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Say I have two random variables $X$ and $Y$ defined on a probability space ($\Omega,\mathbb{F},\mathbb{P}$).

To prove $X+Y$ is also a random variable, I need $\{\omega:(X+Y)(\omega)\le x\}\in \mathbb{F},\ \forall x\in \mathbb{R}.$

For positive random variables, I can simply write this set as an intersection of two sets, $\{\omega:X(\omega)\le x\}$ and $\{\omega:Y(\omega)\le x\}$, and invoke the property of the sigma algebra.

Now, how do I extend this to arbitrary-valued random variables? Or is there an intersection that is irrespective of the value of the RVs?

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1 Answer 1

up vote 1 down vote accepted

For all $x \in \mathbb{R}$ you have $$ \{\omega \in \Omega \::\: (X+Y)(\omega) < x \} = \bigcup_{q \in \mathbb{Q}} \{\omega \in \Omega \::\: X(\omega) < x-q, Y(\omega) < q \} $$ a countable union of sets of the form $$\{\omega \in \Omega \::\: X(\omega) < x-q\} \cap \{\omega \in \Omega \::\: Y(\omega) < q\} \in \mathbb{F}$$ and we conclude $\{\omega \in \Omega \::\: (X+Y)(\omega) < x \} \in \mathbb{F}$ for all $x \in \mathbb{R}$, which proves $X+Y$ to be a random variable.

Please note, that in general $$ \{\omega \in \Omega \::\: (X+Y)(\omega) \leq x \} \neq \{\omega \in \Omega \::\: X(\omega) \leq x\} \cap \{\omega \in \Omega \::\: Y(\omega) \leq x\}, $$ even if $X$ and $Y$ are assumed positive as there could be an $\omega \in \Omega$ with $X(\omega) = Y(\omega) = x >0$ and then $\omega$ would only be contained in the second one of these sets.

EDIT (suggested by Didier Piau):

Using the result above you also have $$\{\omega \in \Omega \::\: (X+Y)(\omega) \leq x \} = \bigcap_{n \in \mathbb{N}} \{\omega \in \Omega \::\: (X+Y)(\omega) < x +\frac{1}{n} \} \in \mathbb{F}.$$

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The OP seems to prefer $\{X+Y\leqslant x\}$ to $\{X+Y\lt x\}$. You might want to adapt your (excellent) solution to this setting. –  Did Feb 17 '12 at 11:23
I think it's clear that the RHS is contained in the LHS. For the other inclusion let $\omega \in \Omega$ with $(X+Y)(\omega) < x$, then you have $\varepsilon > 0$ with $(X+Y)(\omega) \leq x - \varepsilon$. Now you find a $q \in \mathbb{Q}$ with $q-\frac{\varepsilon}{2} \leq Y(\omega) \leq q$ (because $\mathbb{Q}$ is dense, as you suspected) and this shows $X(\omega) + q \leq (X+Y)(\omega) + \frac{\varepsilon}{2} < x $ which shows that $\omega$ is contained in the RHS. –  Matthias Klupsch Feb 17 '12 at 14:01

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