# Statistics of the Cartesian product of Poisson point processes

Let $\mathcal{P}_\nu^d$ be a Poisson point process with intensity $\nu$ defined over the region $[0,1]^d$. Then given $m$ disjoint bounded Lebesgue measurable regions $R_1,\dots,R_m$ each with Lebesgue measure $|R|$, the distribution over the number of points in each region has a Poisson distribution with intensity $\nu|R|$ independent from the number of points in any other region. Consequently, the number of regions with no samples has a binomial distribution: if $E_k$ is the event that exactly $k$ of the $m$ regions have no samples, then $$\mathbb{P}(E_k)=\text{Bin}(k;m,\exp(-\nu |R|))$$

Now, suppose I instead generate the points as the Cartesian product of samples from $d$ separate one-dimensional Poisson point processes. Specifically, let $X\sim\mathcal{PP}_\nu^d$ mean that $X=X_1\times\dots\times X_d$, where $X_i\sim \mathcal{P}_\nu^1$. Clearly, the distribution over the number of points in each region is no longer independent from the number of points in the other regions.

Does this 'product process' have a name? And more importantly, can I conclude anything about the distribution of the number of regions without samples under this product point process? In particular, if $E_k'$ is the event that exactly $k$ of the $m$ regions have no samples in a set $X$ drawn from the product process, can I determine or upper bound $\mathbb{P}(E_k')$?

• From the description, I guess that if there is at least one of the $X_i = 0$, then there will be no sample inside the unit hypercube. I know the shape of the partitioned region $E_k$ can be very general, but if for the simplest case that it is rectangular, then the probability of no sample inside can be calculated via the above argument. Not so sure about the general case - maybe can be extend by some differential arguments. – BGM May 31 '16 at 4:34

## 1 Answer

Although I haven't been able to find any references to a construction like this, after some more thought, I have established the answer to the last question is affirmative. Let $r_i=\mathrm{sup}\{r\in\mathbb{R}^+:\exists x\,B_r(x)\subset R_i\}$ be the radius of the largest open ball that fits inside $R_i$, and let $r=\underset{i}{\mathrm{min}}\,r_i$. Then for any $k > m\exp(-2\frac{\nu r}{d})$, the probability of the event $E_k'$ that more than $m$ regions do not contain a sample from the product process is bounded from above. $$\mathbb{P}(E_k') \le \exp(-2\frac{\nu r}{d})\frac{1-\exp(-2\frac{\nu r}{d})}{\frac{k^2}{dm^2}+(1-2\frac{k}{m})\exp(-2\frac{\nu r}{d})}$$ I'm sure there is a two-sided variant as well.

Proof: the ball $B_r(x_i)\subset R_i\,\forall\,i$. Let $I_i$ be the indicator for the event that $R_i$ does not contain a sample, and let $I_i'$ be the indicator for the event that $B_r(x_i)$ does not contain a sample; clearly, $\sum_i I_i \le \sum_i I_i'$. Let $x_{i,n}$ be the $n$th component of $x_i$, the center of the ball contained in $r_i$. Define $B_{i,n}=\{|x_{i,n}-x|<\frac{r}{d}\}$, and let $I_{i,n}$ be the event that the interval $B_{i,n}$ does not contain a sample. By construction, $B_{i,1}\times\dots\times B_{i,d}\subset B_r(x_i)$, so $I_i'\le\sum_n I_{i,n}$ and $\sum_i I_i'\le\sum_{i,n} I_{i,n}$. Then $\mathbb{P}(\sum_i I_i\ge k)\le\mathbb{P}(\sum_{i,n} I_{i,n}\ge k)$

$I_{i,n}$ is a Bernoulli random variable with probability $\bar{p}=\exp(-\frac{2\nu r}{d})$. The regions $\{B_{i,n}\}$ are not disjoint, so the variables $\{I_{i,n}\}$ are not independent; however, the event that any two regions $B_{i,n}$ and $B_{i',n}$ have no samples is the same as the event that their union $B_{i,n}\cup B_{i',n}$ lacks a sample; the volume of this region is bounded from below by $2\frac{r}{d}$ and from above by $4\frac{r}{d}$, so $\exp(-\frac{4\nu r}{d})=\bar{p}^2\le\mathbb{E}[I_{i,n}I_{i',n}]\le\bar{p}$. We can then compute the expectation and variance of the sum $\sum_{i,n}I_{i,n}$. $$\mathbb{E}[\sum_{i,n}I_{i,n}]=\sum_{i,n}\mathbb{E}[I_{i,n}]=md\bar{p}$$ $$\mathrm{var}[\sum_{i,n}I_{i,n}]=\sum_{n}\mathrm{var}[\sum_{i}I_{i,n}]= d(\sum_{i}\mathrm{var}[I_{i,n}]+2\sum_i\sum_{i'<i}\mathbb{E}[I_{i,n}I_{i',n}]-\mathbb{E}[I_{i,n}]\mathbb{E}[I_{i',n}])$$ $$\therefore \mathrm{var}[\sum_{i,n}I_{i,n}]\le dm^2(\bar{p}-\bar{p}^2)$$

Then for any $\alpha>\bar{p}$ Cantelli's inequality gives the desired result. $$p(\sum_i I_i > \alpha m)\le \frac{dm^2 (\bar{p}-\bar{p}^2)}{dm^2 (\bar{p}-\bar{p}^2) + (\alpha-d\bar{p})^2m^2}=\frac{(\bar{p}-\bar{p}^2)}{\frac{\alpha^2}{d}+(1-2\alpha )\bar p}$$