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!

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.
