Consider a Poisson variable with parameter $\lambda$: $$ X \sim Pois(\lambda) $$ For an arbitrary $\delta > 0.5$ and $x \in (1, \infty)$ is there a simple way to select $\lambda$ such that we always have: $$ P(X \geq x) > \delta $$
Here is an effort: $$ \sum_{k=x}^{\infty}\frac{\lambda^k e^{-\lambda} }{k!} \geq \delta $$ Simplfiying this equation is tricky. Since we know that $\delta>0.5$, $x$ should be close to $\lambda$ or smaller. Wondering if I can find a simple expression.
$$ \sum_{k=x}^{\infty}\frac{\lambda^k e^{-\lambda} }{k!} \geq \delta $$
We other approach is to use the Chernoff bound: $$ P(X \leq x) \leq \frac{e^{-\lambda} (e \lambda)^x}{x^x}, \text{ for } x < \lambda. $$ Using the second inequality, we have: $$ P(X > x) = P(X \geq x+1) \geq 1 - \frac{e^{-\lambda} (e \lambda)^x}{x^x}, \text{ for } x < \lambda. $$ Hence: $$ P(X \geq x) \geq 1 - \frac{e^{-\lambda} (e \lambda)^{x-1}}{(x-1)^{x-1}}, \text{ for } x < \lambda-1. $$ We set: $$ 1 - \frac{e^{-\lambda} (e \lambda)^{x-1}}{(x-1)^{x-1}} \geq \delta \Rightarrow 1 - \delta \geq \frac{e^{-\lambda} (e \lambda)^{x-1}}{(x-1)^{x-1}} $$
$$ \Rightarrow \log(1 - \delta) \geq -\lambda + (x-1)(1+ \log \lambda) - (x-1) \log (x-1) $$
Simplifying it a little more: $$ \Rightarrow \log(1 - \delta) \geq (x-1) \left[(1+ \log \lambda) - \log (x-1) \right] - \lambda $$