Bayes Theorem

The probability of an Event ''A'' conditional on another event ''B'' is generally different from the probability of ''B'' conditional on ''A''. However, there is a definite relationship between the two, and Bayes' theorem is the statement of that relationship. As a formal and Frequentist Probability discuss these debates at greater length. STATEMENT OF BAYES' THEOREM Bayes' theorem relates the conditional and marginal probabilities of Stochastic Events ''A'' and ''B'': : \begin{align}
	P(''A''''B'') Is The	"http://wwwinformationdelightinfo/information/entry/conditional_probability" class="copylinks">Conditional Probability of ''A'', given ''B'' It is also called the Posterior Probability because it is derived from or depends upon the specified value of ''B''
	''L''(''A''''B'') Is The	"http://wwwinformationdelightinfo/information/entry/Likelihood_function" class="copylinks">Likelihood of ''A'' given fixed ''B'' Although in this case the relationship <math>P(B A) = L(A B)</math>, in other cases Likelihood ''L'' can be multiplied by a constant factor, so that it is proportional to, but does not equal Probability ''P''

In words: the posterior probability is proportional to the product of the prior probability and the likelihood.

:<math>P(AB) rac{P(A \cap B)}{P(B)}</math>
:<math>P(BA) rac{P(A \cap B)}{P(A)} \!</math>
:<math>P(AB)\, P(B) P(A \cap B) = P(BA)\, P(A) \!</math>
:<math>P(AB) rac{P(BA)\,P(A)}{P(B)} \!</math>
:<math>P(B) P(A\cap B) + P(A^C\cap B) = P(BA) P(A) + P(BA^C) P(A^C)\,</math>
:<math>P(AB) rac{P(B A)\, P(A)}{P(BA) P(A) + P(BA^C) P(A^C)} \!</math>
:<math>P(A IB) rac{P(B A_i)\, P(A_i)}{\sum_j P(BA_j)\,P(A_j)} , \!</math>
:<math>O(AB) O(A) \cdot \Lambda (AB) </math>
Where <math>O(AB) rac{P(AB)}{P(A^CB)} \!</math> are the ''odds'' of ''A'' given ''B'',
While <math>\Lambda (AB) rac{L(AB)}{L(A^CB)} = rac{P(BA)}{P(BA^C)} \!</math> is the likelihood ratio
:<math> F(xy) rac{f(x,y)}{f(y)} = rac{f(yx)\,f(x)}{f(y)} \!</math>
:<math> F(xy) rac{f(yx)\,f(x)}{\int_{-\infty}^{\infty} f(yx)\,f(x)\,dx}
''f''(''x''''y'') Is The Posterior Distribution Of ''X'' Given ''Y'' ''y'',
''f''(''y''''x'')&nbsp &nbsp''L''(''x''''y'') is (as a function of ''x'') the likelihood function of ''X'' given ''Y''=''y'',
:<math>E P "X\mathcal{G}]" class="copylinks" target="_blank">= rac{E_Q[rac{dP}{dQ} X \mathcal{G} }{E_Q[rac{dP}{dQ}\mathcal{G}]}</math>
:<math> P(AB,C) rac{P(A) \, P(BA) \, P(CA,B)}{P(B) \, P(CB)} </math>
:<math> P(AB,C) rac{P(A,B,C)}{P(B,C)} = rac{P(A,B,C)}{P(B) \, P(CB)} = </math>
:<math> rac{P(CA,B) \, P(A,B)}{P(B) \, P(CB)} = rac{P(A) \, P(BA) \, P(CA,B)}{P(B) \, P(CB)} </math>
:<math>P(AB) rac{P(B A) P(A)}{P(B)} = rac{075 imes 05}{0625} = 06</math>
{ Class "wikitable"

	Pr(''+''''N''), Or The Probability That The Test Is Positive, Given That The Employee Is Not A Drug User This Is 001, Since The Test Will Produce A	"http://wwwinformationdelightinfo/information/entry/false_positive" class="copylinks">False Positive for 1% of non-users
	:<math>\begin{align}P(D+) &	rac{P(+ D) P(D)}{P(+)} \
	&	rac{P(+ D) P(D)}{P(+ D) P(D) + P(+ N) P(N)} \
	:<math> F(r N	10, m=7) =
	Rac {f(m	7 r, n=10) \, f(r)} {\int_0^1 f(m=7r, n=10) \, f(r) \, dr} \!</math>
	From This We See That From The Prior Probability Density Function ''f''(''r'') And The Likelihood Function ''L''(''r'')&nbsp	&nbsp''f''(''m''&nbsp=&nbsp7''r'', ''n''&nbsp=&nbsp10), we can compute the posterior probability density function ''f''(''r''''n''&nbsp=&nbsp10, ''m''&nbsp=&nbsp7)
	Under The Assumption Of Random Sampling, Choosing Voters Is Just Like Choosing Balls From An Urn The Likelihood Function ''L''(''r'')&nbsp	&nbsp''P''(''m''&nbsp=&nbsp7''r'', ''n''&nbsp=&nbsp10,) for such a problem is just the probability of 7 successes in 10 trials for a Binomial Distribution
	:<math> P( M	7 r, n=10) = {10 \choose 7} \, r^7 \, (1-r)^3 </math>
	:<math> \int 0^1 P( M	7r, n=10) \, f(r) \, dr = \int_0^1 {10 \choose 7} \, r^7 \, (1-r)^3 \, 1 \, dr = {10 \choose 7} \, rac{1}{1320} \!</math>
	:<math> F(r N	10, m=7) =
	In The Situation Where The Prize Is Behind The Red Door, The Host Is Free To Pick Between The Green Or The Blue Door At Random Thus, <math>P(BA R)	1/2</math>
	In The Situation Where The Prize Is Behind The Green Door, The Host Must Pick The Blue Door Thus, <math>P(BA G)	0</math>
	In The Situation Where The Prize Is Behind The Blue Door, The Host Must Pick The Green Door Thus, <math>P(BA B)	1</math>
	P(A RB) &	rac{P(B A_r) P(A_r)}{P(B)} & = rac{rac 1 2 rac 1 3}{rac 1 2} & = rac 1 3
	P(A GB) &	rac{P(B A_g) P(A_g)}{P(B)} & = rac{0 rac 1 3}{rac 1 2} & = 0
	P(A BB) &	rac{P(B A_b) P(A_b)}{P(B)} & = rac{1 rac 1 3}{rac 1 2} & = rac 2 3

	APPAREL
	BABY
	BEAUTY
	BOOKS
	CAR TOYS
	CELL PHONES
	DVD'S
	ELECTRONICS
	GOURMET FOOD
	GROCERIES
	HEALTH & PERSONAL
	HOME & GARDEN
	JEWELRY
	MUSIC
	MUSIC INSTRUMENTS
	OFFICE PRODUCTS
	SOFTWARE
	SPORTING GOODS
	TOOLS & HARDWARE
	TOYS
	VIDEO GAMES
	SHOPPING HOME

	MORE SHOPPING...

	CATEGORIES ABOUT BAYES THEOREM

	probability theory

	mathematical theorems

Information About

Bayes Theorem