Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show: If $X$ and $Y$ are random variables on a probability space $(\Omega, F, \mathbb P)$, then so is $X+Y$.

The definition of a random variable is a function $X: \Omega \to \mathbb R$ with the property that $\{\omega\in\Omega: X(\omega)\leq x\}\in F$ for each $x\in\mathbb R$.

and further more, how to approach $X+Y$ and $\min\{X,Y\}$?

share|improve this question
    
This was asked (and answered) pretty recently on the site. –  Did Feb 17 '13 at 7:34
add comment

2 Answers

up vote 3 down vote accepted

There are a number of ways to do it. A standard trick for proving things like this is by noticing that $\{X + Y < x\} = \displaystyle \bigcup_{r \in \mathbb{Q}} \{X < r\} \cap \{Y < x - r\} $, and that showing that this is in $F$ is enough to show that $X + Y$ is measurable. Then use the properties of $X$, $Y$, and $\sigma-$algebras to deduce that this set is measurable.

share|improve this answer
    
Why it is sufficient to union through the rationals? –  JFK Feb 17 '13 at 5:54
1  
Obvioulsy whenever $X(\omega) < r$ and $Y(\omega) < x-r$, $X(\omega)+Y(\omega) < x$. On the other hand, if $X(\omega)+Y(\omega) = z < x$, take some rational $r$ with $X(\omega) < r < X(\omega) + x - z$, and you have $Y(\omega) = z - X(\omega) < x - r$. –  Robert Israel Feb 17 '13 at 6:28
add comment

In fact, a random variable is a measurable function from $\Omega$ to $\mathbb{R}$. $$\{X+Y>x\}=\{X>x-Y\}=\bigcup_{q\in \mathbb{Q}}\{X>q>x-Y\}=\bigcup_{q\in \mathbb{Q}}(\{X>q\ \}\bigcap\{Y>x-q\})=\bigcup_{q\in \mathbb{Q}}(\{X\le q\ \}^c\bigcap\{Y\le x-q\}^c) $$ Since $ \{X\le q\ \}^c\in \mathcal{F}\ ,\ \{Y\le x-q\}^c\in \mathcal{F} $, $\mathbb{Q}$ is countable and $\mathcal{F}$ is a $\sigma-field$, we obtain $\{X+Y>x\}=\{X+Y\le x\}^c\in \mathcal{F}$. Then $$ \{\omega:X(\omega)+Y(\omega)\le x\}\in \mathcal{F}$$ $$ \{\omega:min\{X(\omega),Y(\omega)\}\le x\}=\{\omega:X(\omega)\le x\}\bigcup\{\omega:Y(\omega)\le x\}\in \mathcal{F}$$ By definition, $X+Y,and\ min\{X,Y\}$ are both random variables.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.