Chain Rule for Probability
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Theorem
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$ such that $\map \Pr B > 0$.
The conditional probability of $A$ given $B$ is:
- $\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
Proof
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Suppose it is given that $B$ has occurred.
Then the probability of $A$ having occurred may not be $\map \Pr A$ after all.
In fact, we can say that $A$ has occurred if and only if $A \cap B$ has occurred.
So, if we know that $B$ has occurred, the conditional probability of $A$ given $B$ is $\map \Pr {A \cap B}$.
It follows then, that:
- if we don't actually know whether $B$ has occurred or not
- but we know its probability $\map \Pr B$
we can say that:
- The probability that $A$ and $B$ have both occurred is the conditional probability of $A$ given $B$ multiplied by the probability that $B$ has occurred.
Hence:
- $\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): conditional probability
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- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 1.6$: Conditional probabilities: $(19)$
- 1988: Dominic Welsh: Codes and Cryptography ... (previous) ... (next): Notation