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If I have two probability spaces :

$\\\Omega_1=\{w^1_1,w^1_2,w^1_3\}$ with $P_1$ defined to be $P_1(w^1_1)=P_1(w^1_2)=P_1(w^1_3) = 1/3$ and $\Omega_2=\{w^2_1,w^2_2,w^2_3\}$ with $P_2$ defined to be $P_2(w^2_1)=P_2(w^2_2)=1/5$ , $P_2(w^2_3)=3/5.$

I am asked to find the (or a) smallest probability space I can use to define a coupling between $(\Omega_1,P_1)$ and $(\Omega_2,P_2)$.

Now, I already have the answer to the question, but I was hoping for a formal way to tell what the size of the minimal probability space would be, and also what methods there are of finding it other than brute forcing the solution or estimating what the size of that space and probabilities for each of the $\omega\in\Omega$ should be for it to work.

Per request the answer is :

$\Omega=\{w_1,w_2,w_3,w_4,w_5\} P(w_1)=1/5, P(w_2)=1/3, P(w_3)=1/3-1/5, P(w_4)=1/3-1/5, P(w_5)=1/5$

The method I used is guessing and intuition is the only way I could explain why $\Omega$ can't be smaller than 5.

Thanks a million!

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I already have the answer to the question... Then write it down here, with the method you use (size 5). –  Did Feb 22 '13 at 5:59
    
The problem is that I had no real method, but I edited it in. (Once you get the answer it's obvious why it works, but not how to get it nor how to prove it's the smallest one). –  NBP Feb 22 '13 at 13:35
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All you have to do now is to show size 4 is not possible. Assume $\Omega$ has size 4 and try to reach a contradiction. –  Did Feb 22 '13 at 14:36
    
@Did Even once I did that, I dislike the notion of having to brute force the answer like this, I didn't feel like any direction of trying to build that $\Omega$ could be more fruitful than the other. As afortementioned my main concern about this exercise is that it seems up in the air as to ways of getting to the solution. –  NBP Feb 22 '13 at 14:44
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I am not sure there is a lot of structure behing this problem so what you call brute force approach looks all right to me. A deep (whatever that means) property of the triplets (5,5,5) and (3,3,9) might be at work behind the scene but at the moment I fail to see what this could be. –  Did Feb 24 '13 at 20:14

1 Answer 1

For the sake of having an answer, I reproduce a comment:

I am not sure there is a lot of structure behing this problem so what you call brute force approach looks all right to me. A deep (whatever that means) property of the triplets (5,5,5) and (3,3,9) might be at work behind the scene but at the moment I fail to see what this could be.

A reformulation of the problem is that one looks for a polynomial $T(x,y)=\sum\limits_{i,j}t_{i,j}x^iy^j$ with the minimal number of nonzero coefficients $t_{i,j}$ such that $T(x,1)=5+5x+5x^2$ and $T(1,y)=3+3y+9y^2$.

Does this make the problem any easier or its structure more apparent? I am not sure.

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