# What does it mean to multiply 2 random variables that have different sample space?

Here's a problem from Harvard's Stats 110 class:

Let $Z \sim N(0,1)$ and let $S$ be a "random sign" independent of $Z$, i.e., $S$ is $1$ with probability $1/2$ and $-1$ with probability $1/2$. Show that $SZ \sim N(0,1)$.

Here is the proof, given by professor Blitzstein:

Condition on $S$ to find the CDF of $SZ$: \begin{align*} P(SZ \leq x) &= P(SZ \leq x | S = 1)P(S=1) + P(SZ \leq x | S = (-1))P(S=-1) \\ &= P(Z \leq x)\frac{1}{2} + P(Z \geq -x)\frac{1}{2} \\ &= P(Z \leq x)\frac{1}{2} + P(Z \leq x)\frac{1}{2} \\ &= \Phi(x) \end{align*}

I'm having a hard time understanding what it means to multiply two random variables (here $S$ and $Z$) that are from different sample spaces. Since RVs are functions, then isn't the only way to multiply them that they share the same domain?

At first I thought that I could define $S$ on the same domain as $Z$, having half of the outcomes map to $1$ and the other half map to $-1$, but doing this, its possible to get $SZ$ not to be normal (for eg. map all outcomes that map to the positive axis on $S$ to $1$, and all the outcomes that map to the negative axis on $S$ to $-1$. Then $SZ \geq 0$.

I guess the problem in having $S$ defined on the same space as $Z$ is that they no longer are independent, right? But then how can you multiply them? Do we need to redefine both random variables so that they are defined on the cross product of the spaces?

They are jointly defined on a sample space that is the Cartesian product of their respective sample spaces. In this particular case, they are jointly defined on the product $\{-1, 1\} \times \mathcal{R}$.
Independence means that the PDF of the event $(s, \_) \in \{-1, 1\} \times \mathcal{R}$, where $\_$ means "anything", is independent of the outcome of the second variable (and likewise for the second element).
• Independence follows quite naturally if you take the Cartesian product yes, but my question was if you define $S$ on the same space as $Z$, so that half (arbitrarily chosen) of the outcomes in the space of outcomes of $Z$ are mapped to $1$, and the other half to $-1$. – samlaf Jul 18 '17 at 17:36