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Latest Proof: Double Negation Elimination on 2 September 2010 by Prime.mover

Top 10 Wanted Proofs

Gelfond-Schneider Theorem --Joe 00:57, 11 August 2008 (UTC)
Hausdorff Maximality Principle --Matt Westwood 18:38, 8 November 2008 (UTC)
Frobenius's Theorem --Matt Westwood 18:38, 8 November 2008 (UTC)
Fundamental Theorem of Galois Theory --Matt Westwood 18:38, 8 November 2008 (UTC)
Wedderburn's Theorem --Matt Westwood 18:38, 8 November 2008 (UTC)
Whitney Immersion Theorem --Matt Westwood 08:09, 18 January 2009 (UTC)
Whitney Embedding Theorem --Matt Westwood 08:09, 18 January 2009 (UTC)
Topological h-Cobordism Theorem --Matt Westwood 08:09, 18 January 2009 (UTC)
Thurston's Geometrization Conjecture --Matt Westwood 08:09, 18 January 2009 (UTC)
Tartaglia's Formula --Matt Westwood 06:39, 15 March 2009 (UTC)

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Proof of the Week

Bayes' Theorem

Theorem

Let $\Pr$ be a probability measure on a probability space $\left({\Omega, \Sigma, \Pr}\right)$ .

Let $\Pr \left({A | B}\right)$ denote the conditional probability of $A$ given $B$ .

Let $\Pr \left({A}\right) > 0$ and $\Pr \left({B}\right) > 0$ .

Then:

$\Pr \left({B | A}\right) = \frac {\Pr \left({A | B}\right) \Pr \left({B}\right)} {\Pr \left({A}\right)}$

Generalized Versions

There are other more or less complicated ways of saying very much the same thing, all of which can be derived from the basic version with the help of other fairly elementary results.

For example:

Let $\left\{{B_1, B_2, \ldots}\right\}$ be a partition of the event space $\Sigma$ .

Then, for any $B_i$ in the partition:

$\Pr \left({B_i | A}\right) = \frac {\Pr \left({A | B_i}\right) \Pr \left({B_i}\right)} {\Pr \left({A}\right)} = \frac {\Pr \left({A | B_i}\right) \Pr \left({B_i}\right)} {\sum_j \Pr \left({A | B_j}\right) \Pr \left({B_j}\right)}$

Proof

From the definition of conditional probabilities, we have:

$\Pr \left({A | B}\right) = \frac{\Pr \left({A \cap B}\right)} {\Pr \left({B}\right)}$

$\Pr \left({B | A}\right) = \frac{\Pr \left({A \cap B}\right)} {\Pr \left({A}\right)}$

After some algebra:

$\Pr \left({A | B}\right) \Pr \left({B}\right) = \Pr \left({A \cap B}\right) = \Pr \left({B | A}\right) \Pr \left({A}\right)$

Dividing both sides by $\Pr \left({A}\right)$ (we are told that it is non-zero), the result follows:

$\Pr \left({B | A}\right) = \frac {\Pr \left({A | B}\right) \Pr \left({B}\right)} {\Pr \left({A}\right)}$

$\blacksquare$

Proof of Generalized Version

Follows directly from the Total Probability Theorem:

$\Pr \left({A}\right) = \sum_i \Pr \left({A | B_i}\right) \Pr \left({B_i}\right)$

$\blacksquare$

Note

This result is also known as Bayes' Formula.

The formula: