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Let $X$ be a r.v. on a given probability space and let $a \in \mathbb{R}$. Show that $aX$ is a r.v.

I need to check if my proof is correct.

Proof:

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

So $aX$ will be the random function such that $aX : \Omega \rightarrow \mathbb{R}$ with the property that $\{\omega \in \Omega : aX(\omega) \leq x\} \in \mathcal{F}$ for each $x \in \mathbb{R}$.

Updated Proof

Let $Y = aX$ be a random variable. Then $Y : \Omega \rightarrow \mathbb{R}$ such that $\{\omega \in \Omega : Y(\omega) \leq x \} \in \mathcal{F}$. This is implies that $\{\omega \in \Omega : aX(\omega) \leq x \} \in \mathcal{F}$ which implies that $\{\omega \in \Omega : X(\omega) \leq \frac{x}{a} \} \in \mathcal{F}$. Therefore $aX$ is a random variable.

Corrected Proof

Let $Y = aX$. Then $Y : \Omega \rightarrow \mathbb{R}$ such that $\{\omega \in \Omega : Y(\omega) \leq x \}$. This is implies that $\{\omega \in \Omega : aX(\omega) \leq x \}$ which implies that $\{\omega \in \Omega : X(\omega) \leq \frac{x}{a} \} \in \mathcal{F}$ since $\frac{x}{a} \in \mathbb{R}$. Therefore $aX$ is a random variable.

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    It's not clear you've established anything in the second part about aX. You need to show $aX$ obeys the definition. Maybe write $Y=aX,$ then write the definition in terms of Y and then show you can get from the fact that $X$ obeys the definition to the fact that Y does – 2017-01-12
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    @spaceisdarkgreen please can you check my new proof. – 2017-01-18
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    You are more on the right track now, but you still have it a little bit backwards. You start by assuming $Y=aX$ is a random variable, but this is what you need to prove. You go on to show this means that $X$ is a random variable. So you have shown that if $Y$ is a random variable and $Y=aX$ then $X$ is a random variable. This is actually true and your proof is correct, but backwards from what you intended. The forwards proof is almost exactly the same. I posted a proof below. I think if you try again and compare you'll get it. – 2017-01-18

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We know that $X$ is a random variable. This means that for any $x\in\mathbb{R}$ the set $\{\omega\in\Omega\mid X(\omega) < x\}\in \mathcal{F}.$

Let $Y = aX.$ To establish that $Y$ is an RV we must show that for any $y\in\mathbb{R}$ the set $\{\omega\in\Omega\mid Y(\omega) < y\}\in \mathcal{F}.$

Since $Y(\omega) = aX(\omega),$ we can write $$\{\omega\in\Omega\mid Y(\omega) < y\} = \{\omega\in\Omega\mid aX(\omega) < y\} = \{\omega\in\Omega\mid X(\omega) < y/a\} .$$

Since $y/a\in\mathbb{R},$ by the definition of "$X$ is a random variable," $$ \{\omega\in\Omega\mid X(\omega) < y/a\}\in \mathcal{F}.$$

So we have shown that $$ \{\omega\in\Omega\mid Y(\omega) < y\}\in \mathcal{F}.$$

This means $Y$ is a random variable.

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    Cool! So the errors in my proof were that in the first sentence I should not say that $Y$ is a random variable. In the second sentence, I should not claim that set is $\mathcal{F}-$ measurable. I just need to add that since $\frac{x}{a}$ is $\mathbb{R}$ then that satisfies that $aX$ is a random variable. – 2017-01-18
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    @Mika'il. Yes, you need to *show* the set is $\mathcal{F}$-measurable. Looking at your corrected proof, it's unclear what the second sentence means (You say then '$Y$ is a map such that (a set)' A set is not a statement. If you meant to say the set is measureable, that's wrong, since we can't assume it. It's what we're trying to prove.) You need to start with the fact that for any $x$, $\{\omega \in \Omega \mid X(\omega) < x\}$ is $\mathcal{F}$-measurable, and then prove that the corresponding set for $Y,$ $\{\omega \in \Omega \mid Y(\omega) < y\}$ is $\mathcal{F}$-measurable for any $y.$ – 2017-01-18