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I know that: $$P(c|o) = \frac{P(o|c) P(c)}{P(o)}$$

My question is why can't one calculate $P(c|o)$ directly and not use this formula?

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What do you mean 'calculate it directly'? This formula is meant only to put the (typically unknown) posterior distribution, P(c | o) in this case, in terms of the (typically known) sampling distribution and prior distributions. –  Nick Nov 21 '11 at 16:10

2 Answers 2

Often you can calculate the conditional probability directly. For example, if you're given the following table of frequencies

        | Positive | Negative |
--------+----------+----------+
Group A | 100      | 200      |
Group B | 500      | 200      |    

Then you can easily calculate P(positive|A) = 1/3 and P(A|positive) = 1/6.

Bayes' rule still applies, although it's not the easiest way to compute the conditional probabilities. A quick calculation shows that

  • P(A) = 3/10
  • P(positive) = 6/10

and you can check that

1/6 = P(A|positive) = P(positive|A) P(A) / P(positive) = 1/3 * (3/10) / (6/10)

However, in a more complicated situation it may not be obvious what the conditional probabilities are. In this case you can use Bayes' rule to describe a conditional probability in terms of its inversion, if you also know the unconditional probabilities. One way to view Bayes' rule is simply as a method to calculate one of the quantities given the other three (although that's not the most sophisticated point of view).

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Bayes' formula should be written using the law of total probability as $$ P(B|A) = \frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^c)P(B^c)} $$ which emphasizes that the numerator is one of the terms in the denominator. If $P(B|A)$ can be computed directly, then there is no need to use Bayes' formula. But, when $P(B|A)$ is not easy to compute directly but $P(A|B)$ and $P(A|B^c)$ can be determined more easily, then Bayes' formula is used to "turn the conditioning around, getting $P(B|A)$ from $P(A|B)$".

Example: The experiment consists of picking one coin at random from three coins, two of which are fair and one is biased, turning up Heads with probability $p \neq \frac{1}{2}$, and tossing it. Let $A$ be the event that the coin turn up Heads, and $B$ the event that the coin picked is fair. What is $P(B|A)$? This is a bit tricky to get directly, while $P(A|B) = \frac{1}{2}$, $P(A|B^c) = p$, $P(B) = \frac{2}{3}$, and $P(B^c) = \frac{1}{3}$ are all known. Thus, $$P(B|A) = P\{\text{fair}\mid \text{Heads}\} = \frac{\frac{1}{2}\times\frac{2}{3}}{\frac{1}{2}\times\frac{2}{3} + p\times\frac{1}{3}} = \frac{1}{1+p}. $$

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