}{2k}) \int_0^t E(Z_s^{p-1})ds-pk \int_0^tE(Z_s^p)ds $

    so, your p-th moment is the solution of the above deterministic ODE which can be solved step-by-step starting $p=1$.

  • 3 ) Exact distribution of the yield: in general it's far to be simple to get exact distribution/ simulation for diffusions (and more for multidimensional SDE's -- example: Heston's joint SDE has an exact pdf which computation requires Malliavin calculus).

    So, in my opinion unless you are considering simulations with large time step it would be useless to deal with the exact distribution of your SDE.

  • $\begingroup$ would you like to check out if OP found the right equation for $Z_t$? $\endgroup$
    – SBF
    Jul 16, 2011 at 19:02

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