Let $\{X_n\}_{n=1}^\infty$ is a succession of indepedent random variables, such that for all $n\geq 1$, $\mathbb E(X_n) =0$ and $\mathbb E(|X_n|) = 1$,

Prove or disprove that $\mathbb P(\lim \inf_{n} X_n < 0) > 0.$

I tried to handle it like this

Let us consider the succession of events $A_n = \{X_n < 0\}$. Since $(X_n)_n$ are independent then $(A_n)_n$ and $A^c_n = \{X_n \geq 0\}$ are also indepedent events. We have \begin{align*} \mathbb P(\lim \inf_{n} X_n < 0) &= \mathbb P(\lim \inf_{n} A_n)\\ & = \mathbb P(A_n \, \text{ e.v.})\\ &= 1 - \mathbb P\big((A_n \, \text{ e.v.})^c\big)\\ &= 1 - \mathbb P(A_n^c \, \text{ i.o.}\big). \end{align*} On the other hand, we have \begin{align*} \mathbb P(A_n^c) &= \mathbb P(X_n \geq 0)\\ & = ... \end{align*}

Here some recall of the notations used and my intention is to use the second BorelCantelli lemma that also I recall it here

First let's recall some definitions. Let $(A_n)_n$ be a sequence of events, we define \begin{align*} A_{n} \text{ infinitely often (i.o.) } &\equiv\left\{\omega: \omega \text { is in infinitely many } A_{n}\right\}\equiv \limsup _{n} A_{n} \equiv \bigcap_{m}^{\infty} \bigcup_{n=m}^{\infty} A_{n} \end{align*}

Note that $$ \mathbb {I}_{A_{n} \,i.o. }=\lim_{n} \sup \mathbb{I}_{A_{n}} $$ Similarly, \begin{align*} A_{n}\text{ eventually (e.v.) } \equiv\left\{\omega: \omega \text { is in } A_{n} \text { for all large } n\right\} \equiv \liminf _{n} A_{n} \equiv \bigcup_{m} \bigcap_{n=m}^{\infty} A_{n}. \end{align*} Note that $$ \mathbb{I}_{A_{n} \,e.v.} =\liminf _{n} \mathbb{I}_{A_{n}} $$ Also we have $\left(A_{n} \text { e.v.}\right)^{c}=\left(A_{n}^{c} \text { i.o. }\right)$. Moreover recall the second Borel-Cantelli Lemma:

If the events $(A_n)_n$ are independent, then $\sum_{n} \mathbb{P}(A_{n})=\infty$ implies $\mathbb{P}(A_{n} \text{ i.o.})=1$

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    $\begingroup$ What have you tried? $\endgroup$ Mar 24, 2020 at 11:00
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    $\begingroup$ Please look to my update $\endgroup$ Mar 24, 2020 at 11:02
  • $\begingroup$ I am not familiar with the notation $e.v.$, what do you mean by it? $\endgroup$ Mar 24, 2020 at 11:14
  • $\begingroup$ ah sorry e.v. to note eventually and i.o. infinitely often. I will update with a recall of Second Borela-Cantelli Lemma $\endgroup$ Mar 24, 2020 at 11:16
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    $\begingroup$ I am familiar with the notation 'i.o', but not with 'e.v'. $\endgroup$ Mar 24, 2020 at 11:17

1 Answer 1


Design $X_n \in \{-b_n, 0, b_n\}$ for $n \in \{1, 2 ,3, ...\}$ to satisfy

(i) $X_n \neq 0$ finitely often (with prob 1);

(ii) $E[X_n]=0$ for all $n$;

(iii) $E[|X_n|]=1$ for all $n$.

  • $\begingroup$ I am wondering how $\mathbb E[|X_n|] = 1$, I think it's $2/3$ since there is equiprobability between the realizations of $X_n$, meaning $\mathbb P[X_n=-b_n] = \mathbb P[X_n=0] = \mathbb P[X_n = b_n] = 1/3$. In this case $E[|X_n]| = 2/3$. $\endgroup$ Mar 25, 2020 at 9:12
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    $\begingroup$ Who said $P[X_n=-b_n]=P[X_n=0] = P[X_n=b_n]$? Surely not! You can never get $P[\{X_n \neq 0\} f.o.]=1$ that way. I mentioned "Design $X_n$" and so there are a few things left to design (such as the PMF). (And the specific $b_n$ values). I think the structure $X_n \in \{-b_n, 0, b_n\}$ should be suggestive enough to complete the steps. $\endgroup$
    – Michael
    Mar 25, 2020 at 16:04
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    $\begingroup$ Anyway, even if $P[X_n=-b_n]=P[X_n=0]=P[X_n=b_n]=1/3$, you cannot claim $E[|X_n|]=2/3$ without specifying the value of $b_n$. You would only get $2/3$ in the special case when $b_n=1$ for all $n$. $\endgroup$
    – Michael
    Mar 25, 2020 at 16:10
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    $\begingroup$ In your terminology, you want $$P[\{X_n=0\} \: \mbox{eventually}] = 1$$ where I believe you are using "$\{X_n=0\}$ eventually" to mean that "$X_n=0$ for all but a finite number of indices $n \in\{1, 2, 3, ...\}$." (This is equivalent to my condition (i) above). $\endgroup$
    – Michael
    Mar 27, 2020 at 17:16

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