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Let $X_1, X_2, \ldots, X_n$ be independent random variables, where $X_k \sim Unif(0,k)$.

Let $Y_n = X_{(1)} =\min\{X_1, X_2, \ldots, X_n\}$

Show that $Y_n$ converges in distribution to $Y \sim exp(1)$ (the exponential distribution with mean 1).

I will let $F_k(y) = \mathbb{P}(X_k \leq y)$ be the CDF of $X_k$. Then, I know the following: $$F_{Y_n}(y) = \mathbb{P}(\text{At least one } X_k \leq y) = 1-\prod_{k = 1}^n(1-F_k(y)).$$

As far as I can tell, this can be rewritten as $$F_{Y_n}(y) = \cases{0~~\text{ if } y < 0\\\\ 1-\prod_{k =1}^n(1-y/k)~~\text{ if } 0 \leq y < 1\\\\1 ~~\text{ if } y \geq 1}$$

If $y \geq 1$, then $F_1(y) = 1$, which makes the product $\prod_{k = 1}^n(1-F_k(y)) = 0.$

This is where I have been stuck now. How can I proceed to show that $F_{Y_n}(y) \rightarrow Y \sim \text{exp}(1)$ in distribution?



ADDED As we have somewhat concluded via the comments and answers, it seems that perhaps each $X_k$ is mean to be independent AND identically distributed as Unif(0,n).

The text I pulled this question from is unpublished (course notes for some class I believe), so occasionally the notation is a bit ambiguous.

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    $\begingroup$ Er... it doesn't. Let $k=1$; clearly all values are smaller than 1. The distribution of the smallest, $X_{(1:n)}$, cannot converge to a number more than 2.7 times as large as the upper bound on the domain of the variable. Are you sure you transcribed everything correctly? And how does this not depend on $k$? $\endgroup$
    – Glen_b
    Mar 2, 2017 at 5:16
  • $\begingroup$ Sort of what I was thinking........the original statement speaks of $X_{(1)}$ converging to "exp(1)" in distribution. I guess it is possible that they mean a random variable having exponential distribution with rate parameter 1. $\endgroup$
    – Vladhagen
    Mar 2, 2017 at 5:22
  • $\begingroup$ No, $X_{(1:n)}$ doesn't converge to $\text{Exp}(1)$; its mean is $1$, how could that work? (unless $n=k$ I guess). However, $\frac{n}{k}X_{(1:n)}$ should. Oh, I should define my notation; that $(1:n)$ subscript means "the first order statistic of $n$"; that saves trouble when $n$ is changing. $\endgroup$
    – Glen_b
    Mar 2, 2017 at 5:22

1 Answer 1

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If you mean $X_1, \ldots, X_n \overset{iid}{\sim} \text{Uniform}(0,n)$, then $1-\prod_{k = 1}^n(1-F_k(y))$ turns into $$ 1-\prod_{k =1}^n(1-y/n) = 1 - (1-y/n)^n \to 1 - e^{-y}. $$ This is the CDF of an Exponential(1) random variable. Perhaps this is what you meant? They are independent, but they are also identical. And their distribution does depend on how big your sample size is.

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  • $\begingroup$ This seems to be the most likely scenario. The distribution is changing for each sampling I guess, depending on the sample size. $\endgroup$
    – Vladhagen
    Mar 2, 2017 at 6:02

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