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If we know that two events $A$ and $B$ are independent, can we say that $A$ and $B$ are also conditionally independent given an arbitrary event $C$?

$$P(A\cap B) = P(A)P(B) \overset{?}{\Rightarrow} P(A\cap B|C) = P(A|C)P(B|C)$$

Parna
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2 Answers2

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No, Toss an unbiased coin twice. Let $A$ be the event that the first toss is a head, let $B$ be the event that the second toss is a head and let $C$ be the event that the results of both tosses are the same (that is $C=\{HH,TT\}$).

Then $A$ and $B$ are independent but $\mathbb P(A\cap B|C) = \mathbb P(A|C) = \mathbb P(B|C) = \frac 12$.

Tim
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Hint: Consider the example of $${{\text{Event }A}\atop \begin{array}{|c|c|c|} \hline\strut& & \\\hline \strut& & \\\hline \strut\Large\color{red}{\bullet} & \Large\color{red}{\bullet} & \Large\color{red}{\bullet}\\\hline \end{array}}\qquad {{\text{Event }B}\atop \begin{array}{|c|c|c|} \hline\strut& & \Large\color{blue}{\bullet}\\\hline \strut\;\;&\;\; & \Large\color{blue}{\bullet}\\\hline \strut & & \Large\color{blue}{\bullet}\\\hline \end{array}}\qquad {{\text{Event }C}\atop \begin{array}{|c|c|c|} \hline\strut& & \\\hline \strut\;\;&\Large\color{green}{\bullet} & \\\hline \strut & \Large\color{green}{\bullet} & \Large\color{green}{\bullet}\\\hline \end{array}}$$

Zev Chonoles
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  • $P(A) = P(B) = \frac{1}{3}$, $P(A\cap B) = \frac{1}{9}$, $P(A|C)= \frac{2}{3}$, $P(B|C) = \frac{1}{3}$, $P(A\cap B|C) = \frac{1}{3}$. Would you please confirm my calculations for the probabilities? – Parna Jun 10 '13 at 10:14
  • Yup, that's right. – Zev Chonoles Jun 10 '13 at 10:16
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    You might want to add the similar (quite illuminating) pictures for the example of size 4 explained in the other answer. – Did Jun 10 '13 at 10:25