# 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!

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 $$X \sim \frac{1}{2}(\delta_{-1} + \delta_{1})$$.

• Thank you for your help Commented Nov 11, 2021 at 11:33

This is definitely a fiddly question. If you don't want full answers just read this plan of attack:

1. Calculate the CDF of $$X_n$$ by working out $$\mathbb{P}[X_n < x]$$
2. Differentiate the CDF to reach the PDF
3. 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.