$X_1$, $X_2$, . . . are iid random variables having pdf

$$f_X(x) =\frac{2}{x^3}I_{(1,\infty)}(x)$$


$$V_n=\sqrt{n}\cdot \text{min}{\{X_1, X_2, . . . , X_n}\}$$

Consider the sequence $V_1$, $V_2$, . . . and give the pmf or pdf of the limiting distribution.

I first note that the cdf of $X$ is given by

$$ F_{X}(x)= \begin{cases} 1-\frac{1}{x^2} & x \gt 1 \\ 0 & x\leq 1 \\ \end{cases} $$

We have

$$\begin{align*} F_{V_n}(v) &=\mathsf P(V_n\leq v)\\\\ &=\mathsf P(\sqrt{n}\cdot \text{min}{\{X_1, X_2, . . . , X_n}\}\leq v)\\\\ &=\mathsf P\left(\text{min}{\{X_1, X_2, . . . , X_n}\}\leq \frac{v}{\sqrt{n}}\right)\\\\ &=1-\mathsf P\left(X_1\gt \frac{v}{\sqrt{n}}\right)^n\\\\ &=1-\left(1-\mathsf P\left(X_1 \leq \frac{v}{\sqrt{n}}\right)\right)^n\\\\ &=1-\left(1-\left(1-\frac{1}{\left(\frac{v}{\sqrt{n}}\right)^2}\right)\right)^n\\\\ &=1-\left(\frac{n}{v^2}\right)^n \end{align*}$$

Altogether, we have

$$ F_{V_n}(t)= \begin{cases} 1-\left(\frac{n}{v^2}\right)^n & v\gt \sqrt{n} \\ 0 & v\leq \sqrt{n} \\ \end{cases} $$

and so for all $v\in\mathbb{R}$

$$\lim_{n\rightarrow\infty} F_{V_n}(t)=1$$

Since there is not a valid cdf equal to $1$ except at points of discontinuity, a limiting distribution does not exist.

Is this a valid solution?

  • 1
    $\begingroup$ Sure about the text of the exercise? Since $X_i\geqslant1$ almost suerly, for every $i$, $V_n\geqslant\sqrt n$ almost surely, for every $n$, hence, with no computations, one knows that $V_n\to\infty$ almost surely. $\endgroup$
    – Did
    Sep 25, 2018 at 21:20
  • $\begingroup$ Yes, I'm sure. Some of the solutions don't have limiting distributions so this seems to be one of them. I'm just not sure what good notation would be for the cdf if I did include it. $\endgroup$
    – Remy
    Sep 25, 2018 at 21:24
  • $\begingroup$ (1) The notation $F_{V_n}(v)$ suddenly changed to $F_{T_n}(t)$. (2) For each fixed $v$, $v \leq \sqrt{n}$ holds for all sufficiently large $n$ and we know that this implies $F_{V_n}(v) = 0$. This also matches Did's comment that $V_n \geq \sqrt{n}$ almost surely. What is your rationale for concluding that the limit is $1$? $\endgroup$ Sep 27, 2018 at 0:03
  • $\begingroup$ Whoops, my mistake. So essentially, my thinking was we would expect the minimum to get infinitely close to $1$, and so $v\rightarrow\sqrt{n}$ and so $\frac{n}{v^2}\rightarrow 1$. Wasn't sure how to take the limit from there since we have $\lim_{n\rightarrow\infty} 1-\left(\frac{n}{v^2}\right)^n$. Here, it looks like we have convergence in probability inside a limit. $\endgroup$
    – Remy
    Sep 27, 2018 at 0:07
  • 1
    $\begingroup$ Yes, as Did pointed out in his/her comment. $\endgroup$ Sep 27, 2018 at 1:50

1 Answer 1



$F_{T_n}(v)= \begin{cases} 1-\left(\frac{n}{v^2}\right)^n & v\gt \sqrt{n} \\ 0 & v\leq \sqrt{n} \\ \end{cases}$

Logically speaking, the right way to understand $F_{T_n}$ is the following: $ \forall n, \ \exists v, \ s.t. \ c v > \sqrt{n} \Rightarrow F_{T_n}(t) = 1-\left(\frac{n}{v^2}\right)^n $

For a fixed $n \ $, if the $v$ in your hands can't make the inequality hold: $\frac{n}{v^2} < 1$, then you just assign a $0$ to $F_{T_n}(v)$. In this way, $F_{T_n}$ never goes beyond $1$ because that's how you assign the probabilities (how you design this distribution)

  • $\begingroup$ I have edited my attempted solution. $\endgroup$
    – Remy
    Sep 26, 2018 at 23:44

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