Convergence in Distribution for uniformly distributed RVs So I was wondering if someone could help me understand the following exercise a little bit more, cause I'm currently very confused.
Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables on the $\mathbb{P}$-Space $([-1,1], \mathcal{B}[-1,1], \mathcal{U}[-1,1])$, where $\mathcal{U}[-1,1]$ is the uniform distribution. Calculate the PDF and CDF of $X_n$ when:
$X_n(t) = \left\{\begin{array}{1l} t^{\frac{1}{n}}, & t>0 \\
         -\vert t \vert^{\frac{1}{n}}, & t \leq 0 \end{array}\right. .$
and show that $X_n \xrightarrow{\mathcal{D}}X$ with $X \text{~} \frac{1}{2} \delta_{-1} + \frac{1}{2} \delta_{1}$.
So my first question is if I'm understanding the first parts correctly about calculating the PDF because I thought for all $n$ the PDF is given by $\frac{1}{2},\ -1 \leq t \leq 1$ and $0$ otherwise but with this approach there is no $n$ dependency anywhere. Thank you for help!
 A: I will write $\mathbf{P} = \mathcal{U}[-1,1]$ for the probability measure on $[-1, 1]$ given by the uniform distribution. Then the CDF of $X_n$ is the function $F_{X_n}$ defined by
$$ F_{X_n}(x) = \mathbf{P}(X_n \leq x) = \mathbf{P}(\{t \in [-1, 1] : X_n(t) \leq x \}) $$
Now noting that $t \mapsto X_n(t) = \operatorname{sgn}(t) |t|^{1/n}$ is strictly increasing, it follows that
$$ X_n(t) \leq x
\quad \Leftrightarrow \quad
t \leq X_n^{-1}(x) = \operatorname{sgn}(x)|x|^n. $$

So
$$ F_{X_n}(x)
= \mathbf{P}(t \leq \operatorname{sgn}(x)|x|^n)
= \begin{cases}
1, & \text{if $x \geq 1$} \\
\frac{1+\operatorname{sgn}(x)|x|^n}{2}, & \text{if $-1 \leq x < 1$} \\
0, & \text{if $x < -1$}
\end{cases} $$
Differentiating the CDF $F_{X_n}$ then gives the PDF
$$ f_{X_n}(x) = \frac{\mathrm{d}}{\mathrm{d}x} F_{X_n}(x) = \begin{cases}
\frac{n}{2}|x|^{n-1}, & \text{if $|x| < 1$} \\
0, & \text{if $|x| \geq 1$}
\end{cases} $$
Finally, letting $n \to \infty$,
$$ \lim_{n\to\infty} F_{X_n}(x) = \begin{cases}
1, & \text{if $x \geq 1$} \\
\frac{1}{2}, & \text{if $-1 < x < 1$} \\
0, & \text{if $x \leq -1$}
\end{cases} $$
This coincides with the CDF of $\frac{1}{2}(\delta_{-1} + \delta_{1})$ on a dense subset of $\mathbb{R}$. (In fact, they coincide on all of $\mathbb{R}\setminus\{-1\}$.) Therefore the desired claim follows.
Alternatively, note that
$$ \lim_{n\to\infty} X_n(t) = \operatorname{sgn}(t) =: X(t) $$
for any $t \in [-1, 1]$. Since $X_n$ converges to $X$ everywhere, $X_n$ converges in distribution to $X$. Now it is not hard to check that $X \sim \frac{1}{2}(\delta_{-1} + \delta_{1})$.
A: This is definitely a fiddly question. If you don't want full answers just read this plan of attack:

*

*Calculate the CDF of $X_n$ by working out $\mathbb{P}[X_n < x]$

*Differentiate the CDF to reach the PDF

*To show $X_n \to X$ weakly (in distribution) you just need the CDFs to converge at continuity points, hence work out the CDF of $X$ first.

Part 1
For $x > 0$ $\mathbb{P}[X_n < x] = \frac{1}{2} + \frac{x^n}{2}$
For $x < 0 $ $\mathbb{P}[X_n < x] = \frac{1-(-x)^n}{2}$
Part 2
I will leave the differentiating that yields the PDF down to you!
Part 3
The CDF of $X$ should clearly be $1$ at $1$, and $\frac{1}{2}$ everywhere else. Remember you only need to check the limits at continuity points.
