Let $\lambda(t)$ be a CIR process, i.e. the strong solution of the SDE $$ \mathrm{d}\lambda(t)=\kappa(\theta-\lambda)\mathrm{d}t+\sigma\sqrt{\lambda(t)}\mathrm{d}W^1(t) $$ The integrated CIR is defined by $\Lambda(t):=\int_0^t\lambda(s)\mathrm{d}s$. Let $W^2(t)$ be another Brownian motion independent of the Brownian motion $W^1(t)$ driving the intensity $\lambda(t)$. I want to know whether the processes $$ X_1:=\left(\lambda(t), \Lambda(t), W^1\left(\Lambda(t)\right), W^2\left(\Lambda(t)\right)\right) $$ and $$ X_2:=\left(\lambda(t), \Lambda(t), \int_0^t\sqrt{\lambda(t)}\mathrm{d}W^1(t), \int_0^t\sqrt{\lambda(t)}\mathrm{d}W^2(t)\right) $$ follow the same law. Since all components are continuous it should suffice to show that all joint finite distributions are equal. Unfortunately, I'm not able to verify this.

My first approach was to make use of the Dubins-Schwarz theorem which states that there are two Brownian motions $B_1$ and $B_2$ such that $$ W^i\left(\int_0^t\lambda(s)\mathrm{d}s\right)=\int_0^t\sqrt{\lambda(t)}\mathrm{d}B^i(t) $$ However, then I am not sure, if these new Brownian motions $B^1$ and $B^2$ are independent and how they are correlated with $W^1$ still driving $\lambda(t)$. Also, I'm not sure to which $\sigma$-algebra the $B_i$ are adapted.


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