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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!

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Does $1_{a,b}(\omega)$ denote the indicator that $\omega$ is either $a$ or $b$? –  Sasha May 30 '12 at 16:28
    
Yes, that's correct. –  Justin May 30 '12 at 16:29
1  
Note that in your attempt the sets $\sigma(X)$ and $\sigma(Y)$ are not $\sigma$-fields to start with. –  Thomas E. May 30 '12 at 16:45
    
@ThomasE., I agree...it appears that the post was edited and some of my formatting was lost. I had the sets {a,b}, etc. delineated as such. –  Justin May 30 '12 at 16:52
    
In that case, the sets are correct. Btw notice that while inside dollars you have to type \{ to make the regular { symbol appear. I edited the set brackets accordingly, if you want you can review it after it's visible. –  Thomas E. May 30 '12 at 16:56
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1 Answer

up vote 2 down vote accepted

Firstly, there are 16 subsets of $\Omega$. $X$ maps $\Omega$ into $\left\{ 2, 3 \right\}$. The inverse mapping gives $\sigma$-field generated by $X$: $$ \sigma(X) = \left\{ \emptyset, \Omega, \{a,b\}, \{c,d\}\right\} $$ $Y$ maps $\Omega$ into $\{2,3\}$ with generated $\sigma$-field being $$ \sigma(Y) = \left\{ \emptyset, \Omega, \{a,c\}, \{b,d\}\right\} $$

Now $(X,Y)$ maps $\Omega$ into $\{(3,2),(3,3),(2,2),(2,3)\}$. The $\sigma$-field, generated by $(X,Y)$ is all subsets of $\Omega$. Clearly $P_1$ and $P_2$ do not agree there.

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It might be clearer to write "$(X,Y)$ maps $\Omega$ into $\{(3,2), (3,3), (2,2), (2,3)\}$, to avoid mixing up sets and ordered pairs. –  Nate Eldredge May 30 '12 at 17:18
    
@NateEldredge Agreed and edited. Thanks for the feedback. –  Sasha May 30 '12 at 17:20
    
What confuses me is if $\sigma(X,Y)$ consists of all subsets (which would include {a} and {b,c,d}, correct?), then why does neither appear in either $\sigma(X)$ or $\sigma(Y)$? –  Justin May 31 '12 at 1:04
    
$\{a\}$ in $\sigma(X,Y)$ results as inverse image of $(3,2)$. Inverse image of $(3,3)$ is $\{b\}$. But the inverse image of $X$ of $3$ is $\{a, b\}$. –  Sasha May 31 '12 at 4:40
    
@Sasha - what would F(x,y) look like in this case? –  Justin May 31 '12 at 15:56
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