Let $\gamma\in[0,1]$, $M>0$, $n\in\mathbb{Z}^{+}$ and $$ P\leq \exp\left(\frac{M}{5}\right)\exp\left(-n\gamma x + \frac{nx^2}{M-5x}\right),\qquad \text{for }x\in[0, M/5). $$ I want to show that I can reformulate an upper bound that is independent of $x$, by minimising it in $[0,M/5)$. In particular, I want to show that $$ P\leq \exp\left(\frac{M}{5}\right)\exp\left(-n\frac{M\gamma^2}{12}\right). $$ $P$ denotes a probability and can only output values between $0$ and $1$. $P$ is independent of $x$.

This new upper bound works when I check it using some graphing tools. However, I want to derive it analytically from the info I have here. Also, the value $\exp(-nM\gamma^2/12)$ is not the local minimum point of the original bound in $[0,M/5)$. It is just a value large enough for us to justify this uniform bound. Can anyone provide any insights to this question?


1 Answer 1


This doesnt seem right since for $x = M/5-\epsilon$ the RHS is arbitrarily large.

  • $\begingroup$ I see. I was trying to simplify the question and I miss out some important points. I have added those in. Thanks for pointing out $\endgroup$
    – Nicolas
    Sep 29, 2023 at 17:40
  • $\begingroup$ We still have the same issue. Are you sure you are not looking to find that $P$ is bounded above by the expression you showed for some interval around $0$? if you include values arbitrarily close to $x=M/5$ then the upper bound is unbounded (and thus trivial) $\endgroup$
    – dmh
    Sep 29, 2023 at 18:32

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