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My discrete math text says, if $E_1$ and $E_2$ are independent events, and $E_1 | S$ and $E_2 | S$ are independent events, then $\text{P}(E_1 \cap E_2 | S) = \text{P}(E_1 | S) \cdot \text{P}(E_2 | S)$. But, what's the proof?

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    $\begingroup$ if $E_1$ and $E_2$ are...? $\endgroup$ – TheRealFakeNews Feb 5 '13 at 7:25
  • $\begingroup$ @AlanH independent events. Edited. Thanks. $\endgroup$ – f.nasim Feb 5 '13 at 7:27
  • $\begingroup$ Do you understand what independent mean ? $\endgroup$ – Damien L Feb 5 '13 at 7:35
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    $\begingroup$ What is the definition of "$E | S$"? Your text really says this is an event?? $\endgroup$ – Did Feb 5 '13 at 7:50
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    $\begingroup$ yes, and giving the correct statement of the problem is also important, if you expect an answer. $\endgroup$ – UwF Feb 5 '13 at 9:13
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Conditional probabilities are defined by

$P(A|B)=\frac{P(A\cap B)}{P(B)}$ (where you assume $P(B)>0$).

Two events $A,B$ are called independent, if $P(A\cap B)=P(A)P(B)$.

If your text says that $E_1$ and $E_2$ are independent conditionally on $S$, then that simply means that they are independent w.r.t. the probability $P(\cdot|S)$, i.e. the relation that you want to prove,

$P(E_1\cap E_2|S)=P(E_1|S)P(E_2|S)$.

If your text says that all three events $E_1,E_2,S$ are independent (and $P(S)>0$), then you have

$P(E_1\cap E_2|S) =\frac{P(E_1\cap E_2\cap S)}{P(S)}=\frac{P(E_1)P(E_2)P(S)}{P(S)}=\frac{P(E_1)P(S)}{P(S)}\frac{P(E_2)P(S)}{P(S)}=\frac{P(E_1\cap S)}{P(S)}\frac{P(E_2\cap S)}{P(S)}=P(E_1|S)P(E_2|S)$

Please go back and read what exactly your text says. Then it should be easy to figure out the answer.

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  • $\begingroup$ The problem statement clearly reads "...Assuming that $E_1$ and $E_2$ are independent events and that $E_1 | S$ and $E_2 | S$ are independent events,...". But, in the solution to the problem it reads "...Because of the assumed independence of $E_1$, $E_2$, and $S$, we have $P(E_1 \cap E_2 | S) = P(E_1 | S) \cdot P(E_2 | S)$,...". Now, it is clear $E_1$, $E_2$, and $S$ are independent actually; the text never said $E_1$ and $E_2$ are independent conditionally on $S$. So your second solution is the correct answer. Thanks. $\endgroup$ – f.nasim Feb 5 '13 at 14:12
  • $\begingroup$ I doubt very much that anybody would use the mysterious statement in your question to mean that $(E_1,E_2,S)$ are independent. $\endgroup$ – Did Feb 5 '13 at 15:27

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