Approximating Returns and Moments in a Mean-Reverting Log-Price Diffusion
Summary
The document studies a stochastic process whose log price follows a mean-reverting diffusion. It transforms the original price equation into an Ornstein–Uhlenbeck form for the log price, then asks about the conditional distribution and moments of the one-period proportional price change. The replies suggest using a first-order Taylor approximation to relate a small log return to a simple return, and using Itô’s formula with powers of the log process to derive moment equations recursively.
The discussion also points to the Fokker–Planck equation as a route to a density, while cautioning that an exact distribution for the simple yield may be difficult and may be unnecessary for small time steps. These are methodological suggestions rather than a worked numerical example. The approximation depends on a sufficiently small interval, and the displayed discussion does not provide a complete density or final expectation and standard deviation for the requested return. Care is also needed to distinguish the log-price distribution from that of the price return.
Key ideas
- A logarithmic transformation turns the stated price diffusion into a mean-reverting process for log price.
- For short intervals, a first-order expansion relates simple returns to log returns.
- Itô’s formula can generate recursive equations for moments of the log process.
- A Fokker–Planck equation may be used to derive a density, though the exact return distribution can be difficult.
- The small-step approximation does not supply an exact distribution for larger intervals.
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# Obtaining characteristics of stochastic model solution
# Obtaining characteristics of stochastic model solution
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.
## Answer by SBF (score 4, accepted)
https://quant.stackexchange.com/a/1453
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.
## Answer by Beer4All (score 3)
https://quant.stackexchange.com/a/1454
- 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, $E\left( Z_t^p \right)=E\left( Z_0^p \right)+\int_0^t\left(\frac{\sigma^2}{2}+ E(Z_s^{p-1})\cdot{pk(\theta-\frac{\sigma^2}{2k})-E(Z_s^p)\cdot pk} \right)ds$ finally: $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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.