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I am thinking whether it is possible to have a random variable $X$ defined on the probability space $(\Omega, \mathscr{F}, \mathbb{P})$ such that $\mathbb{P}(|X|>N)=1$ as $N\to +\infty$. To me, Borel measurability of $X$ alone does not seem to prevent this from being true. The definition of $X$ also allows it to take on values of $\pm \infty$. Please correct me if I am wrong, or give an example if this can be true.

Thank you.

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3 Answers 3

up vote 5 down vote accepted

The confusion, if any, stems from not specifying the target space $E$ of the arrow $X:\Omega\to E$.

If $E$ is the real line, obviously $P(|X|\ge N)\to0$ when $N\to+\infty$, but if $E$ is the extended real ine, this need not be true anymore since $P(X=+\infty)$ and $P(X=-\infty)$ may be any nonnegative real numbers whose sum does not exceed $1$. In the second case, the sum $P(X=+\infty)+P(X=-\infty)$ might even be $1$, this would happen if $P(|X| \mbox{finite})=0$ and $P(|X|=+\infty)=1$.

Both topological spaces $E$ are endowed with a natural Borel sigma-algebra. Some people call random variable any random variable with values in the (non extended) real line, some consider the extended real line, and, although I prefer one choice over the other, both seem legitimate to me.

A common situation where the extended real line enters the picture is when a sequence $(\xi_t)$ of random variables indexed by $t\in T$ is given, for example with $T=\mathbb{N}$, and one defines $X$ as the smallest index $t$ such that an event like $[\xi_t\in B]$ happens, for a given measurable subset $B$ of the common target space of the random variables $\xi_t$. Then the natural target space of $X$ is the extended line because $X=+\infty$ on the event $[\forall t\in T, \xi_t\notin B]$, and it may well happen that $P(X=+\infty)>0$.

However, I can think of no natural situation where a random variable $X$ such that $P(|X|=+\infty)=1$ is involved.

The relevant wikipedia paragraph explains this concisely and competently.

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thanks for your answer. Is there any distinction between $X(\omega)=+\infty$ and $X$ undefined? Is the former a limiting process or simply a designation, given the target space $E$ chosen to be the extended real line? –  Qiang Li Jun 5 '11 at 16:30
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If ${\rm P}(|X| > N) = 1$, then $1-{\rm P}(|X| > N) = 0$; that is, ${\rm P}(|X| \leq N) = 0$. But the distribution function of $|X|$ necessarily tends to $1$ as $N \to \infty$, that is ${\rm P}(|X| \leq N) \to 1$ as $N \to \infty$, a contradiction. While a random variable is often allowed to take the value $+\infty$ or $-\infty$ with probability $0$, it is not allowed to take these values with positive probability.

EDIT: I considered real-valued random variables, but in general random variables need not be real-valued, hence Jonas' answer may be good as well.

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thank you for your answer. How do you make it a rigorous argument that when the r.v. is real-valued, it would be allowed to take the value $\pm \infty$ with probability 0? –  Qiang Li Jun 5 '11 at 16:32
If $X$ is a (real-valued) random variable and $A$ a set with ${\rm P}(A)=0$, you can define $\tilde X$ by $\tilde X(\omega) = X(\omega)$ if $\omega \notin A$, $\tilde X(\omega)=\infty$ if $\omega \in A$. This will not affect measurability, expectations (with respect to ${\rm P}$), or anything essential. This is just as in measure theory, where modifying a function $f$ on a set of $\mu$-measure $0$, does not affect measurability, integrals (with respect to $\mu$), or anything essential. In this context, recall that ${\rm E}(X) = \int_\Omega{Xd{\rm P}}$ (which corresponds to $\int_X {fd\mu }$). –  Shai Covo Jun 6 '11 at 6:34
@Qiang Li: Further, see Section 1.3 (Random Variables) on p. 13 in web.mit.edu/hmsallum/www/GradSchool/sc505notes.pdf, as well as the related thread math.stackexchange.com/questions/39610/… –  Shai Covo Jun 6 '11 at 6:46
+1 and thank you. –  Qiang Li Jun 8 '11 at 14:36
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Of course this is true if $\mathbb P(X=\pm\infty)=1$. That is the only case. When you have a decreasing sequence of events, the limit of the probabilities is the probability of the limit. Therefore $\displaystyle{\lim_{N\to\infty}\mathbb{P}(|X|>N)=\mathbb{P}(X=\pm\infty)}$.

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