How can I find the CDF and show convergence of $n \cdot \min\{X_1,X_2,... \}$

Question

Let $X_1,X_2,\ldots$ be independent r.v. and uniformly distributed on $[0,1]$. Let $n\in\mathbb N$.

How can I determine the cumulative distribution function (CDF) $F_n$ of the random variables $n \min\{X_1,X_2,\ldots,X_n \}$ and to what CDF does $F_n$ converge for $n \to \infty$?

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My attempt: I wasn't sure how to really start because the factor "$n$" is making things difficult so I have a few questions upfront:

1. When dealing with problems like this, is my strategy to find the probability density function (PDF) first and only then try to figure out the CDF or can I go straight to determining the CDF?

2. What is the correct notation for the CDF? I am used to seeing $F_X(x)$=$P(X\le x)$ which makes sense to me since it specifies the distribution "$X$" and the value "$x$". Is $F_n$ just an equivalent notation?

For the CDF my guess would be:

$$F_n=\begin{cases}0 & \text{for} & x <0 \\ ? & \text{for} & 0 \le x\le 1 \\ ? & \text{for} & x>1\end{cases} $$

I just don't know how to deal with the $n$ factor.

Answer 1

You can use the fact that $\min(X_i, i \in I) > t$ if and only if $X_i > t$ for all $i\in I$. That's why it is first suggested to find the CDF: $$ \begin{align} \Bbb P(n\cdot \min(X_i, i\leq n) \leq t) &= 1 -\Bbb P(n\cdot \min(X_i, i\leq n) > t) \\ &= 1 - \Bbb P\left(X_i > \frac tn, i\leq n\right) \\ &= 1 - \prod_{i = 1}^n \Bbb P\left(X_i> \frac tn\right) \end{align} $$ due to the independence of $X_i$. Just tell me if you can figure out the rest by yourself.