# what is the negation of the statement

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.

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

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.