Find the min distribution of independent uniform distribution

61 Views Asked by At

Let $\Omega$ be the space of sequences $\omega = (\omega_1,...,\omega_n)$, where $\omega_i \in [0,1].$ Let $P_n$ be the probability distribution corresponding to the homogeneous sequence of independent trials, each $\omega_i$ having uniform distribution on $[0,1].$ Let $\eta_n=\min \omega_i, 1\leq i \leq n$. Find $P_n (\eta_n \leq t)$ and $\lim_{n \to \infty} P_n (n\eta_n \leq t) $.

I know how to solve the first question. If $0 \leq t <1$, then I can use the complement and independence to get $P_n (\eta_n \leq t) = 1-(1-t)^n$. So in sum, my answer is:

$P_n (\eta_n \leq t)=0$ if $t<0$,

$P_n (\eta_n \leq t)=1-(1-t)^n$ if $0 \leq t<1$,

$P_n (\eta_n \leq t)=1$ if $t \geq 1$.

But I am not sure the second question. My attempt is $\lim_{n \to \infty} P_n (n\eta_n \leq t) = \lim_{n \to \infty} [1-(1-t/n)^n]= 1- e^{-t}$. But I don't know how to analyze the region of t.