I'm planning to use a discrete-time stochastic process defined in the following paper:

  • Nicolau, J. (2002). Stationary Processes That Look Like Random Walks—The Bounded Random Walk Process in Discrete and Continuous Time. Econometric Theory, 18(1), 99–118. https://doi.org/10.1017/S0266466602181060

On page 109 of the paper (in PROPOSITION 5), the author states:

... and has a stationary density of the form: $$\begin{equation} \bar{p}(x) \propto \sigma^{-2} \exp \left\{-\frac{2 e^{k}}{\sigma^{2}}\left(\frac{e^{-\alpha_{1}(x-\tau)}}{\alpha_{1}}+\frac{e^{\alpha_{2}(x-\tau)}}{\alpha_{2}}\right)\right\} \end{equation}$$

This actually relates to the continuous-time version of the stochastic process.

Typical parameter values are: $$k = -5, \alpha_1 = \alpha_2 = 3, \tau = 100, \sigma=4$$

As far as I can tell, the area under this curve does not sum to 1 by definition. The author's example plot of the function (assuming you can resolve the bizarre x-axis range) also looks like it might sum to less than one:

enter image description here

My question is, is this a pdf and should the area under it sum to one? Or is the author saying this function is simply an indication of the "form" (i.e. shape) of the pdf, but is not the actual pdf? In which case, if I want the pdf, presumably I should divide by the integral, assuming I have a way to calculate it.

I suspect this is simply a matter of my interpretation of the what the equation above represents rather than a flaw in the paper.


1 Answer 1


Here, the symbol $\propto$ indicates that the density $\bar p$ is proportional to the quantity on the right. In other words, there exists a constant $C$ such that $$\bar p(x) = C\sigma^{-2} \exp \left(-\frac{2 e^{k}}{\sigma^{2}}\left(\frac{e^{-\alpha_{1}(x-\tau)}}{\alpha_{1}}+\frac{e^{\alpha_{2}(x-\tau)}}{\alpha_{2}}\right)\right)$$ Since $\bar p$ is a density, that constant $C$ is actually given by $$C = \left(\int_{-\infty}^\infty\sigma^{-2} \exp \left(-\frac{2 e^{k}}{\sigma^{2}}\left(\frac{e^{-\alpha_{1}(x-\tau)}}{\alpha_{1}}+\frac{e^{\alpha_{2}(x-\tau)}}{\alpha_{2}}\right)\right)dx \right)^{-1}$$ This notation is often seen in MCMC algorithms such as Metropolis Hastings, where we want to draw samples from a distribution known only up to a constant. In these cases, knowing the "shape" of the function is often enough and we don't need to compute the normalizing constant $C$. Perhaps it is also the case for the application you have in mind.

  • 2
    $\begingroup$ Excellent thanks, that's what I suspected but wasn't sure. I only plan to present an example of this pdf to contrast it to that of a different process. So I will simply compute a numerical estimate of this integral and scale my pdf plot by it. $\endgroup$
    – Bill
    Commented Apr 3, 2022 at 18:48
  • $\begingroup$ This is a different question, but am I committing a sin by suggesting that the stationary distribution of the continuous-time version of this process is a "good indication" of that of the discrete-time process? I assumed the continuous-time pdf was derived to provide insight into the discrete-time process, simply because it is analytically easier to derive. $\endgroup$
    – Bill
    Commented Apr 3, 2022 at 18:57
  • 1
    $\begingroup$ I don't really know, I'd have to look into it in more details... I suspect there should be no problem at all but maybe I'm overlooking something important, so don't want to give bad advice. If you want to know for sure, feel free to ask a separate question ;) $\endgroup$ Commented Apr 3, 2022 at 19:06
  • 1
    $\begingroup$ Thanks. If I don't figure it out I'll ask a separate question... $\endgroup$
    – Bill
    Commented Apr 3, 2022 at 19:21

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