Integrated CIR Distributions and Conditional Independence

# Law of an integrated CIR Process as sum of Independent Random Variables


# Law of an integrated CIR Process as sum of Independent Random Variables












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 6.3.4.1

Best regards

## Answer by kurtosis (score 2)

https://quant.stackexchange.com/a/56000

> 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).