I have that:
P(n.m_{n} <= x)≈1-exp(-x)
I would like to understand how:
P(m_{n} <= x)=1-exp(-nx)
Thank you!
$P[n \cdot m_n \leq x) \approx 1-e^{-x}$
Assuming $n >0$
$P\left[m_n \leq \dfrac{x}{n}\right]\approx 1-e^{-x}$
$u = \dfrac{x}{n},~x = n u$
$P\left[m_n \leq u\right] \approx 1- e^{-n u}$
$u$ is a dummy variable so just replace it by $x$
$P[m_n \leq x] \approx 1 - e^{- n x}$