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.
Best Answer
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.