Define $E_1,E_2,\ldots, E_i,\ldots E_n$ as i.i.d. exponentials with parameter $\lambda$. These define processes on some interval $[0,\delta]$ (think of $\delta$ as very small, it will come into play later) in the following way
$I_i(t):=\mathbb 1_{\{E_{i}\leq t\}}$.
One could couple each one of the $I_i$'s with a Poisson Process with parameter $\lambda$ in the following way. Define recursively
$N_i^1 := E_i$, $N_i^k := N_i^{k-1} + \mathcal E^k$,
with $\mathcal E^k\sim \textrm{exp}(\lambda)$. Then $(N_i^j)_{j\geq1}$ are the points of a Poisson Process with parameter $\lambda$ (essentially, $E_i$ is the first point of the coupled Poisson Process). Call $N_i(t)$ the corresponding Poisson Process. It is clear that the stochastic domination
$I_i(t)\preceq N_i(t)\qquad\forall t\in[0,\delta]$
holds. Indeed, $N_i$ and $I_i$ share their first jump, but $N_i(\cdot)$ may have more after that in the interval $[0,T]$.
Now, by summing the processes $I_i$'s and the Poisson Processes $N_i(\cdot)$'s over $i\in\{1,\ldots,n\}$ we get the stochastic domination
$\sum_{i=1}^nI_i(t) \preceq \overline N(t)\qquad t\in[0,\delta]$,
where $\overline N(t)$ is a Poisson Process with parameter $n\lambda$.
I am interested in finding the opposite inequality, possibly holding only with high probability. In other words, given the sequence $(I_i(t))_{i\geq1}$, is it possible to couple $\sum_{i=1}^nI_i(t)$ with a Poisson Process $\underline N(\cdot)$ (with some unknown rate, depending on $n$, but which should be less than $n\lambda$) such that
$\lim_{n\rightarrow+\infty}\mathbb P(\sum_{i=1}^nI_i(t)\geq \underline N(t)) = 1$?
Furthermore, the rate $\underline \lambda$ of $\underline N(\cdot)$ should be sufficiently close to $\lambda$, for small $\delta$ (otherwise the trivial choice $\underline \lambda = 0$ would be a solution).