Why is $p(C_k|x)=frac{p(x | C_k, theta_k) p(C_k)}{p(x)}$ a valid application of Bayes' theorem?
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Why is $p(C_k|x)=frac{p(x | C_k, color{red}{theta_k}) p(C_k)}{p(x)}$ a valid application of Bayes' theorem?
bayes-theorem
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
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Why is $p(C_k|x)=frac{p(x | C_k, color{red}{theta_k}) p(C_k)}{p(x)}$ a valid application of Bayes' theorem?
bayes-theorem
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
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add a comment |
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0
$begingroup$
Why is $p(C_k|x)=frac{p(x | C_k, color{red}{theta_k}) p(C_k)}{p(x)}$ a valid application of Bayes' theorem?
bayes-theorem
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
$endgroup$
Why is $p(C_k|x)=frac{p(x | C_k, color{red}{theta_k}) p(C_k)}{p(x)}$ a valid application of Bayes' theorem?
bayes-theorem
bayes-theorem
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
asked Dec 9 '18 at 18:25
mavaviljmavavilj
2,71111034
2,71111034
add a comment |
add a comment |
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