# Convergence in Probability of Positive Random Variables

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.

• It's not clear what is the first $E$. Moreover, the $E$ you defined by $\{|X_n-X|>\varepsilon \}$ depend on $n$ (so, it would be better to write it as $E_n$). And I don't understand why the fact that $X_n\to X$ in probability implies that there is $N$ s.t. $\{|X_n-X|>\varepsilon \}=E$ for all $n\geq N$...
– Surb
Commented Apr 2, 2022 at 10:18
• Just to conclude, what you have done at the end is good but not so well written. You have that $|X_n-X|<\varepsilon \implies X>X_n-\varepsilon >-\varepsilon$, therefore $\forall \varepsilon >0$, $\mathbb P\{|X_n-X|\leq \varepsilon \}\leq \mathbb P\{X>-\varepsilon \}$. Since for all $\varepsilon >0$, $$\mathbb P\{|X_n-X|\leq \varepsilon \}\underset{n\to \infty }{\longrightarrow }1,$$ you get that $\mathbb P\{X>-\varepsilon \}=1$ for all $\varepsilon >0$, and thus $\mathbb P\{X\geq 0\}=1$ as wished.
– Surb
Commented Apr 2, 2022 at 10:23
• Much appreciated! Can you post this as an answer? I'd like to vote this answer as the solution. Commented Apr 2, 2022 at 11:05
• If $X_n \to X$ in probability, then $X_{n_k} \to X$ almost surely for some subsequence. The result follows. Commented Apr 4, 2022 at 20:06

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 $$P(X<-\eta)=0,\,\forall \eta\implies P(X<0)=0$$