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EVIDENCE AND CHANGING BELIEFS Bayesian inference uses aspects of the Scientific Method , which involves collecting Evidence that is meant to be consistent or inconsistent with a given Hypothesis . As evidence accumulates, the degree of belief in a hypothesis changes. With enough evidence, it will often become very high or very low. Thus, proponents of Bayesian inference say that it can be used to discriminate between conflicting hypotheses: hypotheses with a very high degree of belief should be accepted as true and those with a very low degree of belief should be rejected as false. However, detractors say that this inference method may be biased due to initial beliefs that one needs to hold before any evidence is ever collected. : An example of Bayesian inference is :: ''For billions of years, the sun has risen after it has set. The sun has set tonight. With very high probability (''or'' 'I strongly believe that' ''or'' 'it is true that') the sun will rise tomorrow. With very low probability (''or'' 'I do not at all believe that' ''or'' 'it is false that') the sun will not rise tomorrow.'' Bayesian inference uses a numerical estimate of the degree of belief in a hypothesis before evidence has been observed and calculates a numerical estimate of the degree of belief in the hypothesis after evidence has been observed. Bayesian inference usually relies on degrees of belief, or subjective probabilities, in the induction process and does not necessarily claim to provide an objective method of induction. Nonetheless, some Bayesian statisticians believe probabilities can have an objective value and therefore Bayesian inference can provide an objective method of induction. See Scientific Method . Bayes' theorem adjusts probabilities given new evidence in the following way: | ||
|   | <Math>P(EH 0)</math> Is Called The '' | "http://wwwinformationdelightinfo/information/entry/conditional_probability" class="copylinks">Conditional Probability '' of seeing the evidence <math>E</math> given that the hypothesis <math>H_0</math> is true It is also called the '' Likelihood Function '' when it is expressed as a function of <math>H_0</math> given <math>E</math> |
|   | <Math>P(E)</math> Is Called The '' | "http://wwwinformationdelightinfo/information/entry/marginal_probability" class="copylinks">Marginal Probability '' of <math>E</math>: the probability of witnessing the new evidence <math>E</math> under all mutually exclusive hypotheses It can be calculated as the sum of the product of all probabilities of mutually exclusive hypotheses and corresponding conditional probabilities: <math>\sum P(EH_i)P(H_i)</math> |
|   | <Math>P(H 0E)</math> Is Called The '' | "http://wwwinformationdelightinfo/information/entry/posterior_probability" class="copylinks">Posterior Probability '' of <math>H_0</math> given <math>E</math> |
|   | Multiplying The Prior Probability <math>P(H 0)</math> By The Factor <math>P(EH 0) / P(E)</math> Will Never Yield A Probability That Is Greater Than 1 Since <math>P(E)</math> Is At Least As Great As <math>P(E \cap H 0)</math>, Which Equals <math>P(EH 0) \cdot P(H 0)</math> (see | "http://wwwinformationdelightinfo/information/entry/joint_probability" class="copylinks">Joint Probability ), replacing <math>P(E)</math> with <math>P(E \cap H_0)</math> in the factor <math>P(EH_0) / P(E)</math> will yield a posterior probability of 1 Therefore, the posterior probability could yield a probability greater than 1 only if <math>P(E)</math> were less than <math>P(E \cap H_0),</math> which is never true |
|   | The Probability Of <math>E</math> Given <math>H 0</math>, <math>P(EH 0)</math>, Can Be Represented As A Function Of Its Second Argument With Its First Argument Held At A Given Value Such A Function Is Called A | "http://wwwinformationdelightinfo/information/entry/likelihood_function" class="copylinks">Likelihood Function it is a function of <math>H_0</math> given <math>E</math> A ratio of two likelihood functions is called a likelihood ratio, <math>\Lambda </math> For example, |
|   | : <Math>\Lambda | rac{L(H_0E)}{L(\mbox{not } H_0E)} = rac{P(EH_0)}{P(E\mbox{not } H_0)} </math> |
|   | : <Math>P(H 0E) | rac{P(EH_0)P(H_0)}{P(EH_0)P(H_0)+ P(E\mbox{not }H_0)P(\mbox{not }H_0)} = rac{\Lambda P(H_0)}{\Lambda P(H_0) +P(\mbox{not } H_0)}</math> |
|   | : <Math>P(E 1, E 2 H 0) | P(E_1 H_0) imes P(E_2 H_0)</math> |
|   | : <Math>P(E 1,E 2\mbox{not }H 0) | P(E_1\mbox{not }H_0) imes P(E_2\mbox{not }H_0)</math> |
|   | : <Math>P(H 0E 1, E 2) | rac{P(E_1H_0) imes P(E_2H_0)\P(H_0)}{P(E_1) imes P(E_2)}</math> |
|   | : <Math>P(H 0E 1, E 2) | rac{\Lambda_1 \Lambda_2 P(H_0)}{ P(H_0) + P(\mbox{not } H_0) P(H_0) + P(\mbox{not } H_0) } </math>, |
|   | The Datum ''D'' Is The Observation Of A Plain Cookie From The Contents Of The Bowls, We Know That P(''D'' ''H''<sub>1</sub>) | 30/40 = 075 and P(''D'' ''H''<sub>2</sub>) = 20/40 = 05 Bayes' formula then yields |
|   | \begin{matrix} P(H 1 D) & | & rac{P(H_1) \cdot P(D H_1)}{P(H_1) \cdot P(D H_1) + P(H_2) \cdot P(D H_2)} \ \ \ & =& rac{05 imes 075}{05 imes 075 + 05 imes 05} \ \ \ & =& 06 \end{matrix} |
|   | : <Math>\begin{matrix} P(A B) & | & rac{P(B A) P(A)}{P(B A)P(A) + P(B \mbox{not } A)P(\mbox{not }A)} \ \ |
|   | P(AB) & | &rac{099 imes 0001}{099 imes 0001 + 005 imes 0999} \ ~\ &\approx &0019 \end{matrix}</math> |
|   | : <Math>P(AB) | rac{099 imes 0001}{099 imes 0001 + 0001 imes 0999} \approx 05 </math>, |
|   | : <Math>\begin{matrix} P(A \mbox{not } B) & | & rac{P(\mbox{not }B A) P(A)}{P(\mbox{not }B A)P(A) + P(\mbox{not }B \mbox{not } A)P(\mbox{not }A)} \ \ |
|   | P(A\mbox{not }B) & | &rac{001 imes 0001}{001 imes 0001 + 095 imes 0999}\, ,\ ~\ &\approx &00000105\, \end{matrix}</math> |
|   | : <Math>\begin{matrix} P(A \mbox{not } B) & | & rac{P(\mbox{not }B A) P(A)}{P(\mbox{not }B A)P(A) + P(\mbox{not }B \mbox{not } A)P(\mbox{not }A)} \ \ |
|   | P(A\mbox{not }B) & | &rac{001 imes 06}{001 imes 06 + 095 imes 04}\, ,\ ~\ &\approx &00155\, \end{matrix}</math> |
|   | : <Math>P(G E) | rac{P(G) P(E G)}{P(E)}</math> |
|   | : <Math>P(G E) | (03 imes 10) /(03 imes 10 + 07 imes 10^{-6}) = 099999766667</math> |
|   | : <Math> P(m,na) | \begin{pmatrix} n+m \ m \end{pmatrix} a^m (1-a)^n </math> |
|   | : <Math> P(am,n) | rac{p(m,na)\,p(a)}{\int_0^1 p(m,na)\,p(a)\,da} |
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