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Suppose $X \sim Pois(\lambda)$. I'm interested in an upper bound on the ratio, $$\dfrac{P(X \leq n)}{P(X \leq n-1)}\,,\,\,\text{for $n=1,2,3,...$}$$ Observe that, the ratio is $>1$ & as $n \to \infty,\,$ the ratio $\to 1$. Thus, It is interesting to see if there exists some constant $K>1$ (depending on whether $\lambda \leq n$ or $\lambda \geq n$) so that the ratio is $\leq K^{1/n}.$

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  • $\begingroup$ What happens if you write $P(X \le n)$ as $P(X \le n-1) + P(X = n)$? $\endgroup$
    – Jukka Kohonen
    Commented Mar 12, 2021 at 11:08
  • $\begingroup$ @JukkaKohonen The ratio becomes $1+ P(X=n)/P(X \leq n-1)$ $\endgroup$
    – SL_MathGuy
    Commented Mar 12, 2021 at 11:13
  • $\begingroup$ Your ratio is simply asymptotic to $1+P(X=n)$. This is explicit.... $\endgroup$
    – ofer zeitouni
    Commented Mar 12, 2021 at 12:43

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Clearly $\Pr[X\le n-1]\ge \Pr[X=n-1]$ so you have the ratio

$$ 1+\frac{\Pr[X=n]}{\Pr[X\le n-1]}\le 1+\frac{\Pr[X=n]}{\Pr[X=n-1]} =1+\frac{\lambda^n/n!}{\lambda^{n-1}/(n-1)!} =1+\lambda/n. $$ Now $$ 1+\lambda/n \le \exp(\lambda/n). $$ Hence you can take $K=e^\lambda$ in your example.

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  • $\begingroup$ Too bad I couldn't observe that. Thanks. $\endgroup$
    – SL_MathGuy
    Commented Mar 13, 2021 at 7:30

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