# Limiting distribution of first order statistic x

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!

• would you like to include your working so that others don't have to repeat your work? Aug 27 '18 at 0:46
• If this is a homework problem, it would need a self-study tag, or it risks being deleted due to site policy.
– Carl
Aug 27 '18 at 1:36
• This is not homework, I am just practicing! Aug 27 '18 at 12:46
• Please have a look here: stats.meta.stackexchange.com/a/3176. And type out the exam question in the picture.
– Jim
Sep 13 '18 at 21:55

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