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Prove the Multiplication Rule (Conditional Form) with more than two events.

For events $A_1, A_2,\ldots, A_n$ prove that $$ P(A_1 \cap A_2 \cap\ldots\cap A_n)= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_n|A_1 \cap A_2 \cap ... \cap\ A_{n-1}). $$

My first attempt was to try induction and I do get through the first two induction steps but I am not getting the answer when trying to prove for all $n$

Any help would be highly appreciated

Thanks

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3 Answers 3

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You know that the definition of conditional probability is

$$P(B | A) = \frac{P(A\cap B)}{P(A)},$$ so just apply the definition to every term in the right hand side of your equation. Starting with $$\quad P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \cdots\ P(A_n|A_1 \cap A_2 \cap ... \cap\ A_{n-1})$$ and applying the definition of conditional probability to each term we get $$ \ \color{blue}{P(A_1)} \frac{\color{red}{P(A_1 \cap A_2)}}{\color{blue}{P(A_1)}}\frac{\color{green}{P(A_1 \cap A_2 \cap A_3)}}{\color{red}{P(A_1 \cap A_2)}} \cdots \frac{P(A_1 \cap A_2 \cap\cdots \cap A_n) }{\color{purple}{P(A_1 \cap A_2 \cap\cdots \cap A_{n-1})}} $$ And almost every term will cancel out, except $P(A_1 \cap A_2 \cap \cdots \cap A_n)$. Hence

$ P(A_1 \cap A_2 \cap \cdots \cap A_n)= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \cdots\ P(A_n|A_1 \cap A_2 \cap \cdots \cap\ A_{n-1}). $

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Starting with the required result $$ P(A_1 \cap A_2 \cap\ldots\cap A_n)= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_n|A_1 \cap A_2 \cap ... \cap\ A_{n-1}), $$ just look at the last term and use the definition of conditional probability: $P(E|F)=P(E\cap F)/P(F)$. You can rewrite the last term as $$ P(A_n|A_1 \cap A_2 \cap ... \cap\ A_{n-1})= \frac{P(A_n\cap A_1 \cap A_2 \cap ... \cap\ A_{n-1})} {P(A_1 \cap A_2 \cap ... \cap\ A_{n-1})} =\frac{P(A_1 \cap A_2 \cap ... \cap\ A_{n})} {P(A_1 \cap A_2 \cap ... \cap\ A_{n-1})}. $$ Therefore the result is true if $$ P(A_1 \cap A_2 \cap\ldots\cap A_n)= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_{n-1}|A_1 \cap A_2 \cap ... \cap\ A_{n-2})\ \frac{P(A_1 \cap A_2 \cap ... \cap\ A_{n})} {P(A_1 \cap A_2 \cap ... \cap\ A_{n-1})}. $$ Moving terms around, this is equivalent to $$ P(A_1 \cap A_2 \cap\ldots\cap A_{n-1})= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_{n-1}|A_1 \cap A_2 \cap ... \cap\ A_{n-2}) $$ which is precisely the required property when $n$ is replaced by $n-1$. This establishes the basis for induction on $n$. All that remains to show is that the identity holds for $n=2$, which as you know it does.

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Using your notation,

$P(($intersection of all $A_i$'s from $i=1$ to $n)=$ $P(($intersection of all $A_i$'s from $i=1$ to $n-1,$ intersected with $An)= $probability of intersection up to n-1, times probability of An given interaction up to n-1 (this is just using the base case)=$P(A1)P(A2|A1)P(A3|A1 \A2) ..... P(An-1|A1 \A2 \ ....  \ An-2) $(by induction step) times $P(An|A1 \A2 \ ....  \ An-1)$.

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