A classical inequality of Teicher (1955) asserts

**Proposition** (*Teicher*). Let $X \sim \mathrm{Pois}(\lambda)$. Then, $\mathbb P(X \leq [\lambda]) > e^{-1}$.

A modification of his argument will allow us to prove the following.

**Proposition**. For $\lambda \in (0,1)$, let $X_{n\lambda} \sim \mathrm{Pois}(n\lambda)$. Then the sequence $b_n := b_{n,n\lambda} = \mathbb P(X_{n\lambda} < n)$ is monotonically increasing. In particular, $b_n \geq e^{-\lambda}$ for all $n$.

*Proof*. First, note that, by considering a Poisson process with rate 1, we have,
$$
b_{n+1,\mu} = \sum_{x=0}^n \frac{e^{-\mu} \mu^x}{x!} = \int_\mu^\infty \frac{x^n e^{-x}}{n!} \,\mathrm{d}x \,,
$$
for all $n$ and $\mu$. Now,
$$
b_{n+1,n\lambda} - b_{n+1,(n+1)\lambda} = \int_{n\lambda}^{(n+1)\lambda} \frac{x^n e^{-x}}{n!} \,\mathrm{d}x = \int_0^1 (\lambda(y+n))^n \frac{e^{-\lambda(y+n)}}{n!} \lambda \,\mathrm{d}y \,,
$$
where the last equality follows from the substitution $y = (x-n\lambda)/\lambda$.

We can rewrite the last integral as
$$
\frac{e^{-\lambda n}(\lambda n)^n}{n!} \lambda \int_0^1 (1+y/n)^n e^{-\lambda y} \,\mathrm{dy} < \frac{e^{-\lambda n}(\lambda n)^n}{n!} = b_{n+1,n\lambda} - b_{n,n\lambda} \,
$$
where the inequality follows from facts that $(1+y/n)^n < e^y$ and (upon integrating) $e^{1-\lambda} < \lambda^{-1}$, true for any $\lambda \in (0,1)$.

But, then $b_{n,n\lambda} < b_{n+1,(n+1)\lambda}$ which is what was to be shown. Since $b_{1,\lambda} = e^{-\lambda}$, the second part of the proposition statement holds.