I have a Poisson point process with rate $\lambda$. I want to consider a new process that fires an event every time the Poisson point process finds 3 points. How this process can be described? What is the event rate and the variance of this process?
I put on evidence that the second process is not simply a Poisson point process with rate $\frac{\lambda}{3}$.
Denote this process by $X(t)$. The interarrival distribution has $\mathrm{Erlang}(3,\lambda)$ distribution with density $f_J(t) = \lambda\frac{(\lambda t)^2}2 e^{-\lambda t}$. The Laplace-Stieltjes transform of this distribution is \begin{align} \hat F(z) &= \mathbb E[e^{-zJ}]\\ &= \int_0^\infty e^{-zt}\lambda\frac{(\lambda t)^2}2 e^{-\lambda t}\mathsf dt\\ &= \frac{\lambda^3}{(\lambda+z)^3}. \end{align} Let $M(t) = \mathbb E[X(t)]$ be the renewal function. Then the Laplace-Stieltjes transform of $M$ is \begin{align} \hat M(z) = \frac{\hat F(z)}{1-\hat F(z)} = \frac{\frac{\lambda^3}{(\lambda+z)^3}}{1-\frac{\lambda ^3}{(\lambda +z)^3}} = \frac{\lambda ^3}{z^3+3 \lambda z^2+3 \lambda ^2 z}. \end{align} Inversion yields $$ M(t) = \frac{1}{3} \lambda \left(1-e^{-\frac{1}{2} (3 \lambda t)} \left(\sqrt{3} \sin \left(\frac{1}{2} \sqrt{3} \lambda t\right)+\cos \left(\frac{1}{2} \sqrt{3} \lambda t\right)\right)\right). $$ From this question we can derive the variance of a renewal process from the renewal function: \begin{align} \operatorname{Var}(X(t)) &= 2\int_0^t M(t-s)\ \mathsf d M(s) + M(t) - M(t)^2\\ &=2\int_0^t \frac{\lambda ^3}{3 \lambda ^2 (t-s)+3 \lambda (t-s)^2+(t-s)^3}\left(-\frac{3 \lambda ^3}{(\lambda +s)^4}\right)\ \mathsf ds\\ &\quad+\frac{1}{3} \lambda \left(1-e^{-\frac{1}{2} (3 \lambda t)} \left(\sqrt{3} \sin \left(\frac{1}{2} \sqrt{3} \lambda t\right)+\cos \left(\frac{1}{2} \sqrt{3} \lambda t\right)\right)\right)\\ &\quad - \left(\frac{1}{3} \lambda \left(1-e^{-\frac{1}{2} (3 \lambda t)} \left(\sqrt{3} \sin \left(\frac{1}{2} \sqrt{3} \lambda t\right)+\cos \left(\frac{1}{2} \sqrt{3} \lambda t\right)\right)\right)\right)^2. \end{align} Unfortunately, I can't get Mathematica to evaluate the integral. There isn't likely to be a nice closed form for this one.