Show that $X_n \overset{p}{\rightarrow} 0$ as $n \rightarrow \infty$

Question

Let $X_1,X_2 ,\ldots$ be random variables defined by the relations

$P(X_n = 0) = 1−\frac{1}{n}$, $P(X_n = 1) = \frac{1}{2n}$ and $P(X_n = −1) = \frac{1}{2n}$ , $n\ge 1$

Show that:

$X_n \overset{p}{\rightarrow} 0 \quad \text{ as } \quad n \rightarrow \infty$,

$X_n \overset{r}{\rightarrow} 0\quad \text{ as }\quad n \rightarrow \infty$ for any $r>0$

For the first one I did $$\lim_{n \to \infty}P(|X_n \ge \epsilon|)=\lim_{n \to \infty}1/2n=0$$ I used 1/2n as $X_n$ can't be as to be greater than 1.

But for the second one, I'm kind of stuck but I tried:

$$\lim_{n \to \infty}E(|X_n-0|^r)=\lim_{n \to \infty}E(X_n^r)=$$

Answer 1

The random variable $\left\lvert X_n\right\rvert$ takes the value one with probability $1/n$ and $0$ with probability $1-1/n$ hence it is also the case for the random variable $\left\lvert X_n\right\rvert^r$. We deduce that $\mathbb E\left\lvert X_n\right\rvert^r=1/n$.