$U_1$, $U_2$, . . . are random variables, with $U_n\sim\text{uniform}
(0, n)$, and $$T_n= \frac{1}{\sqrt{U_n}}$$Consider the sequence $T_1$, $T_2$, . . . and give the pmf or pdf of
the limiting distribution.
We have
$$\begin{align*}
F_{T_n}(t)
&=\mathsf P\left(\frac{1}{\sqrt{U_n}} \leq t\right)\\\\
&=\mathsf P\left(\sqrt{U_n} \gt \frac{1}{t}\right)\\\\
&=1-\mathsf P\left(\sqrt{U_n}\leq\frac{1}{t}\right)\\\\
&=1-\mathsf P\left(U_n\leq\frac{1}{t^2}\right)\\\\
&=1-F_{U_n}\left(\frac{1}{t^2}\right)\\\\
&=1-\frac{1}{nt^2}
\end{align*}$$
Then
$$ F_{T_n}(t)=
\begin{cases}
1-\frac{1}{nt^2} & t \gt \frac{1}{\sqrt{n}} \\
0 & t \leq\frac{1}{\sqrt{n}} \\
\end{cases} $$
and so
$$ \lim_{n\rightarrow\infty}F_{T_n}(t)=
\begin{cases}
1 & t \gt \frac{1}{\sqrt{n}} \\
0 & t \leq\frac{1}{\sqrt{n}} \\
\end{cases} $$
which is the cdf of a degenerate random variable, $T$, for which $\mathsf P\left(T=\frac{1}{\sqrt{n}}\right)=1$
Thus, $T_1$, $T_2$, . . . converges in distribution to a degenerate random variable having pmf
$$f_T(t)=I_{\{\frac{1}{\sqrt{n}}\}}(t)$$
Is this a valid solution?
Best Answer
$\lim F_{T_{n}}$ cannot depend on $n$. Note that of $t >0$ then $t > \frac 1 {\sqrt n}$ for all $n$ sufficiently large. Hence $F_{T_{n}}(t)=1- \frac 1 {nt^{2}}$ for $n$ sufficiently large. Hence the limit is $1$. The correct limit is $1$ for $t \geq 0$ and $0$ for $t <0$.