Limiting distribution of $V_n=\sqrt{n}\cdot \text{min}{\{X_1, X_2, . . . , X_n}\}$ where $f_X(x) =\frac{2}{x^3}I_{(1,\infty)}(x)$

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

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

Let

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

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?

• 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.
– Did
Sep 25, 2018 at 21:20
• 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.
– Remy
Sep 25, 2018 at 21:24
• (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 $1$? Sep 27, 2018 at 0:03
• 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.
– Remy
Sep 27, 2018 at 0:07
• Yes, as Did pointed out in his/her comment. Sep 27, 2018 at 1:50

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