# A “random variable” with infinite value

A random variable (r.v.) is a (measurable) fucntion from probability space $\Omega$ to R. In our applied problem, the best model would be an extended "r.v." from $\Omega$ to $R\cup{-\infty}$. For such "r.v." the cumulative distribution function can be defined naturally, it will be a right-continuous nondecreasing function with $F(\infty)=1$ but with $F(-\infty)$ not nessesary $0$. Expectation does not exists if $P(X=-\infty) > 0$ but conditional expectation $E[X | X > -\infty]$ makes sence. Next step is to define different types of convergence (in probability, in distribution, etc.). The question is: Is this or similar "probability theory" known? Maybe, it can be derived as a corollary of some more general theory, etc.? I would be happy to develop it myself, but afraid to "reopen" a well-known theory.

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Expectation can still exist if P(X = -∞) > 0, it just won't be finite. –  Ricky Demer Feb 15 '11 at 10:48

The definition you gave is a special case of the definition I'm used to, where the target space is an arbitrary measurable space or topological space instead of the real line. Your extended version fits just fine into that framework. See: http://en.wikipedia.org/wiki/Random_variable#Formal_definition

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While you are at it, you could allow the value $+\infty$ as well...

The resulting object is a random variable from a probability space $(\Omega,\mathcal{F},P)$ to a bona fide measurable space $(E,\mathcal{E})$, in this case $E=\mathbb{R}\cup\{-\infty,+\infty\}$ and $\mathcal{E}$ the Borel $\sigma$-algebra of $E$ endowed with its usual topology (roughly speaking, this is equivalent to seeing $E$ as the closed interval $[0,1]$ endowed with its subspace topology). A description of $\mathcal{E}$ is that $A\subset E$ is in $\mathcal{E}$ iff there exists $B\in\mathcal{B}(\mathbb{R})$ and $I\subset\{-\infty,+\infty\}$ such that $A=B\cup I$.

So you are squarely in probability theory in the most standard sense of the term.

Rereading your post, I feel I should mention that $E(X|X\in\mathbb{R})$ need not be well defined, for the same reason that a real valued random variable need not be integrable.

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Thank you for answer. I would prefer to work with $R\cup\{-\infty}$ only - this is also a measureable space. But I would not say that I am in standard probability theory - most books that I know works with real-valued r.v. Can you recommend a good book where all the standard probability results are proved in this more general settings? For example, here en.wikipedia.org/wiki/Convergence_of_random_variables there are many theorems (like Portmanteau lemma) about convergence of r.v.s - are they valid for r.v. from $\Omega$ to E? –  Bogdan Feb 15 '11 at 13:02
@Bogdan: Take your favorite probability textbook and replace every random variable $X$ with values in $\mathbb{R}\cup\{-\infty,+\infty\}$ you are interested in, by $Y=\arctan(X)$. Then $Y$ takes values in $[-\pi/2,+\pi/2]$, hence you are back to a setting which is even less general than the one needed for real valued random variables, and the convergence theorems of your textbook hold for your $Y$ random variables. –  Did Feb 15 '11 at 13:37
For example, sequence of constant random variables $X_n=-n$ does not converge in Probability to constant $X=-\infty$ (at least with definition in wikipedia),but $Y_n=arctan(X_n)$ converges in probability to $Y=arctan(X)=-\pi/2$. So, this trick already does not help with all theorems involving convergence in probability. Continuous mapping theorem states that $X_n -> X$ implies $g(X_n) -> g(X)$ for cont. function g(.) on metric space, but it is not applicable because $R\cup{-\infty}$ is not a metric space. So, many theorems stop working! Rather than check them one by one, I would want a book.. –  Bogdan Feb 15 '11 at 14:53
@Bogdan: OK, let me expand on my previous answer. Consider the distance $d$ on $E=\mathbb{R}\cup\\{-\infty,+\infty\\}$ defined by $d(x,y)=|\arctan(x)-\arctan(y)|$ if $x$ and $y$ are both in $\mathbb{R}$ and extended in the obvious way to $E\times E$ (or, more generally, $d(x,y)=|\delta(x)-\delta(y)|$ for any bounded increasing function $\delta:\mathbb{R}\to\mathbb{R}$). Then $(E,d)$ is a separable metric space hence (even according to the wikipedia page you refer to), the convergence of $(X_n)$ to $X$ in probability is defined by the condition that $P(d(X_n,X)\ge u) \to 0$ for every .../... –  Did Feb 15 '11 at 16:58
.../... positive $u$. (Hence yes, $X_n=-n$ does converge, almost surely and in probability and in distribution, to $X=-\infty$.) The continuous mapping theorem is still valid, naturally, if the continuity hypothesis refers to the metric $d$, as it should. And so on. As far as convergences in probability, in distribution or almost sure are concerned, everything works fine. But expectations and convergences in $L^p$ are another matter, of course. Books about probability measures on metric spaces would explain this, for example Billingsley's classic Convergence of Probability Measures. –  Did Feb 15 '11 at 16:59