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Question from "Introduction to probability and mathematical statistics" 2nd edition by Bain and Engelhardt

Good evening everyone, I am attempting a problem on limiting distributions. The problem is as seen in the picture above. Could you please help me with part c) of this problem. The answer as per the solutions at the back is

$$ 0 \,\text{for}\, x\le 1 \\ 1-\frac{1}{y}\, \text{for} \,x\gt1$$

I dont quite understand why, I thought it would be $1 \, \text{for} \, x\gt 1$. Thank you!

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    $\begingroup$ would you like to include your working so that others don't have to repeat your work? $\endgroup$ Commented Aug 27, 2018 at 0:46
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    $\begingroup$ If this is a homework problem, it would need a self-study tag, or it risks being deleted due to site policy. $\endgroup$
    – Carl
    Commented Aug 27, 2018 at 1:36
  • $\begingroup$ This is not homework, I am just practicing! $\endgroup$
    – Destro
    Commented Aug 27, 2018 at 12:46
  • $\begingroup$ Please have a look here: stats.meta.stackexchange.com/a/3176. And type out the exam question in the picture. $\endgroup$
    – Jim
    Commented Sep 13, 2018 at 21:55

1 Answer 1

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From the specified distribution, for all $x \geqslant 1$ you have:

$$\begin{equation} \begin{aligned} F_{X_{1:n}}(x) \equiv \mathbb{P}(X_{1:n} \leqslant x) &= 1 - \mathbb{P}(X_{1:n} > x) \\[6pt] &= 1 - \mathbb{P}(\min \{ X_1,...,X_n \} > x) \\[6pt] &= 1 - \prod_{i=1}^n \mathbb{P}(X_i > x) \\[6pt] &= 1 - \prod_{i=1}^n x^{-1} \\[6pt] &= 1 - x^{-n}. \end{aligned} \end{equation}$$

Hence, for all $x \geqslant 1$ you have:

$$\begin{equation} \begin{aligned} F_{X_{1:n}^n}(x) \equiv \mathbb{P}(X_{1:n}^n \leqslant x) &= \mathbb{P}(X_{1:n} \leqslant x^{1/n}) \\[6pt] &= 1 - (x^{1/n})^{-n} \\[6pt] &= 1 - x^{-1}. \end{aligned} \end{equation}$$

This gives us back the original distribution of $X_i$. This distribution does not depend on $n$ so its limiting distribution is just the original distribution.

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