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Hello I am trying to figure out how to fill out a joint distribution table when I am only given the marginal distributions.

$$ \begin{array}{c|ccc|l} _{\large X}\backslash^{\large Y} & 0 & 1 & 2 \\ \hline 0 & \text{?} & \text{?} & \text{?} & 0.5 \\ 1 & \text{?} & \text{?} & \text{?} & 0.25 \\ 2 & \text{?} & \text{?} & \text{?} & 0.25 \\ \hline & 0.5 & 0.25 & 0.25 \end{array} $$

Basically I don't need answers as much as the marginal distributions relate to finding the unknown numbers.

Michael Hardy
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faschijp
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    There are multiple ways to fill the table, so without further information there is no one answer. If however you know the two distributions are independent, then each entry of the table is the product of the corresponding marginals. – angryavian Nov 17 '16 at 21:09

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$$ \begin{array}{c|ccc|l} _{\large X}\backslash^{\large Y} & 0 & 1 & 2 \\ \hline 0 & a & b & (0.5-a-b) & 0.5 \\ 1 & c & d & (0.25-c-d) & 0.25 \\ 2 & (0.5-a-c) & (0.25-b-d) & (a+b+c+d-0.5) & 0.25 \\ \hline & 0.5 & 0.25 & 0.25 \end{array} $$ If we fill in the upper left four cells with $a,b,c,d$, then (since the first row must add up to $0.5$) the third entry in the first row must by $0.5-a-b$, and the other entries are filled in in the same way.

You can put any numbers in place of $a,b,c,d$ that make all entries $\ge 0$ but $\le 1$, and you'll have a probability distribution.

Note that $a$ could be as big as $0.5$, but not bigger, and the table would look like this: $$ \begin{array}{c|ccc|l} _{\large X}\backslash^{\large Y} & 0 & 1 & 2 \\ \hline 0 & 0.5 & 0 & 0 & 0.5 \\ 1 & 0 & d & (0.25-d) & 0.25 \\ 2 & 0 & (0.25-d) & d & 0.25 \\ \hline & 0.5 & 0.25 & 0.25 \end{array} $$ and in this case $d$ could be as small as $0$ or as big as $0.25$

[ to be continued, maybe $\ldots$ ]

Michael Hardy
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