Say, we have a sequence of random variables with $X_n \geq 0$ almost everywhere. Which of the following types of convergence:

  1. almost everywhere: $X_n \xrightarrow{a.e.} X$
  2. in probability: $X_n \xrightarrow{\mathbb{P}} X$
  3. in distribution: $X_n \xrightarrow{d} X$

preserve this property? In other words, which gives us $P(X<0)=0$?

The first case is easy since the union of null sets is null set. But I have troubles proving (or disproving) the second and third.

Any hints are hugely appreciated.

Edit. My attempt:

For 2: I've tried to introduce $X_n$ to $\{X<0 \}$ in order to get something like $\{|X_n-X|>\varepsilon \}$ yet to no avail.

For 3: $P(X<0)=0$ seems to be equivalent with $\forall \varepsilon>0: P(X \leq -\varepsilon)=0$. From $X_n \xrightarrow{d} X$ and if we assume that $F$ is continuous in $-\varepsilon$ (why can we?), we'd get $F_n(-\varepsilon) \xrightarrow{n\to \infty} F(-\varepsilon)$ where $F_n(-\varepsilon)$ are zero for all $n$.

  • 3
    $\begingroup$ All of them. What did you try? $\endgroup$ – Did Feb 16 '14 at 17:03


  1. Convergence in probability: Note that $$\begin{align*} \mathbb{P}(X<- \varepsilon) &= \mathbb{P}(X < - \varepsilon, |X_n-X|>\varepsilon/2) + \mathbb{P}(X<-\varepsilon,|X_n-X| \leq \varepsilon/2) \\ &\leq \mathbb{P}(|X_n-X|>\varepsilon/2)+ \mathbb{P}(X_n<-\varepsilon/2) \end{align*}$$ for all $\varepsilon>0$.
  2. Convergence in distribution: Recall that the distribution function $F(x) := \mathbb{P}(X \leq x)$ is a monotone function. This implies in particular that $F$ has at most countable many discontinuities (see this question for a proof). Consequently, there exists a dense subset $C$ in $(-\infty,0)$ such that $F$ is continuous at any $x \in C$. Now you can apply your argumentation.
  • $\begingroup$ Great hints, many thanks! $\endgroup$ – Leo Mar 9 '14 at 19:11

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