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This problem comes from Grimmett's Probability and Random Processes Problem 1.5.5.

Suppose that A and B are conditionally independent events given $C$, and they are also conditionally independent given $C^c$, and that $0 < P(C) < 1$. Prove that $A$ and $B$ are independent if and only if either $$ P(A|C) = P(A|C^c) $$ or $$ P(B|C) = P(B|C^c) $$

My attempt:

I was able to show that if $P(A|C)=P(A|C^c)$, then $A$ and $B$ are independent. I did it by using the Partioning Theorem on $A$ and $B$, multiplying $P(A)$ and $P(B)$, and showing that it was equal to $P(A \cap B)$.

The reverse direction has been much more difficult for me. Intuitively, the statement makes sense. If $A$ and $B$ are separately independent and conditionally independent given $C$, then you would expect either $A$ and $C$ or $B$ and $C$ to be independent, but I can't seem to find the formalism. My attempt was to partition the intersection and get the result algebraically: $$ P(A)P(B) = P(A \cap B) = P(A \cap B | C)P(C) + P(A \cap B | C^c)P(C^c) $$ $$ = \ldots = \left[P(A|C)-P(A|C^c))\right]P(B \cap C) + P(A | C^c)P(B)$$ I'd like to match terms and say that since $P(BC)$ does not appear in $P(A)P(B)$, then $P(A|C)-P(A|C^c)=0$, but I don't think I can formally.

I may be taking the wrong approach (or an unnecessarily complicated approach). I would appreciate it if anyone could drop a hint on this. Thanks!

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To prove that if $A$ and $B$ are independent, then $P(A|C) = P(A|C^c)$ or $P(B|C) = P(B|C^c)$, you take the difference of $P(A \cap B)$ and $P(A)P(B)$ which we know are equal because $A$ and $B$ are independent.

Write $P(A \cap B)$ conditioned on $C$, and use the fact that $A$ and $B$ are conditionally independent on $C$ to write $$ P(A \cap B) = P(A|C)P(B|C)P(C) + P(A|C^c)P(B|C^c)P(C^c). $$ Next write $P(A)$ and $P(B)$ conditioned on $C$. $$ P(A)P(B) = \left[P(A|C)P(C)+P(A|C^c)P(C^c)\right]\left[P(B|C)P(C)+P(B|C^c)P(C^c)\right] $$ $$ = P(A|C)P(B|C)P(C)^2 + \left[P(A|C)P(B|C^c)+P(B|C)P(A|C^c)\right]P(C)P(C^c)+P(A|C^c)P(B|C^c)P(C^c)^2 $$ Now we subtract the two $$ P(A \cap B) - P(A)P(B) = $$ $$ P(A|C)P(B|C)P(C)(1-P(C)) + P(A|C^c)P(B|C^c)P(C^c)(1-P(C^c)) - \left[P(A|C)P(B|C^c)+P(B|C)P(A|C^c)\right]P(C)P(C^c) $$ $$ = P(C)P(C^c)\left[P(A|C)P(B|C)+P(A|C^c)P(B|C^c)-P(A|C)P(B|C^c)-P(B|C)P(A|C^c)\right]$$ $$ = P(C)P(C^c)\left[(P(A|C)-P(A|C^c)(P(B|C)-P(B|C^c))\right] = 0$$ Since $0 < P(C) < 1$, we must have either $P(A|C)=P(A|C^c)$ or $P(B|C)=P(B|C^c)$.

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I think that $A$ and $B$ are conditionally independent on $C$ does not imply that $A$ and $B$ are conditionally independent on the complementary set of $C$,which means that we cannot derive

$P(AB)=P(A|C)P(B|C)P(C)+P(A|C^c)P(B|C^c)P(C^c). C^c$ represents the complimentary set of $C$.

An example is, we toss flip three times, $A=(HHH, THH, HHT), B=(HHH,TTH,HHT,TTT), C=(HHH, HTH, THH,TTH)$, and $C^c=(HHT,HTT,THT,TTT)$. In this example, $A$ and $B$ are conditionally independent on $C$, but not on the commplement of $C$.

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