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 $\mathbb{P}(E_k')$?
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. $$p(\sum_i I_i > \alpha m)\le \frac{dm^2 (\bar{p}-\bar{p}^2)}{dm^2 (\bar{p}-\bar{p}^2) + (\alpha-d\bar{p})^2m^2}=\frac{(\bar{p}-\bar{p}^2)}{\frac{\alpha^2}{d}+(1-2\alpha )\bar p}$$