Hints find $Z$ so that $n \min\{X_{1},...,X_{n}\}\xrightarrow{d} Z$

Question

Let $n \in \mathbb N$ while $(X_{i})_{i=1}^{n}$ are independent and uniformly distributed random variables on $[0,1]$ and define $M_{n}:=\min\{X_{1},...,X_{n}\}$.

Find a $Z$ so that $n M_{n}\xrightarrow{d} Z$

Solution:

If we can show $\lim_{n \to \infty}P(nM_{n}>z)=P(Z>z)$ then it immediately follows by the definition of the cdf that $\lim_{n \to \infty}F_{X_{n}}=F_{Z}$

$P(nM_{n}> z)=P(nX_{1} > z,...nX_{n}>z)=\prod_{i=1}^{n}P(nX_{1}>z)=(P(nX_{i}>z))^{n}=(\int_{\frac{z}{n}}^{1}1dx)^n=(1-\frac{z}{n})^{n}1_{z \in [0,n]}$

For $\lim_{n \to \infty}(1-\frac{z}{n})^{n}=e^{-z}$

So we can choose $Z$~$\exp({-z})$ where $z \in [0, n]$ such that

$nM_{n} \xrightarrow{d} Z$

Is this correct?

Answer 1

For each $n$, compute $P(nM_n > z)$.

$$P(nM_n > z) = (1 - \frac{z}{n})^n \cdot 1_{z \in [0, n]}.$$

Then $$P(Z > z) = \lim_{z \to \infty} P(n M_n > z) = \cdots$$ and hopefully you recognize the resulting distribution.