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Could anyone tell me what is the negation of the statement:

$\forall \delta >0, \forall M>0$ there exists $N\in \mathbb N$ such that

$\text{Probability}( \text{ at least } M \text{ number of events occur in time step } [0,N] )>1- \delta$.

Thanks for helping.

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  • $\begingroup$ Try changing $\forall$ by $\exists$ and $\exists$ by $\forall$. $\endgroup$
    – A. Salguero-Alarcón
    Apr 24, 2020 at 13:17
  • $\begingroup$ and change $>$ to $\leq$ $\endgroup$
    – Loobear23
    Apr 24, 2020 at 13:19
  • $\begingroup$ so $\exists \delta >0$ and $\exists M>0$ such that $\text{Probability}( \text{ at least } M \text{ number of events occur in time step } [0,N] )\le 1- \delta$. $\endgroup$
    – Balbichi
    Apr 24, 2020 at 13:19
  • $\begingroup$ $\exists \delta>0, \exists M>0, \forall N \in N$ such that Probability( at least M number of events occur in time step [0,N]) $\le 1-\delta$ $\endgroup$
    – Mathsmerizing
    Apr 24, 2020 at 13:21

1 Answer 1

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In general, to negate a statement we swap $\forall$s and $\exists$s while negating the expression on the RHS of the ‘such that’. So your statement $$\forall \delta>0,\ \forall M>0 \exists\ N\in\mathbb{N}\ : \mathbb{P}\left( \text{At least M events in } [0,N]\right)>1-\delta,$$ has negation $$\forall\ N\in\mathbb{N} \exists \delta>0\ \text{ and }\exists\ M>0: \mathbb{P}\left( \text{At least M events in } [0,N]\right)\leq 1-\delta.$$

This says that there is an exception; for any $N\in\mathbb{N}$ we are given, we can find a $\delta>0$ and $M>0$ such that the statement does not hold. Imagine it like playing a game: in the first instance, you are given a $\delta>0$ and an $M>0$ and your challenge is to find an $N\in\mathbb{N}$ such that the probability is true. Well then negating the statement is taking the reverse role in the game: someone is giving you an $N$ and you have to give them a $\delta,M$ such that they can’t win.

Hope this helps, stay safe

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