Limiting Distribution of $T_n= \frac{1}{\sqrt{U_n}}$ where $U_n\sim\text{uniform} (0, n)$

Question

$U_1$, $U_2$, . . . 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?

Answer 1

$\lim F_{T_{n}}$ 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 $t <0$.

Answer 2

Almost. Your last 2 or 3 equations have the value of the limit as $n\to\infty$ depending on $n$. On the left hand side of your $=$ signs $n$ is a bound variable; your equations make it seem like you want $n$ to be a free variable on the right hand side of the $=$ signs.

What you want to say is that the limit distribution is the degenerate distribution that concentrates all its mass at $0$, so its cdf is $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$. This cdf does not have a density function. You need check the convergences at continuity points of $F$, that is, for all $x\ne0$.