Convergence in Probability of Positive Random Variables

Question

We have to show that for a positive sequence of random variables $X_n \geq 0$ and a random variable $X$ s.th. $X_n \overset{\mathbb{P}}{\rightarrow}X$ it follows that $\mathbb{P}(X \geq 0) = 1$. Additionally, let $\mathbb{P}(E) = 0$ be the set with probability of zero.

Here is my argument (i have the feeling it's not correct):

$$ X_n \overset{\mathbb{P}}{\rightarrow}X := lim_{n \rightarrow \infty} \mathbb{P}(|X_n - X| > \epsilon) = 0, \text{ for } \epsilon > 0. $$

This implies that there exists an $N \in \mathbb{N}$ such that $\forall n \geq N$:

$$ \{|X_n - X| > \epsilon\} = E $$

otherwise $\mathbb{P}(\{|X_n - X| > \epsilon\}) > 0$. Or considering the compliment: there exists an $N \in \mathbb{N}$ such that $\forall n \geq N$:

$$ \{|X_n - X| \leq \epsilon\} = E^c $$

thus in the set with positive probability. Thus we can rewrite the equation:

\begin{align*} |X_n - X| \leq \epsilon &\Leftrightarrow -\epsilon \leq X_n - X \leq \epsilon \\ &\Leftrightarrow -\epsilon - X_n \leq - X \leq \epsilon - X_n \\ &\Leftrightarrow \epsilon + X_n \geq X \geq X_n - \epsilon \\ \end{align*}

Thus $X \geq X_n - \epsilon > 0 \Rightarrow \mathbb{P}(X \geq 0) = 1$, because $X_n \geq 0$.

Is that a correct argument? I have the feeling that I used an argument for almost sure convergence.

Answer 1

You can also show that $P(X<0)=0$ to arrive at the conclusion. Let $\eta,\varepsilon>0$. Then $$\begin{aligned}\{X<-\eta\}&=(\{X<-\eta\}\cap \{X_n>\varepsilon\})\cup(\{X<-\eta\}\cap \{X_n\leq \varepsilon\})\subset\\ &\subset \{|X-X_n|>\varepsilon\}\cup\{|X_n-X| \geq \eta\}\end{aligned}$$ and $$P(X<-\eta)\leq P(|X-X_n|>\varepsilon)+P(|X_n-X| \geq \eta)\to 0$$ and finally $$P(X<-\eta)=0,\,\forall \eta\implies P(X<0)=0$$