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If I have a conditional probability matrix for binary variables $A,B,C$ with entries of the form $$ P(A | B \cap C) = \left( \begin{matrix} P(a_1 | b_1 \cap c_1) & P(a_1| b_1 \cap c_2) & P(a_1|b_2 \cap c_1) & P(a_1 | b_2 \cap c_2) \\ P(a_2 | b_1 \cap c_1) & P(a_2| b_1 \cap c_2) & P(a_2|b_2 \cap c_1) & P(a_2 | b_2 \cap c_2) \end{matrix} \right) $$

how can I use this matrix to obtain the matrix $$ P(A | C) = \left( \begin{matrix} P(a_1 | c_1) & P(a_1| c_2) \\ P(a_2 | c_1) & P(a_2| c_2) \end{matrix} \right) $$

where $A$ is assumed to be independent of $B$?

Clearly in general $P(X | Y \cap Z) = P(X|Z)$ under the assumption that $X$ and $Y$ are independent, but I am confused as to how the entries of the matrices above are related.

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How can $P(A|B \cap C) = P(A|C)$ if they have different dimensions? –  TMM May 6 '12 at 15:43
    
Ambiguous notation, I was referring to conditional probabilities in general there not the matrices. –  David Park May 6 '12 at 15:47
    
One cannot deduce the second matrix from the first one, some information is lacking, which is the joint distribution of (B,C). All one can say is that each $P(a_i\mid c_j)$ is a barycenter of $P(a_i\mid b_1\cap c_j)$ and $P(a_i\mid b_2\cap c_j)$. –  Did May 6 '12 at 15:58

1 Answer 1

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One needs the values of the conditional probabilities $P(b_1\mid c_j)$ and $P(b_2\mid c_j)=1-P(b_1\mid c_j)$, since $$ P(a_i\mid c_j)=P(a_i\mid b_1\cap c_j)P(b_1\mid c_j)+P(a_i\mid b_2\cap c_j)P(b_2\mid c_j). $$

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