Given:
$\Omega=\{a,b,c,d\}$
$P_1: a=\frac{1}{6}, b=\frac{1}{3}, c=\frac{1}{3}, d=\frac{1}{6}$
$P_2: a=\frac{1}{3}, b=\frac{1}{6}, c=\frac{1}{6}, d=\frac{1}{3}$
$X(\omega)=2+1_{a,b}(\omega), Y(\omega)=3-1_a(\omega)-1_c(\omega)$
Problem:
Define the $\sigma$-field for $X$, $Y$ and $(X,Y)$, show that $P_1$ and $P_2$ agree on $\sigma(X), \sigma(Y)$ but not $\sigma(X,Y)$, and derive the CDF for all three under both $P_1$ and $P_2$.
Attempted solution:
The real problem I'm having is that I cant see the $\sigma$-field of $(X,Y)$. I believe I have everything for $X$ and $Y$ as follows:
$\sigma(X(\omega))=\{\emptyset,\Omega,\left\{{a,b}\right\},\left\{c,d\right\} \}$
$\sigma(Y(\omega))=\{\emptyset,\Omega,\left\{a,c\right\},\left\{b,d\right\}\}$
Showing the probabilities agree is easy enough, so no need to type it here. For the CDFs, I have:
$F_1(X) = F_2(X)= \begin{cases} 0 & \text{for }x < 2 \\ \frac{1}{2} &\text{for } 2 \leqslant x < 3 \\ 1 &\text{for } x \geqslant 3 \end{cases}$
and similarly for $Y$.
So, how do I handle $(X,Y)$? I know that the $\sigma$-field must be a subset of the space, but with $2$ random variables, I can't seem to make sense of it.
Any help would be appreciated. Thanks!