I'm reading a paper which states the following (I'll just write the troubling parts):
The probability that $n < N(T)$, which we represent by $P(n<N(T))$ ... because $N(T) = \lceil \lambda T\rceil$ we get... $$P \Big(n < \lceil \lambda T\rceil \Big) = P(\lambda > \frac{n}{T} \Big)$$
I'm not really sure, but if I understand correctly the ceiling function,
$$n < \lceil \lambda T \rceil$$ is $$n-1 < \lambda T$$
and so
$$P(n < \lceil \lambda T \rceil) = P(n - 1 < \lambda T) = P \Big( \lambda > \frac{n-1}{T} \Big)$$
If the paper is right, do I missed something?
This is the paper (p. 452)
First line of p. 452 "... then $N(t)$ equals the smallest integer greater than $\lambda t$, which we represent by the symbol $\lceil \lambda t \rceil$" gives a slightly unusual definition of the "ceiling" brackets. Normally, the ceiling of $(0,1]$ is $1$. Here, $\lceil x \rceil$ for $x \in [0,1)$ is $1$. (The smallest integer greater than $0$ is $1$ and the smallest integer greater than $1$ is $2$.) When we use the ceiling brackets in this answer, we are using the paper's definition.
This means $\lambda T \in \left[ \left\lceil \lambda T \right\rceil - 1, \lceil \lambda T \rceil \right)$. Let's write this as $\lceil \lambda T \rceil = \lambda T + \varepsilon$ for some $\varepsilon \in (0,1]$. Then $n < \lceil \lambda T \rceil$ means $$ n < \lambda T + \varepsilon $$ for some $\varepsilon \in (0,1]$. Subtracting $\varepsilon$ and dividing by $T$ (assuming $T > 0$), we have $$ \frac{n - \varepsilon}{T} < \lambda \text{.} $$
This isn't quite $\frac{n}{T} < \lambda$. It is an approximation with error (strictly) less than $1/T$. Their subsequent approximations are about as precise, so this isn't an awful approximation.