$X_1$, $X_2$, . . . 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?
Best Answer
HINT:
$F_{T_n}(v)= \begin{cases} 1-\left(\frac{n}{v^2}\right)^n & v\gt \sqrt{n} \\ 0 & v\leq \sqrt{n} \\ \end{cases}$
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)