It is known (see for example Joshi-Chan "Fast and Accureate Long Stepping Simulation of the Heston SV Model" available at SSRN) that for a CIR process defined as :

$$dY_t= \kappa(\theta -Y_t)dt+ \omega \sqrt{Y_t}dW_t$$ $Y_0=Y$ together with a correct set of constraints on parameters' value.

Then the law of $\int_0^T Y_t dt$ conditionaly on $Y_0,Y_T$ can be seen as the sum of three (rather complicated) independent random variables (see proposition 4 eq 2.10 in Joshi Chan article)

NB : The original result is coming from Glassermann and Kim "Gamma Expansion of the Heston Stochastic Volatility Model" available at SSRN, but I'm more used to Joshi Chan's expression.

So here is my question :

Does the integrated CIR process itself by any chance has such a representation in the form of the sum of independent random variables ?

PS: The Laplace transform has a known closed-form expression but I couldn't infer directly from this such a representation.

Edit : As Tal has opened a bounty on this here is the Laplace transform of the integrated CIR process :

$$\mathcal{L}\left\{\int_0^t Y_s ds\right\}(\lambda)=\mathbb{E}\left[e^{-\lambda\int_0^t Y_s ds}\right]=e^{-A_\lambda(t)-Y_0.G_\lambda(t)}$$ with $A_\lambda(t)=-\frac{2\kappa.\theta }{\omega^2}. \mathrm{Ln}\left[\frac{2\gamma.e^{(\gamma+\kappa).t/2}}{\gamma.(e^{t.\gamma}+1)+\kappa.(e^{t.\gamma}-1)}\right]$ and $G_\lambda(t)=\frac{2.\lambda.(e^{t.\gamma}-1)}{\gamma.(e^{t.\gamma}+1)+\kappa.(e^{t.\gamma}-1)} $ where $\gamma=\sqrt{\kappa^2+\omega^2.\lambda}$

This is coming from Chesnay, Jeanblanc-Picqué, Yor "Mathematical Methods for Financial Markets" Proposition

Best regards

  • $\begingroup$ Well the CIR can be written as non central chi-squared random variable if $\nu=\frac{4\kappa \theta}{\sigma^2} \in \mathbb{N}$. So a finite sum of such process sould also be a non central Chi-square. Now if we could find stabiling transformation such that $\nu_n% tend to $\vu$. We could get a converging series. However, this remind me alot of the cramer-von-mises statistics series, which a sum of chi-squared with variance equal to the the reciprocal of the square of an odd number. That distribution is unknown. $\endgroup$
    – Drmanifold
    Commented Dec 13, 2013 at 23:15
  • $\begingroup$ Can I ask you, what use case do you have in mind for the determining some law of integrated exponential martingales? Asian commodity options? Perpetuities and annuities? $\endgroup$ Commented Apr 17, 2018 at 5:36
  • $\begingroup$ Very old post no use today for me but I was interested at the time by stochastic volatility models where volatilty would be driven by a CIR process. Inverting Fourier transform could give many interesting computationnal possibilities for such dynamics in pricing applications $\endgroup$
    – TheBridge
    Commented Apr 17, 2018 at 8:01
  • $\begingroup$ Very interesting. Being on the equities side, I had not thought of that use case. The problem looks related to the Yor process, which is defined as the sum of lognormals, which is difficult to evaluate since its distribution is not lognormal. Also, I doubt you’ll be able to express the law of integrated CIR as the sum of independent variables since, like an O-U, it’s increments are not independent. Dassios and Jayalaxschmi explore integrated CIR processes as they relate to arithmetic Asian options: eprints.lse.ac.uk/2851/1/…. $\endgroup$ Commented Apr 17, 2018 at 16:34
  • $\begingroup$ I'm actually very interested in this, but not in the realm of finance. I study bioinformatics (namely phylogenetics), and Lepage et al 2006 used CIR to model mutation rates over time. I am now trying to sample mutation rates for each branch of an evolutionary tree, which requires integrating $dY_t$ (given $Y_0$) over some $t$ $\endgroup$ Commented Jun 22, 2018 at 17:50

1 Answer 1


Does the integrated CIR process itself by any chance has such a representation in the form of the sum of independent random variables?

I think the answer to this is clearly "no." The CIR process is (as @DavidAddison points out in comments above) like an Ornstein-Uhlenbeck process. The mean-reverting property of the O-U (and the CIR) mean that the process is serially-correlated. Therefore, you would be hard-pressed to find a sum of independent random variables to represent $\int_0^T Y_t dt$.

To flesh this out more: we know the distribution of $Y_T$ if we are given only $Y_0$. We can also find the distribution of $Y_{T/2}$. However, $Y_{T/2}$ and $Y_T$ are not independent, they are correlated.

Perhaps we could interpolate more frequently, since that will be needed to approximate the interval (numerically or taking a limit). If we interpolate more frequently, however, we run into the same problem: the samples are serially-correlated to one another. Thus the boxes we create to approximate our integral cannot be independent of one another.

How, then, did Chan and Joshi (and Glasserman and Kim before them) decompose $\int_0^T Y_t dt$? By conditioning on $Y_0$ and $Y_T$, they have a squared Ornstein-Uhlenbeck bridge. Just as with a Brownian bridge, that give them more information about the distribution of values in $(0,T)$. In this case, that information is enough to write down the integrated CIR process as a sum of three independent variables (which are themselves sums of independent variables having different distributions).

For more on this, Theorem 2.1 of Glasserman and Kim is informative (and has references to the decomposition which was originally proven by Pitman and Yor).

  • $\begingroup$ Thanks for this interesting contribution, I don't set your answer as a solution to the question because I don't think you have proven the fact that a serially correlated process admits no such represetation for the integrated process over the interval [0,T] (I have no clue whatsoever on this even though this appears plausible), moreover you say yourself that with the bridged process $Y_t$ knowing $Y_T$ that this is possible but the bridge process seems also very serially correlated to me. Finally the result obtained in Joshi Chan has no link to the increments of Y_t itself unless mistaken. $\endgroup$
    – TheBridge
    Commented Jul 31, 2020 at 9:10
  • $\begingroup$ The same situation occurs in a Brownian bridge: because you know the start and end point, the middle point (or points) are related and serially correlated. Nonetheless, it is that extra knowledge of the endpoint which allows you to decompose those midpoints. If Pitman and Yor did not need the endpoint, they would not have assumed it and would have, instead, proven a more general theorem. I believe you are looking for something not yet proven. $\endgroup$
    – kurtosis
    Commented Jul 31, 2020 at 10:14
  • $\begingroup$ As you can see the problme is open for more than 8 years indeed but I would have accepted your answer if it had demonstrated that no such decomposition can exist which is not proven only conjecured based on the serial correlation argument. This line of argument at least shows how difficult the task is. Best regards and thanks for your answer. $\endgroup$
    – TheBridge
    Commented Jul 31, 2020 at 14:58

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